Appendix F — Solutions to Selected Exercises

This appendix collects complete solutions to the starred (\(\star\)) exercises of every chapter. A starred exercise is one whose statement carries the marker \(\star\); its full worked solution appears here. Solutions are organized by chapter in book order, and each is headed by the slug of the exercise it solves together with a plain-language title, so that “Solution to Exercise 6.8” can be located both from the cross-reference in the chapter and from the title in this appendix. Results are cited by their slugs, for example Theorem 6.3 or Equation 6.9, and the notation is that of Notation.

How to use these solutions. These are worked derivations, not just final answers, and they are most useful as a learning companion rather than a first port of call. The intended workflow is: attempt the exercise yourself first; if you are stuck on the opening move, read only the Strategy line that begins the harder solutions, which names the main idea or theorem without giving the algebra away, then try again; and only afterward read the full derivation to check every step. Reading a solution before attempting the exercise trades a small short-term gain for a large loss in understanding, because the hardest part of most of these problems is choosing the first move, and that is exactly the skill the exercises build.

Reading aids. To lower the barrier for a reader who does not have the chapter open alongside, the harder solutions carry three optional scaffolds. A Strategy line states the single idea that unlocks the problem. A Uses line lists, in plain language, the one or two prior results the solution leans on, so that a slug such as Theorem 3.6 is paired with a spoken name. Scattered Common mistake notes flag the conceptual confusions that recur across the book (positivity versus zero probability, non-collapsibility versus effect modification, collapsibility versus transportability, naive versus sandwich variance, latent versus observed correlation). None of these scaffolds is load-bearing for the mathematics; a fluent reader can skip them, and a beginner can lean on them.

Coverage. Every chapter from 1 through 30 contributes at least one starred exercise, and every starred exercise in the book has its full solution here. If you reach a starred exercise in a chapter and cannot find its solution in this appendix, that is a defect to report; the appendix is meant to be complete for the current book structure.

F.1 Chapter 1: Linear Algebra I

F.1.1 Solution to Exercise 1.12: The dimension formula for a sum of two subspaces

This exercise from Chapter 1 asks for the inclusion-exclusion identity \(\dim(U+W)=\dim U+\dim W-\dim(U\cap W)\).

Strategy. Build one basis of \(U+W\) by starting from a basis of the overlap \(U\cap W\) and extending it two ways, once into \(U\) and once into \(W\). Counting the basis vectors then gives the formula directly; the only real work is checking that the combined list is linearly independent.

Uses. The basis-extension lemma (Lemma 1.8: any independent set extends to a basis) and the fact that a spanning independent set is a basis.

Let \(k=\dim(U\cap W)\) and choose a basis \(\{\mathbf u_1,\dots,\mathbf u_k\}\) of \(U\cap W\). By the basis-extension lemma (Lemma 1.8) extend it to a basis \(\{\mathbf u_1,\dots,\mathbf u_k,\mathbf a_1,\dots,\mathbf a_m\}\) of \(U\), so \(\dim U=k+m\), and separately to a basis \(\{\mathbf u_1,\dots,\mathbf u_k,\mathbf b_1,\dots,\mathbf b_n\}\) of \(W\), so \(\dim W=k+n\). We claim \[ \mathcal B=\{\mathbf u_1,\dots,\mathbf u_k,\ \mathbf a_1,\dots,\mathbf a_m,\ \mathbf b_1,\dots,\mathbf b_n\} \] is a basis of \(U+W\).

Spanning. Every element of \(U+W\) is \(\mathbf u+\mathbf w\) with \(\mathbf u\in U\) and \(\mathbf w\in W\). Writing \(\mathbf u\) in the \(U\)-basis and \(\mathbf w\) in the \(W\)-basis exhibits \(\mathbf u+\mathbf w\) as a combination of \(\mathcal B\), so \(\mathcal B\) spans \(U+W\).

Independence. Suppose \[ \sum_{i=1}^k \alpha_i\mathbf u_i+\sum_{r=1}^m \beta_r\mathbf a_r+\sum_{s=1}^n \gamma_s\mathbf b_s=\mathbf 0 . \] The idea is to isolate the part of this relation that lives only in the \(W\)-basis, so that we can pin it down inside the overlap \(U\cap W\) and then invoke \(W\)-basis independence. Accordingly, set \(\mathbf z=\sum_s \gamma_s\mathbf b_s\) (the \(\mathbf b\)-only piece). The displayed equation gives \(\mathbf z=-\sum_i\alpha_i\mathbf u_i-\sum_r\beta_r\mathbf a_r\in U\), while \(\mathbf z\in W\) by construction, so \(\mathbf z\in U\cap W\). Thus \(\mathbf z=\sum_i\delta_i\mathbf u_i\) for scalars \(\delta_i\), and \(\sum_s\gamma_s\mathbf b_s-\sum_i\delta_i\mathbf u_i=\mathbf 0\). Since \(\{\mathbf u_1,\dots,\mathbf u_k,\mathbf b_1,\dots,\mathbf b_n\}\) is a basis of \(W\) and hence linearly independent, every \(\gamma_s=0\) (and every \(\delta_i=0\)). The original relation then reads \(\sum_i\alpha_i\mathbf u_i+\sum_r\beta_r\mathbf a_r=\mathbf 0\), and the independence of the \(U\)-basis forces every \(\alpha_i=0\) and \(\beta_r=0\). Hence \(\mathcal B\) is independent.

Therefore \(\dim(U+W)=k+m+n=(k+m)+(k+n)-k=\dim U+\dim W-\dim(U\cap W)\). \(\square\)

F.1.2 Solution to Exercise 1.13: When the Gram matrix is invertible

This Chapter 1 exercise establishes when the Gram matrix \(\mathbf X^\top\mathbf X\) is invertible, the linear algebra behind the very first step of least squares.

Uses. The rank-nullity theorem (Theorem 1.3) and the definiteness of the Euclidean norm (\(\|\mathbf v\|_2=0\) only for \(\mathbf v=\mathbf 0\)).

(a) If \(\mathbf X\mathbf v=\mathbf 0\) then \(\mathbf X^\top\mathbf X\mathbf v=\mathbf X^\top\mathbf 0= \mathbf 0\), so \(\mathcal N(\mathbf X)\subseteq\mathcal N(\mathbf X^\top\mathbf X)\). Conversely, if \(\mathbf X^\top\mathbf X\mathbf v=\mathbf 0\), then \[ 0=\mathbf v^\top\mathbf X^\top\mathbf X\mathbf v=(\mathbf X\mathbf v)^\top(\mathbf X\mathbf v)=\|\mathbf X\mathbf v\|_2^2, \] and the definiteness of the Euclidean norm gives \(\mathbf X\mathbf v=\mathbf 0\), that is \(\mathbf v\in\mathcal N(\mathbf X)\). Hence \(\mathcal N(\mathbf X^\top\mathbf X)=\mathcal N(\mathbf X)\).

(b) Both \(\mathbf X\) and the \(p\times p\) matrix \(\mathbf X^\top\mathbf X\) have \(p\) columns, so the rank-nullity theorem (Theorem 1.3) gives \[ \operatorname{rank}(\mathbf X^\top\mathbf X)=p-\dim\mathcal N(\mathbf X^\top\mathbf X) =p-\dim\mathcal N(\mathbf X)=\operatorname{rank}(\mathbf X), \] using part (a) for the middle equality. A \(p\times p\) matrix is invertible if and only if its rank is \(p\), so \(\mathbf X^\top\mathbf X\) is invertible if and only if \(\operatorname{rank}(\mathbf X)=p\), that is, if and only if \(\mathbf X\) has full column rank.

Why a health researcher should care: full column rank is exactly the condition that lets you estimate a treatment effect uniquely. If two columns of \(\mathbf X\) are collinear (say a covariate and a near-duplicate of it), the rank drops below \(p\), \(\mathbf X^\top\mathbf X\) is singular, and infinitely many coefficient vectors \(\boldsymbol\beta\) fit the data equally well. The estimated drug effect is then not identified from the data alone; only a modeling choice picks one value. This is the algebraic root of the collinearity warnings that recur when a regression adjustment is used inside an indirect comparison.

(c) The normal equations \(\mathbf X^\top\mathbf X\boldsymbol\beta=\mathbf X^\top\mathbf y\) form a square linear system with coefficient matrix \(\mathbf X^\top\mathbf X\). A square system has a unique solution for every right-hand side if and only if its matrix is invertible, which by (b) happens exactly when \(\mathbf X\) has full column rank \(p\). This is the algebraic condition under which ordinary least squares has a unique solution (Theorem 6.1). \(\square\)

F.2 Chapter 2: Linear Algebra II

F.2.1 Solution to Exercise 2.13: The parallelogram law and the polarization identity

This Chapter 2 exercise proves that a norm comes from an inner product only if it obeys the parallelogram law, and that when it does, the inner product is recovered from the norm by polarization.

Uses. Bilinearity of an inner product, and the definition \(\|\mathbf x\|^2=\langle\mathbf x,\mathbf x\rangle\).

Let \(\langle\cdot,\cdot\rangle\) be the inner product inducing \(\|\mathbf x\|^2=\langle\mathbf x,\mathbf x\rangle\). Expanding bilinearly, \[ \begin{aligned} \|\mathbf u+\mathbf v\|^2&=\|\mathbf u\|^2+2\langle\mathbf u,\mathbf v\rangle+\|\mathbf v\|^2,\\ \|\mathbf u-\mathbf v\|^2&=\|\mathbf u\|^2-2\langle\mathbf u,\mathbf v\rangle+\|\mathbf v\|^2 . \end{aligned} \] Adding the two lines, the cross terms cancel and give the parallelogram law \(\|\mathbf u+\mathbf v\|^2+\|\mathbf u-\mathbf v\|^2=2\|\mathbf u\|^2+2\|\mathbf v\|^2\). Subtracting the second line from the first gives \(\|\mathbf u+\mathbf v\|^2-\|\mathbf u-\mathbf v\|^2=4\langle\mathbf u,\mathbf v\rangle\), which is the polarization identity \(\langle\mathbf u,\mathbf v\rangle=\tfrac14(\|\mathbf u+\mathbf v\|^2-\|\mathbf u-\mathbf v\|^2)\).

Uniqueness. Polarization writes the inner product as an explicit function of the norm alone. Hence two inner products that induce the same norm have the same value on every pair \((\mathbf u,\mathbf v)\), so the inner product is uniquely determined by its norm.

The \(\ell^1\) norm. If \(\|\cdot\|_1\) on \(\mathbb R^2\) came from an inner product it would satisfy the parallelogram law. Take \(\mathbf u=(1,0)\), \(\mathbf v=(0,1)\). Then \(\|\mathbf u\|_1=\|\mathbf v\|_1=1\), while \(\mathbf u+\mathbf v=(1,1)\) and \(\mathbf u-\mathbf v=(1,-1)\) both have \(\ell^1\) norm \(2\). The law would require \(2^2+2^2=2\cdot 1^2+2\cdot 1^2\), that is \(8=4\), which is false. So \(\|\cdot\|_1\) violates the parallelogram law and cannot arise from any inner product. \(\square\)

F.2.2 Solution to Exercise 2.14: Four equivalent characterizations of an orthogonal projection

This Chapter 2 exercise shows that four properties of an idempotent matrix \(P\) (a matrix with \(P^2=P\), that is, a projection) are equivalent. Because a projection splits space into “what it keeps” (its range \(\mathcal C(P)\)) and “what it kills” (its null space \(\mathcal N(P)\)), the four properties are four ways to say “the split is at right angles.”

Strategy. Prove the equivalences as a short cycle of implications rather than all pairs. The one clever step is (iii) \(\Rightarrow\) (i): to turn the size inequality \(\|P\mathbf v\|\le\|\mathbf v\|\) into an orthogonality statement, probe it with a one-parameter family of test vectors and read off the sign of an inner product as the parameter changes sign.

Uses. The projection theorem (Theorem 2.5) and the Pythagorean identity for orthogonal vectors.

Here are the four properties and the plain-language meaning of each, followed by the roadmap of implications we prove.

Label	Property	In words
(i)	\(P=P^\top\)	the projection matrix is symmetric
(ii)	\(P\) is the orthogonal projection onto \(\mathcal C(P)\)	it drops each vector straight down onto the range
(iii)	\(\|P\mathbf v\|\le\|\mathbf v\|\) for all \(\mathbf v\)	projecting never lengthens a vector
(iv)	\(\mathcal C(P)\perp\mathcal N(P)\)	range and null space meet at right angles

We prove the cycle (i) \(\Rightarrow\) (ii) \(\Rightarrow\) (iv) \(\Rightarrow\) (i), which makes (i), (ii), (iv) equivalent, and separately (ii) \(\Leftrightarrow\) (iii), which folds (iii) into the same equivalence class.

Throughout, \(P^2=P\). For any idempotent, \(\mathbb R^n=\mathcal C(P)\oplus\mathcal N(P)\) (a direct sum: every vector splits uniquely as a range part plus a null-space part): indeed \(\mathbf v=P\mathbf v+(\mathbf v-P\mathbf v)\) with \(P\mathbf v\in\mathcal C(P)\) and \(P(\mathbf v-P\mathbf v)=P\mathbf v-P^2\mathbf v=\mathbf 0\), so \(\mathbf v-P\mathbf v\in\mathcal N(P)\); and if \(\mathbf w\in\mathcal C(P)\cap\mathcal N(P)\) then \(\mathbf w=P\mathbf u\) for some \(\mathbf u\) gives \(\mathbf w=P\mathbf u=P^2\mathbf u=P\mathbf w=\mathbf 0\). On \(\mathcal C(P)\), \(P\) acts as the identity (\(P(P\mathbf u)=P\mathbf u\)).

We prove (i) \(\Rightarrow\) (ii) \(\Rightarrow\) (iv) \(\Rightarrow\) (i) and (ii) \(\Leftrightarrow\) (iii).

(i) \(\Rightarrow\) (ii). If \(P=P^\top\) and \(P^2=P\), then for \(\mathbf m=P\mathbf u\in\mathcal C(P)\) and \(\mathbf w\in\mathcal N(P)\), \(\langle\mathbf m,\mathbf w\rangle=\langle P\mathbf u,\mathbf w\rangle=\langle\mathbf u,P^\top\mathbf w\rangle =\langle\mathbf u,P\mathbf w\rangle=0\), so \(\mathcal C(P)\perp\mathcal N(P)\). Combined with the direct-sum decomposition, \(P\) is the projection onto \(\mathcal C(P)\) along its orthogonal complement, that is the orthogonal projection onto \(\mathcal C(P)\) (Theorem 2.5).

(ii) \(\Rightarrow\) (iv). The orthogonal projection onto a subspace \(\mathcal M\) has range \(\mathcal M\) and kernel \(\mathcal M^\perp\), so \(\mathcal C(P)=\mathcal M\perp\mathcal M^\perp=\mathcal N(P)\).

(iv) \(\Rightarrow\) (i). Suppose \(\mathcal C(P)\perp\mathcal N(P)\). Decompose any \(\mathbf x,\mathbf y\) as \(\mathbf x=\mathbf m_x+\mathbf w_x\), \(\mathbf y=\mathbf m_y+\mathbf w_y\) with \(\mathbf m\in\mathcal C(P)\), \(\mathbf w\in\mathcal N(P)\). Then \(P\mathbf x=\mathbf m_x\), \(P\mathbf y=\mathbf m_y\), and by orthogonality \(\langle P\mathbf x,\mathbf y\rangle=\langle\mathbf m_x,\mathbf m_y+\mathbf w_y\rangle =\langle\mathbf m_x,\mathbf m_y\rangle=\langle\mathbf m_x+\mathbf w_x,\mathbf m_y\rangle =\langle\mathbf x,P\mathbf y\rangle\). Hence \(P=P^\top\).

(ii) \(\Rightarrow\) (iii). For the orthogonal projection, decompose \(\mathbf v=\mathbf m+\mathbf w\) with \(\mathbf m\perp\mathbf w\); then \(\|P\mathbf v\|^2=\|\mathbf m\|^2\le\|\mathbf m\|^2+\|\mathbf w\|^2= \|\mathbf v\|^2\) by the Pythagorean theorem, so \(\|P\mathbf v\|\le\|\mathbf v\|\).

(iii) \(\Rightarrow\) (i). Assume \(\|P\mathbf v\|\le\|\mathbf v\|\) for all \(\mathbf v\) and use the direct sum \(\mathbb R^n=\mathcal C(P)\oplus\mathcal N(P)\). We want to force \(\langle\mathbf m,\mathbf w\rangle=0\) for a range vector \(\mathbf m\) and a null-space vector \(\mathbf w\). The trick is to feed the inequality a test vector that keeps \(P\mathbf v\) fixed while sliding \(\mathbf v\) along the null-space direction, so that the inequality becomes a statement about a single scalar \(t\); the vector \(\mathbf v=\mathbf m+t\mathbf w\) does exactly this, because \(P\) annihilates \(\mathbf w\) and so \(P\mathbf v=\mathbf m\) for every \(t\). Fix then \(\mathbf m\in\mathcal C(P)\) and \(\mathbf w\in\mathcal N(P)\) and test on \(\mathbf v=\mathbf m+t\mathbf w\). Since \(P\mathbf v=\mathbf m\), the contraction inequality \(\|\mathbf m\|^2\le\|\mathbf m+t\mathbf w\|^2=\|\mathbf m\|^2+2t\langle\mathbf m,\mathbf w\rangle +t^2\|\mathbf w\|^2\) gives \(0\le 2t\langle\mathbf m,\mathbf w\rangle+t^2\|\mathbf w\|^2\) for every \(t\in\mathbb R\). For small \(t\) the quadratic term \(t^2\|\mathbf w\|^2\) is negligible next to the linear term \(2t\langle\mathbf m,\mathbf w\rangle\), so the sign of the right-hand side is controlled by the sign of \(t\langle\mathbf m,\mathbf w\rangle\); forcing it to stay nonnegative from both sides of \(t=0\) squeezes the inner product to zero. Concretely, dividing by \(t>0\) and letting \(t\downarrow 0\) forces \(\langle\mathbf m,\mathbf w\rangle\ge 0\); dividing by \(t<0\) (which flips the inequality) and letting \(t\uparrow 0\) forces \(\langle\mathbf m,\mathbf w\rangle\le 0\). Hence \(\langle\mathbf m,\mathbf w\rangle=0\) for all such \(\mathbf m,\mathbf w\), that is \(\mathcal C(P)\perp\mathcal N(P)\), which is (iv) and therefore (i). This closes the cycle, so (i) through (iv) are equivalent. \(\square\)

Why this matters for the rest of the book: the hat matrix \(H=\mathbf X(\mathbf X^\top\mathbf X)^{-1}\mathbf X^\top\) of Chapter 6 is exactly such a symmetric idempotent, so this equivalence is what guarantees that the least-squares fitted values \(H\mathbf y\) are the orthogonal (perpendicular) drop of \(\mathbf y\) onto the column space, and hence that the residuals \(\mathbf y-H\mathbf y\) are orthogonal to every fitted value. That perpendicularity is the geometric content of the normal equations and, later, of the projection view of covariate adjustment: a matching or standardization weight that is not a genuine orthogonal projection can lengthen instead of shrink, which is one way an adjustment can inflate rather than reduce error.

F.3 Chapter 3: Linear Algebra III

F.3.1 Solution to Exercise 3.11: \(A^\top A\) and \(AA^\top\) share their nonzero spectrum

This Chapter 3 exercise shows that the two Gram matrices \(A^\top A\) and \(AA^\top\), though generally of different sizes, carry exactly the same nonzero eigenvalues, and deduces that the singular values of \(A\) and \(A^\top\) agree.

Strategy. The multiplication-by-\(A\) map \(\mathbf v\mapsto A\mathbf v\) carries an eigenvector of \(A^\top A\) to an eigenvector of \(AA^\top\) for the same eigenvalue; showing this map and its mirror \(\mathbf u\mapsto A^\top\mathbf u\) are both injective gives matching eigenspace dimensions.

Uses. The spectral theorem (Theorem 3.2), in particular that the trace equals the sum of eigenvalues.

Let \(\lambda\neq 0\) be an eigenvalue of \(A^\top A\) with eigenvector \(\mathbf v\neq\mathbf 0\), so \(A^\top A\mathbf v=\lambda\mathbf v\). Apply \(A\): \[ (AA^\top)(A\mathbf v)=A(A^\top A\mathbf v)=A(\lambda\mathbf v)=\lambda(A\mathbf v). \] Moreover \(A\mathbf v\neq\mathbf 0\), since \(A\mathbf v=\mathbf 0\) would give \(A^\top A\mathbf v=\mathbf 0=\lambda\mathbf v\) and hence \(\mathbf v=\mathbf 0\) (as \(\lambda\neq 0\)), a contradiction. So \(A\mathbf v\) is an eigenvector of \(AA^\top\) for the same \(\lambda\). Thus the linear map \(\mathbf v\mapsto A\mathbf v\) sends the \(\lambda\)-eigenspace \(E_\lambda(A^\top A)\) into \(E_\lambda(AA^\top)\).

This map is injective on \(E_\lambda(A^\top A)\): if \(A\mathbf v=\mathbf 0\) with \(\mathbf v\in E_\lambda\), the same computation gives \(\mathbf v=\mathbf 0\). The reverse injection is the mirror image, and because it is the crux of the argument it is worth writing out rather than dispatching as “by symmetry.” Let \(\mathbf u\neq\mathbf 0\) satisfy \(AA^\top\mathbf u=\lambda\mathbf u\). Apply \(A^\top\): \[ (A^\top A)(A^\top\mathbf u)=A^\top(AA^\top\mathbf u)=A^\top(\lambda\mathbf u)=\lambda(A^\top\mathbf u), \] so \(A^\top\mathbf u\in E_\lambda(A^\top A)\); and \(A^\top\mathbf u\neq\mathbf 0\), since \(A^\top\mathbf u =\mathbf 0\) would give \(AA^\top\mathbf u=\mathbf 0=\lambda\mathbf u\) and hence \(\mathbf u=\mathbf 0\). Thus \(\mathbf u\mapsto A^\top\mathbf u\) injects \(E_\lambda(AA^\top)\) into \(E_\lambda(A^\top A)\). Two subspaces each injecting linearly into the other have equal dimension, so \(\mathbf v\mapsto A\mathbf v\) is an isomorphism between them. Equal dimensions for every nonzero \(\lambda\) means \(A^\top A\) and \(AA^\top\) have exactly the same nonzero eigenvalues with the same multiplicities.

The singular values of \(A\) are \(\sigma_i=\sqrt{\lambda_i}\) for the nonzero eigenvalues \(\lambda_i\) of \(A^\top A\); those of \(A^\top\) are the square roots of the nonzero eigenvalues of \((A^\top)^\top A^\top =AA^\top\). Since these eigenvalue lists coincide, the nonzero singular values of \(A\) and of \(A^\top\) coincide. Finally, \(\|A\|_F^2=\sum_{i,j}a_{ij}^2=\operatorname{tr}(A^\top A)=\sum_i\lambda_i(A^\top A) =\sum_i\sigma_i^2\), the trace being the sum of eigenvalues (Theorem 3.2).

A concrete instance. Take the \(2\times 3\) matrix \(A=\begin{pmatrix}1&0&0\\0&2&0\end{pmatrix}\). Then \(A^\top A=\operatorname{diag}(1,4,0)\) is \(3\times 3\) with eigenvalues \(\{1,4,0\}\), while \(AA^\top=\operatorname{diag}(1,4)\) is \(2\times 2\) with eigenvalues \(\{1,4\}\): the two share the nonzero spectrum \(\{1,4\}\), differing only in the extra zero that the larger matrix must carry. The injection \(\mathbf v\mapsto A\mathbf v\) is visible here, for the eigenvector \(\mathbf v=(0,1,0)^\top\) of \(A^\top A\) at \(\lambda=4\) maps to \(A\mathbf v=(0,2)^\top\), an eigenvector of \(AA^\top\) at the same \(\lambda=4\). The singular values are \(\sigma_1=\sqrt4=2\) and \(\sigma_2=\sqrt1=1\), and indeed \(\|A\|_F^2=1^2+2^2=5=\sigma_1^2+\sigma_2^2\). \(\square\)

F.3.2 Solution to Exercise 3.12: Weyl’s eigenvalue perturbation bounds

This Chapter 3 exercise proves Weyl’s inequalities, which sandwich each eigenvalue of a sum \(A+B\) between the corresponding eigenvalue of \(A\) shifted by the smallest and the largest eigenvalue of \(B\).

Strategy. Everything comes from the variational (min-max) description of an eigenvalue as an optimized Rayleigh quotient. Once eigenvalues are written that way, the eigenvalues of a sum inherit the additivity of the quotient, and the lower bound is the upper bound applied to a cleverly chosen second pair.

Uses. The Courant-Fischer min-max theorem (Theorem 3.6): for a symmetric \(M\), the \(k\)-th smallest eigenvalue is \[ \lambda_k(M)=\min_{\dim S=k}\ \max_{\mathbf 0\neq\mathbf x\in S}R_M(\mathbf x), \qquad R_M(\mathbf x)=\frac{\mathbf x^\top M\mathbf x}{\mathbf x^\top\mathbf x}. \] Read this in words: to find \(\lambda_k\), look at every \(k\)-dimensional subspace \(S\), record the largest Rayleigh quotient attainable inside \(S\), and then take the smallest such record over all choices of \(S\). The min-max form is the only fact about eigenvalues we need.

The Rayleigh quotient is additive, \(R_{A+B}(\mathbf x)=R_A(\mathbf x)+R_B(\mathbf x)\), simply because the numerator splits: \(\mathbf x^\top(A+B)\mathbf x=\mathbf x^\top A\mathbf x+\mathbf x^\top B\mathbf x\), and the two pieces are divided by the same \(\mathbf x^\top\mathbf x\). This additivity is what lets a bound on \(A\) and a bound on \(B\) combine into a bound on \(A+B\).

Upper bound. Let \(S^\star\) be a \(k\)-dimensional subspace attaining \(\lambda_k(A)\), so \(\max_{\mathbf x\in S^\star}R_A(\mathbf x)=\lambda_k(A)\). For every \(\mathbf x\neq\mathbf 0\), \(R_B(\mathbf x)\le\lambda_n(B)\) (the largest eigenvalue is the maximum Rayleigh quotient). Hence \[ \lambda_k(A+B)\le\max_{\mathbf x\in S^\star}R_{A+B}(\mathbf x) =\max_{\mathbf x\in S^\star}\bigl(R_A(\mathbf x)+R_B(\mathbf x)\bigr) \le\lambda_k(A)+\lambda_n(B), \] the first inequality because \(\lambda_k(A+B)\) is the minimum over all \(k\)-dimensional subspaces and \(S^\star\) is one of them.

Lower bound. The idea is that a lower bound on \(\lambda_k(A+B)\) is the same as an upper bound read from the opposite direction: recover \(A\) from \(A+B\) by adding \(-B\), then apply the upper bound we just proved to the pair \((A+B)\) and \((-B)\). Using \(\lambda_n(-B)=-\lambda_1(B)\) (negating a symmetric matrix reverses and sign-flips its spectrum, so the largest eigenvalue of \(-B\) is minus the smallest of \(B\)), \(\lambda_k(A)=\lambda_k((A+B)+(-B))\le\lambda_k(A+B)+\lambda_n(-B)=\lambda_k(A+B)-\lambda_1(B)\), that is \(\lambda_k(A+B)\ge\lambda_k(A)+\lambda_1(B)\). Together, \[ \lambda_k(A)+\lambda_1(B)\le\lambda_k(A+B)\le\lambda_k(A)+\lambda_n(B). \]

Consequences. If \(B\succeq 0\) then \(\lambda_1(B)\ge 0\), so the lower bound gives \(\lambda_k(A+B)\ge\lambda_k(A)\) for every \(k\): adding a positive semidefinite matrix cannot decrease any eigenvalue. In general, \(\lambda_1(B)\ge-\|B\|_2\) and \(\lambda_n(B)\le\|B\|_2\) (the spectral norm of a symmetric matrix is \(\max_i|\lambda_i(B)|\)), so the two bounds give \(-\|B\|_2\le\lambda_k(A+B)-\lambda_k(A)\le\|B\|_2\), that is \(|\lambda_k(A+B)-\lambda_k(A)|\le\|B\|_2\) for every \(k\).

A numerical check. Let \(A=\operatorname{diag}(1,3)\), so \(\lambda_1(A)=1,\lambda_2(A)=3\), and perturb by \(B=\begin{pmatrix}0&0.1\\0.1&0\end{pmatrix}\), whose eigenvalues are \(\pm 0.1\), so \(\|B\|_2=0.1\). Then \(A+B=\begin{pmatrix}1&0.1\\0.1&3\end{pmatrix}\) has eigenvalues \(\tfrac12(4\pm\sqrt{4.04})=0.99501\) and \(3.00499\). Each moved by \(0.00499\) from the unperturbed value, comfortably inside the guaranteed band \(|\Delta\lambda|\le\|B\|_2=0.1\); and the two-sided Weyl bounds \(\lambda_1(A)+\lambda_1(B)=0.9\le 0.99501\le 1.1=\lambda_1(A)+\lambda_2(B)\) hold as promised.

Why this matters later: this is the statement that eigenvalues are stable, they move by no more than the size of the perturbation. A covariance matrix estimated from data is the true covariance plus a small error matrix \(B\); Weyl’s bound guarantees that its eigenvalues (hence its condition number and its positive-definiteness) cannot jump under that error. That stability underlies the robust and sandwich variance arguments of Chapter 5, where inverses of estimated information matrices must stay well-behaved, and the whitening steps of weighted least squares, which require the working covariance to stay positive definite under estimation noise. \(\square\)

F.4 Chapter 4: Probability and Distributions

F.4.1 Solution to Exercise 4.11: Building a normal vector with a prescribed covariance

This Chapter 4 exercise constructs a bivariate normal vector with a given correlation \(\rho\) by taking the symmetric square root of the target covariance matrix, and reads off when the construction degenerates.

Uses. The spectral theorem (Theorem 3.2) to form \(\boldsymbol\Sigma^{1/2}\), and the rule that a linear image \(\mathbf X=B\mathbf Z\) of a standard normal has covariance \(BB^\top\) (Theorem 4.9).

Write \(\boldsymbol\Sigma=\begin{pmatrix}1&\rho\\\rho&1\end{pmatrix}\) with \(\rho\in(-1,1)\). Its eigenvalues solve \(\det(\boldsymbol\Sigma-\lambda I)=(1-\lambda)^2-\rho^2=0\), giving \(\lambda_{\pm}=1\pm\rho\) with orthonormal eigenvectors \(\mathbf q_{+}=\tfrac{1}{\sqrt2}(1,1)^\top\) and \(\mathbf q_{-}=\tfrac{1}{\sqrt2}(1,-1)^\top\). The spectral decomposition (Theorem 3.2) is \(\boldsymbol\Sigma=Q\Lambda Q^\top\) with \(Q=\tfrac{1}{\sqrt2}\begin{pmatrix}1&1\\1&-1\end{pmatrix}\) and \(\Lambda=\operatorname{diag}(1+\rho,1-\rho)\). Taking nonnegative square roots of the eigenvalues, \[ \boldsymbol\Sigma^{1/2}=Q\Lambda^{1/2}Q^\top =\frac12\begin{pmatrix}\sqrt{1+\rho}+\sqrt{1-\rho}&\sqrt{1+\rho}-\sqrt{1-\rho}\\[2pt] \sqrt{1+\rho}-\sqrt{1-\rho}&\sqrt{1+\rho}+\sqrt{1-\rho}\end{pmatrix}, \] which is symmetric positive semidefinite and satisfies \(\boldsymbol\Sigma^{1/2}\boldsymbol\Sigma^{1/2} =\boldsymbol\Sigma\). Let \(\mathbf Z\sim\mathcal N_2(\mathbf 0,I_2)\) and set \(\mathbf X=\boldsymbol\Sigma^{1/2}\mathbf Z\). By Theorem 4.9 (linear images of a standard normal), \(\mathbb E[\mathbf X]=\mathbf 0\) and \(\operatorname{Cov}(\mathbf X)=\boldsymbol\Sigma^{1/2}I_2\boldsymbol\Sigma^{1/2}=\boldsymbol\Sigma\), so \(\mathbf X\) has the required covariance.

The matrix fails to be positive definite exactly when an eigenvalue vanishes, that is \(1+\rho=0\) or \(1-\rho=0\), that is \(\rho=\pm 1\) (excluded from the open interval; the boundary is the degenerate case). At \(\rho=1\) the eigenvalue \(1-\rho=0\) has eigenvector \(\mathbf q_{-}=\tfrac1{\sqrt2}(1,-1)^\top\). The factor \(2\) that appears here is worth spelling out, since it comes from the length of the difference vector, not from anywhere mysterious. Write \(X_1-X_2=(1,-1)\mathbf X\); since \((1,-1)=\sqrt2\,\mathbf q_{-}^\top\) (the raw contrast is \(\sqrt2\) times the unit eigenvector), we have \(X_1-X_2=\sqrt2\,\mathbf q_{-}^\top\mathbf X\), and therefore \[ \operatorname{Var}(X_1-X_2)=(\sqrt2)^2\,\mathbf q_{-}^\top\boldsymbol\Sigma\,\mathbf q_{-} =2\,\mathbf q_{-}^\top\boldsymbol\Sigma\,\mathbf q_{-}=2\,(1-\rho), \] using \(\boldsymbol\Sigma\mathbf q_{-}=(1-\rho)\mathbf q_{-}\) and \(\mathbf q_{-}^\top\mathbf q_{-}=1\). At \(\rho=1\) this is \(2(1-1)=0\), so \(X_1=X_2\) almost surely and the distribution concentrates on a line. For \(\rho\in(-1,1)\) both eigenvalues are positive, \(\boldsymbol\Sigma\succ 0\), and the pair is genuinely two-dimensional. When \(\rho=0\) the covariance is \(I_2\) and the two coordinates are uncorrelated; for jointly normal variables this forces independence (Lemma 4.3), consistent with the conditional law \(X_1\mid X_2\) of Theorem 4.15 having mean \(\rho X_2=0\) and variance \(1-\rho^2=1\), free of \(X_2\). \(\square\)

F.4.2 Solution to Exercise 4.12: The intersection axiom fails without positivity

This Chapter 4 exercise asks for a distribution that satisfies the two premises of the graphoid intersection axiom yet violates its conclusion, showing that the axiom needs a positivity hypothesis.

Uses. The intersection axiom (G5) of Theorem 4.5, restated here so it can be checked without turning back: for a probability law that is everywhere positive, \[ \bigl(X\perp Y\mid Z,W\bigr)\ \text{and}\ \bigl(X\perp W\mid Z,Y\bigr) \quad\Longrightarrow\quad X\perp (Y,W)\mid Z . \] Here positivity means every configuration carries strictly positive probability, \(p(x,y,z,w)>0\) for every combination of values in the supports; it is exactly this hypothesis we will violate.

Strategy. Make three of the variables perfect copies of a single random bit, so that conditioning on any one of them freezes the others and both pairwise premises hold trivially, while the unconditioned trio stays perfectly dependent, breaking the conclusion. The variable held out in the conclusion, \(Z\), must be degenerate, so that conditioning on it supplies no information and cannot itself freeze \(X\).

The construction. Let \(U\sim\mathrm{Ber}(1/2)\) and set \(X=Y=W=U\) (all three equal to the same bit), with \(Z\equiv 0\) a constant. The joint law of \((X,Y,W,Z)\) puts mass \(1/2\) on \((0,0,0,0)\) and mass \(1/2\) on \((1,1,1,0)\).

The two premises hold. Condition on \((Z,W)=(0,w)\). Given \(W=w\), both \(X=w\) and \(Y=w\) are frozen to the same constant, and two constants are independent, so \(X\perp Y\mid(Z,W)\). Symmetrically, condition on \((Z,Y)=(0,y)\): given \(Y=y\), both \(X=y\) and \(W=y\) are frozen, so \(X\perp W\mid(Z,Y)\). Each premise conditions on a variable that ties down \(X\), which is why it holds.

The conclusion fails. The conclusion of (G5) would be \(X\perp(Y,W)\mid Z\). But \(Z\equiv 0\) carries no information, so this is just \(X\perp(Y,W)\) unconditionally. It is false: \(X\), \(Y\), \(W\) are the same bit, so \(P(X=1,(Y,W)=(1,1))=\tfrac12\), whereas \(P(X=1)\,P((Y,W)=(1,1))=\tfrac12\cdot\tfrac12=\tfrac14\). Knowing \((Y,W)\) determines \(X\) exactly, so \(X\not\perp(Y,W)\mid Z\). The conclusion of the intersection axiom is violated even though both premises hold.

Which step of a positivity proof breaks. The standard proof of intersection multiplies and divides conditional densities, using the two premises in the form \[ p(x\mid y,z,w)=p(x\mid z,w)\quad\text{and}\quad p(x\mid y,z,w)=p(x\mid y,z), \] to conclude \(p(x\mid z,w)=p(x\mid y,z)\) and then, letting \(y\) and \(w\) range freely, that this common value depends on neither, which is the desired joint independence. Letting \(y\) and \(w\) range freely requires every combination \((y,z,w)\) to have positive probability, so that each conditional density is defined. Here positivity fails: the pair \((Y,W)\) is supported only on \((0,0)\) and \((1,1)\), so a cross-combination such as \(\{Y=1,W=0\}\) has probability zero and \(p(x\mid y=1,z,w=0)\) is undefined. The substitution chain that would propagate independence across all \((y,w)\) therefore cannot be carried out, and the deterministic tie among \(X,Y,W\) slips through the two pairwise premises undetected. \(\square\)

Common mistake (positivity versus zero probability). The tempting but wrong reading is that the intersection axiom is an unconditional law of probability. It is not: two conditional-independence facts do not combine into a joint one unless every relevant cell has positive mass. The same gap appears substantively in causal inference as the positivity (overlap) assumption of Chapters 9 and 10, where a covariate stratum that receives one treatment with probability zero blocks identification exactly as the zero-probability cell blocks the proof here. Zero probability is not a harmless edge case to wave away; it is precisely where deterministic ties hide.

F.5 Chapter 5: Statistical Inference

F.5.1 Solution to Exercise 5.8: A nonregular model where the score identity breaks

This Chapter 5 exercise exhibits a model, the uniform on \((0,\theta)\), whose moving support violates the regularity conditions, so the mean-zero-score identity and asymptotic normality both fail.

Strategy. Everything hinges on one fact: the support \((0,\theta)\) depends on the parameter, so differentiating under the integral sign is illegal and drops a boundary term. Trace that single defect through the score, the estimator, and its limit law.

Uses. The mean-zero-score identity Lemma 5.1 and the asymptotic-normality theorem Theorem 5.6, both of which assume the regular-model conditions of Definition 5.6.

Let \(X_1,\dots,X_n\) be independent \(\mathrm{Uniform}(0,\theta)\), with density \(f(x;\theta)=\theta^{-1} \mathbf 1\{0<x<\theta\}\).

(a) On the support, \(\log f(x;\theta)=-\log\theta\), so the naive score is \(\partial_\theta\log f=-1/\theta\), a nonzero constant. Its expectation is \(-1/\theta\neq 0\), so the score-mean-zero identity Lemma 5.1 fails. The reason is that the interchange of differentiation and integration underlying that identity requires a \(\theta\)-independent support; here the support \((0,\theta)\) moves with \(\theta\). Making this explicit with the Leibniz rule for a variable upper limit, \(\dfrac{\mathrm d}{\mathrm d\theta}\displaystyle\int_{0}^{b(\theta)}f(x,\theta)\,\mathrm dx =f(b(\theta),\theta)\,b'(\theta)+\displaystyle\int_{0}^{b(\theta)}\partial_\theta f(x,\theta)\,\mathrm dx\), applied to \(\int_0^\theta\theta^{-1}\,\mathrm dx=1\) with \(b(\theta)=\theta\) and \(f=\theta^{-1}\), gives \[ 0=\frac{\mathrm d}{\mathrm d\theta}\!\int_0^\theta\theta^{-1}\,\mathrm dx =\underbrace{\theta^{-1}\cdot 1}_{\text{boundary term}} +\underbrace{\int_0^\theta(-\theta^{-2})\,\mathrm dx}_{=\,-\theta^{-1}} . \] The naive score \(\partial_\theta\log f=-\theta^{-1}\) captures only the second, inside-the-integral piece; it omits the boundary term \(+\theta^{-1}\) contributed by the moving upper limit. That is exactly why \(\mathbb E[\text{naive score}]=-\theta^{-1}\) instead of zero: the two pieces should cancel, but the naive score keeps only one of them.

(b) The likelihood is \(L(\theta)=\theta^{-n}\mathbf 1\{\theta\ge\max_i X_i\}\), which is decreasing in \(\theta\) on \([\max_i X_i,\infty)\) and zero below \(\max_i X_i\). It is therefore maximized at the smallest admissible value, \(\hat\theta_n=\max_i X_i=:X_{(n)}\). For \(0<t<1\), \[ P\bigl(X_{(n)}\le t\theta\bigr)=\prod_{i=1}^n P(X_i\le t\theta)=t^n, \] so \(P\bigl(n(\theta-X_{(n)})/\theta>s\bigr)=P\bigl(X_{(n)}<\theta(1-s/n)\bigr)=(1-s/n)^n\to e^{-s}\) for each \(s>0\). Hence \(n(\theta-\hat\theta_n)/\theta\xrightarrow{d}\mathrm{Exp}(1)\). The estimation error is of order \(1/n\), not \(1/\sqrt n\), and the limit law is exponential, not normal.

(c) The asymptotic-normality theorem Theorem 5.6 assumes a regular model (Definition 5.6): in particular a common support independent of \(\theta\), twice differentiability of the log-likelihood in a neighborhood of the truth, and a finite positive Fisher information obtained by differentiating under the integral. Every one of these fails here. The support depends on \(\theta\) (so Definition 5.6 is violated and Lemma 5.1 does not hold), the maximizer sits on the boundary rather than at an interior stationary point of a smooth log-likelihood (assumption A1 fails), and there is no quadratic expansion of the log-likelihood about the truth driving a \(\sqrt n\) central limit theorem. The faster \(n\)-rate and the non-normal exponential limit are the signature of this nonregular, boundary problem. \(\square\)

F.5.2 Solution to Exercise 5.9: Sandwich versus model-based variance under overdispersion

This Chapter 5 exercise fits a mean by a deliberately wrong (Poisson) working model and shows that the sandwich variance, unlike the naive model-based variance, stays correct when the data are overdispersed.

Strategy. Treat the Poisson score as an estimating function, not a true likelihood. The M-estimation sandwich then supplies the honest variance, and comparing it to the model-based variance isolates a single inflation factor.

Uses. The M-estimation sandwich variance Definition 5.11 (bread \(\mathbf A=-\mathbb E[\partial_\mu\psi]\), meat \(\mathbf B=\mathbb E[\psi^2]\)).

Two pieces of vocabulary first, since the whole solution turns on them. A working model is a model we fit on purpose even though we do not believe it, using its score as an estimating equation vehicle; here the “working Poisson” is used only to define an estimator of the mean, and we make no claim that the counts are truly Poisson. And we write \(\kappa\) for the overdispersion ratio, the index of dispersion \(\kappa=\operatorname{Var}(X)/\mathbb E[X]=v_0/\mu_0\), the true variance-to-mean ratio; \(\kappa=1\) is the Poisson case, and \(\kappa>1\) is overdispersion (more scatter than Poisson allows). Throughout, \(\mu_0=\mathbb E[X]\) and \(v_0=\operatorname{Var}(X)\), so \(v_0=\kappa\mu_0\).

The working Poisson log-likelihood contribution is \(x\log\mu-\mu\).

(a) Differentiating in \(\mu\) gives the working score \(\psi(x;\mu)=\partial_\mu(x\log\mu-\mu)=x/\mu-1\). The estimating equation \(\sum_i\psi(X_i;\mu)=0\) reads \(\sum_i(X_i/\mu-1)=0\), that is \(\mu=\bar X_n\), so the M-estimator is \(\hat\mu_n=\bar X_n\). Its target \(\mu^\star\) solves the population equation \(\mathbb E[\psi(X;\mu^\star)]=0\), that is \(\mathbb E[X]/\mu^\star-1=0\), so \(\mu^\star=\mathbb E[X]=\mu_0\): the working estimator is consistent for the true mean even though the variance is misspecified.

(b) Using the M-estimation sandwich of Definition 5.11 with bread \(\mathbf A(\mu)=-\mathbb E[\partial_\mu\psi(X;\mu)]\) and meat \(\mathbf B(\mu)=\mathbb E[\psi(X;\mu)^2]\), evaluated at \(\mu_0\): \[ \mathbf A(\mu_0)=-\mathbb E\!\left[-\frac{X}{\mu_0^2}\right]=\frac{\mathbb E[X]}{\mu_0^2}=\frac{\mu_0}{\mu_0^2}=\frac{1}{\mu_0}, \] \[ \mathbf B(\mu_0)=\mathbb E\!\left[\Bigl(\frac{X}{\mu_0}-1\Bigr)^2\right]=\frac{\operatorname{Var}(X)}{\mu_0^2} =\frac{v_0}{\mu_0^2}=\frac{\kappa\mu_0}{\mu_0^2}=\frac{\kappa}{\mu_0}. \] The sandwich asymptotic variance is \[ \frac1n\,\mathbf A(\mu_0)^{-1}\mathbf B(\mu_0)\mathbf A(\mu_0)^{-1} =\frac1n\,\mu_0\cdot\frac{\kappa}{\mu_0}\cdot\mu_0=\frac{\kappa\,\mu_0}{n}. \] The naive model-based variance assumes correct Poisson specification, under which the variance equals the inverse information \(\mathbf A(\mu_0)^{-1}/n=\mu_0/n\).

(c) The ratio of sandwich to naive variance is \(\kappa\), so the naive standard error \(\sqrt{\mu_0/n}\) understates the true standard error \(\sqrt{\kappa\mu_0/n}\) by the factor \(\sqrt\kappa>1\). This is exactly the situation of Corollary 5.3 read in reverse: when the model is correct (\(\kappa=1\)), bread and meat agree, \(\mathbf A=\mathbf B=1/\mu_0=\mathcal I\), and the sandwich collapses to the inverse-information variance \(\mu_0/n\); overdispersion (\(\kappa>1\)) drives meat above bread and inflates the true variance by \(\kappa\), which only the sandwich estimator captures. \(\square\)

F.6 Chapter 6: Linear Regression from First Principles

F.6.1 Solution to Exercise 6.8: The weighted (generalized) Gauss-Markov theorem

This Chapter 6 exercise extends the Gauss-Markov optimality of ordinary least squares to correlated, unequal-variance errors, showing that generalized (weighted) least squares is the best linear unbiased estimator.

Strategy. Do not re-run the whole optimality proof. Instead whiten the model by multiplying through by \(V^{-1/2}\), which turns the correlated errors into spherical ones, apply the ordinary Gauss-Markov theorem to the whitened model, and translate the conclusion back.

Uses. The ordinary Gauss-Markov theorem Theorem 6.3 and the symmetric square root \(V^{-1/2}\) from the spectral theorem Theorem 3.2.

Replace G2 by \(\operatorname{Cov}(\boldsymbol\varepsilon)=\sigma^2 V\) with \(V\succ 0\) known, and define \(\hat{\boldsymbol\beta}_V=(\mathbf X^\top V^{-1}\mathbf X)^{-1}\mathbf X^\top V^{-1}\mathbf y\).

Unbiasedness and covariance. With \(\mathbb E[\mathbf y]=\mathbf X\boldsymbol\beta\), \[ \mathbb E[\hat{\boldsymbol\beta}_V]=(\mathbf X^\top V^{-1}\mathbf X)^{-1}\mathbf X^\top V^{-1}\mathbf X\boldsymbol\beta =\boldsymbol\beta, \] so \(\hat{\boldsymbol\beta}_V\) is unbiased. Writing \(G=(\mathbf X^\top V^{-1}\mathbf X)^{-1}\mathbf X^\top V^{-1}\), \[ \operatorname{Cov}(\hat{\boldsymbol\beta}_V)=G\,(\sigma^2 V)\,G^\top =\sigma^2(\mathbf X^\top V^{-1}\mathbf X)^{-1}\mathbf X^\top V^{-1}\,V\,V^{-1}\mathbf X(\mathbf X^\top V^{-1}\mathbf X)^{-1} =\sigma^2(\mathbf X^\top V^{-1}\mathbf X)^{-1}. \]

Optimality (generalized Gauss-Markov). Factor \(V^{-1}=W^\top W\) with \(W\) invertible (the symmetric square root \(V^{-1/2}\) from the spectral theorem Theorem 3.2 serves, with \(W=W^\top=V^{-1/2}\)). Apply \(W\) to the model: set \(\tilde{\mathbf y}=W\mathbf y\), \(\tilde{\mathbf X}=W\mathbf X\), \(\tilde{\boldsymbol\varepsilon}=W\boldsymbol\varepsilon\). Then \(\tilde{\mathbf y}=\tilde{\mathbf X}\boldsymbol\beta+\tilde{\boldsymbol\varepsilon}\) with \(\mathbb E[\tilde{\boldsymbol\varepsilon}]=\mathbf 0\) and \(\operatorname{Cov}(\tilde{\boldsymbol\varepsilon})=W(\sigma^2 V)W^\top=\sigma^2 W V W^\top =\sigma^2 (W V^{1/2})(W V^{1/2})^\top\). Since \(W=V^{-1/2}\), \(WVW^\top=V^{-1/2}VV^{-1/2}=I\), so the transformed model satisfies the ordinary Gauss-Markov hypotheses G1, G2, G3 with spherical errors of variance \(\sigma^2 I\). The ordinary least squares estimator in the transformed model is \[ (\tilde{\mathbf X}^\top\tilde{\mathbf X})^{-1}\tilde{\mathbf X}^\top\tilde{\mathbf y} =(\mathbf X^\top W^\top W\mathbf X)^{-1}\mathbf X^\top W^\top W\mathbf y =(\mathbf X^\top V^{-1}\mathbf X)^{-1}\mathbf X^\top V^{-1}\mathbf y=\hat{\boldsymbol\beta}_V, \] so \(\hat{\boldsymbol\beta}_V\) is exactly the OLS estimator of the whitened model. By the ordinary Gauss-Markov theorem (Theorem 6.3) it is the best linear unbiased estimator there. Linear unbiased estimators of \(\boldsymbol\beta\) in the original model correspond bijectively to linear unbiased estimators in the transformed model (the bijection \(\mathbf y\mapsto W\mathbf y\) is invertible and carries one class onto the other), and the estimand and the variance are unchanged by the relabeling. Hence \(\hat{\boldsymbol\beta}_V\) is the best linear unbiased estimator under \(\operatorname{Cov}(\boldsymbol \varepsilon)=\sigma^2V\), generalizing Theorem 6.3.

This is the weighted least squares optimality on which the matching-adjusted estimator of Chapter 18 rests. Concretely: MAIC fits an outcome model on a reweighted sample, where the matching weights play the role of \(V^{-1}\), downweighting individuals who are over-represented relative to the target population. The theorem says that, given those weights, the weighted fit is the minimum-variance unbiased estimator, so no other linear reweighting of the same data can do better. The practical consequence for a health technology assessment is that once the weights are fixed by the covariate-balancing step, there is nothing further to gain by tinkering with the estimator; the remaining efficiency question is entirely about the weights themselves (the effective sample size they induce), which is precisely why later chapters study weight variance rather than estimator choice. \(\square\)

F.6.2 Solution to Exercise 6.9: General linear hypothesis equals the nested-model \(F\) test

This Chapter 6 exercise proves that the Wald-style statistic for a general linear hypothesis \(R\boldsymbol\beta=\mathbf q\) is algebraically identical to the extra-sum-of-squares \(F\) statistic from comparing the constrained and unconstrained fits.

Strategy. Solve the constrained least-squares problem with a Lagrange multiplier, express the residual sum of squares gap as a single quadratic form in the constraint discrepancy \(R\hat{\boldsymbol\beta}-\mathbf q\), and match it term for term against the \(F\) statistic.

Uses. The normal equations \(\mathbf X^\top(\mathbf y-\mathbf X\hat{\boldsymbol\beta})=\mathbf 0\) and the sampling theory of Theorem 6.10.

Let the full model have least-squares fit \(\hat{\boldsymbol\beta}\) with residual sum of squares \(\mathrm{RSS}_1=S(\hat{\boldsymbol\beta})\), where \(S(\mathbf b)=\|\mathbf y-\mathbf X\mathbf b\|^2\). The reduced model is the constrained fit minimizing \(S\) over the affine set \(\{\mathbf b:R\mathbf b=\mathbf q\}\); call its minimizer \(\hat{\boldsymbol\beta}_R\) and its residual sum of squares \(\mathrm{RSS}_0=S(\hat{\boldsymbol\beta}_R)\). We show the general-linear-hypothesis statistic Equation 6.10 equals the nested-model statistic Equation 6.9 with \(q=\operatorname{rank}(R)\) degrees of freedom in the numerator.

Constrained minimization. Minimize \(S(\mathbf b)\) subject to \(R\mathbf b=\mathbf q\) with a Lagrange multiplier \(\boldsymbol\lambda\in\mathbb R^q\). Stationarity of \(S(\mathbf b)+2\boldsymbol\lambda^\top(R\mathbf b-\mathbf q)\) gives \(-2\mathbf X^\top(\mathbf y-\mathbf X\mathbf b)+2R^\top\boldsymbol\lambda=\mathbf 0\), that is \(\mathbf X^\top\mathbf X\mathbf b=\mathbf X^\top\mathbf y-R^\top\boldsymbol\lambda\). Since \(\mathbf X^\top\mathbf X\hat{\boldsymbol\beta}=\mathbf X^\top\mathbf y\), \[ \hat{\boldsymbol\beta}_R=\hat{\boldsymbol\beta}-(\mathbf X^\top\mathbf X)^{-1}R^\top\boldsymbol\lambda . \] Impose the constraint \(R\hat{\boldsymbol\beta}_R=\mathbf q\): \(R\hat{\boldsymbol\beta}-R(\mathbf X^\top\mathbf X)^{-1}R^\top\boldsymbol\lambda=\mathbf q\), so with \(A=R(\mathbf X^\top\mathbf X)^{-1}R^\top\) (positive definite, hence invertible, as in Theorem 6.10), \[ \boldsymbol\lambda=A^{-1}(R\hat{\boldsymbol\beta}-\mathbf q). \]

Difference of residual sums of squares. Let \(\mathbf d=\hat{\boldsymbol\beta}_R-\hat{\boldsymbol\beta}=-(\mathbf X^\top\mathbf X)^{-1}R^\top\boldsymbol \lambda\), and write \(\mathbf r=\mathbf y-\mathbf X\hat{\boldsymbol\beta}\) for the unconstrained residual, so \(\mathrm{RSS}_1=\|\mathbf r\|^2\). Evaluate \(S\) at \(\hat{\boldsymbol\beta}_R=\hat{\boldsymbol\beta}+\mathbf d\) by expanding the squared norm: \[ S(\hat{\boldsymbol\beta}+\mathbf d)=\|\mathbf y-\mathbf X(\hat{\boldsymbol\beta}+\mathbf d)\|^2 =\|\mathbf r-\mathbf X\mathbf d\|^2 =\|\mathbf r\|^2-2\,\mathbf d^\top\mathbf X^\top\mathbf r+\mathbf d^\top\mathbf X^\top\mathbf X\mathbf d . \] The middle term vanishes because \(\mathbf X^\top\mathbf r=\mathbf X^\top(\mathbf y-\mathbf X\hat{\boldsymbol \beta})=\mathbf 0\) by the normal equations, so \(S(\hat{\boldsymbol\beta}+\mathbf d)=\|\mathbf r\|^2 +\mathbf d^\top\mathbf X^\top\mathbf X\mathbf d\) and hence \[ \mathrm{RSS}_0-\mathrm{RSS}_1=S(\hat{\boldsymbol\beta}_R)-S(\hat{\boldsymbol\beta}) =\mathbf d^\top\mathbf X^\top\mathbf X\mathbf d . \] Substituting \(\mathbf d\), \[ \mathbf d^\top\mathbf X^\top\mathbf X\mathbf d =\boldsymbol\lambda^\top R(\mathbf X^\top\mathbf X)^{-1}\mathbf X^\top\mathbf X(\mathbf X^\top\mathbf X)^{-1}R^\top\boldsymbol\lambda =\boldsymbol\lambda^\top A\boldsymbol\lambda =(R\hat{\boldsymbol\beta}-\mathbf q)^\top A^{-1}(R\hat{\boldsymbol\beta}-\mathbf q), \] using \(\boldsymbol\lambda=A^{-1}(R\hat{\boldsymbol\beta}-\mathbf q)\) and \(A^\top=A\).

Equality of the statistics. The constrained model adds \(q=\operatorname{rank}(R)\) restrictions, so it has \(p_0=p-q\) free parameters and the nested-model statistic Equation 6.9 is \[ F=\frac{(\mathrm{RSS}_0-\mathrm{RSS}_1)/q}{\mathrm{RSS}_1/(n-p)} =\frac{(R\hat{\boldsymbol\beta}-\mathbf q)^\top A^{-1}(R\hat{\boldsymbol\beta}-\mathbf q)/q}{s^2}, \] where \(s^2=\mathrm{RSS}_1/(n-p)\). This is exactly Equation 6.10. Both statistics therefore coincide as random variables; in particular the sampling distribution \(F_{q,n-p}\) established for Equation 6.10 in Theorem 6.10 (via the whitening identity that \(\mathbf v\sim\mathcal N_q(\mathbf 0, \sigma^2A)\) implies \(\mathbf v^\top A^{-1}\mathbf v/\sigma^2\sim\chi^2_q\)) is the distribution of the nested-model \(F\) as well. \(\square\)

F.7 Chapter 7: Generalized Linear Models

F.7.1 Solution to Exercise 7.8: IRLS for the probit link, and why Fisher scoring differs from Newton

This Chapter 7 exercise derives the iteratively reweighted least squares (IRLS) weights and working response for the probit link, then explains why, because probit is a non-canonical link, Fisher scoring and Newton-Raphson are genuinely different algorithms.

Strategy. The IRLS pieces come straight from the general GLM formulas by plugging in the probit mean and variance. The subtle part is the last: differentiate the score once more and watch a residual-weighted term survive in the observed information but average to zero in the expected information.

Uses. The GLM information weight Equation 7.14, the working response Equation 7.17, and the canonical-link Hessian lemma Lemma 7.1.

For the probit link \(g(\mu)=\Phi^{-1}(\mu)\), the mean is \(\mu=g^{-1}(\eta)=\Phi(\eta)\), so \(\mathrm d\mu/\mathrm d\eta=\Phi'(\eta)\) and \(g'(\mu)=\mathrm d\eta/\mathrm d\mu=1/\Phi'(\eta)\), where \(\Phi'=\phi\) is the standard normal density.

Working weights. For binomial data with denominator \(n_i\) the variance of the proportion is \(\operatorname{Var}(Y_i)=\mu_i(1-\mu_i)/n_i=\Phi(\eta_i)(1-\Phi(\eta_i))/n_i\). The GLM information weight of Equation 7.14 is \(W_i=(\mathrm d\mu_i/\mathrm d\eta_i)^2/\operatorname{Var}(Y_i)\), hence \[ W_i=\frac{\Phi'(\eta_i)^2}{\Phi(\eta_i)(1-\Phi(\eta_i))/n_i} =\frac{n_i\,\Phi'(\eta_i)^2}{\Phi(\eta_i)\bigl(1-\Phi(\eta_i)\bigr)}, \] the stated formula.

Working response. By Equation 7.17, \(z_i=\eta_i+(y_i-\mu_i)g'(\mu_i)=\eta_i+\dfrac{y_i-\Phi(\eta_i)}{\Phi'(\eta_i)}\). Each IRLS step (Theorem 7.3) solves the weighted normal equations \(\mathbf X^\top\mathbf W\mathbf X\boldsymbol\beta =\mathbf X^\top\mathbf W\mathbf z\) with these \(W_i\) and \(z_i\), evaluated at the current iterate.

Observed versus expected information. This is the conceptual punchline, so we write the Hessian out. For a canonical link the Hessian of the log-likelihood is nonstochastic, so the observed Hessian equals its expectation and Fisher scoring coincides with Newton-Raphson (Lemma 7.1). The probit link is non-canonical, and the difference surfaces on differentiating the score once more. Write \(c(\eta)\) for the coefficient factor \[ c(\eta)=\frac{\mathrm d\mu/\mathrm d\eta}{\operatorname{Var}(Y)}=\frac{\phi(\eta)}{\Phi(\eta)\bigl(1-\Phi(\eta)\bigr)} \] (taking \(n_i=1\) for transparency), so the score is \(\mathbf U(\boldsymbol\beta)=\sum_i(y_i-\mu_i)\,c(\eta_i)\,\mathbf x_i\). Differentiating in \(\boldsymbol\beta\), with \(\eta_i=\mathbf x_i^\top\boldsymbol\beta\) and \(\mathrm d\mu_i/\mathrm d\eta_i=\phi(\eta_i)\), the product rule gives \[ \nabla^2\ell=\sum_i\Bigl[(y_i-\mu_i)\,c'(\eta_i)-c(\eta_i)\,\phi(\eta_i)\Bigr]\mathbf x_i\mathbf x_i^\top . \] The observed information \(-\nabla^2\ell\) keeps the data-residual term \((y_i-\mu_i)\,c'(\eta_i)\); the expected information replaces \(y_i-\mu_i\) by its mean \(0\), dropping that term and leaving \(\sum_i c(\eta_i)\phi(\eta_i)\mathbf x_i\mathbf x_i^\top=\sum_i\phi(\eta_i)^2/[\Phi(\eta_i)(1-\Phi(\eta_i))]\,\mathbf x_i\mathbf x_i^\top =\mathbf X^\top\mathbf W\mathbf X\), the Fisher-scoring weight matrix. So the two informations differ by exactly \(\sum_i(y_i-\mu_i)\,c'(\eta_i)\,\mathbf x_i\mathbf x_i^\top\), and everything comes down to whether \(c'(\eta)\) vanishes. It does not, in general: using \(\phi'(\eta)=-\eta\phi(\eta)\) and \(\tfrac{\mathrm d}{\mathrm d\eta}[\Phi(1-\Phi)]=\phi(1-2\Phi)\), the quotient rule gives \[ c'(\eta)=\frac{\phi(\eta)\bigl[-\eta\,\Phi(\eta)(1-\Phi(\eta))-\phi(\eta)(1-2\Phi(\eta))\bigr]}{\bigl[\Phi(\eta)(1-\Phi(\eta))\bigr]^2}, \] which is not identically zero; it is nonzero for every \(\eta\neq 0\), vanishing only at the isolated point \(\eta=0\) where the two bracketed terms cancel. Because the residual term therefore does not drop out, observed and expected information differ, and Fisher scoring (which uses the expected information \(\mathbf X^\top\mathbf W\mathbf X\)) is not identical to Newton-Raphson, in contrast to the canonical case, where \(c'\equiv 0\) makes the residual term disappear. \(\square\)

F.7.2 Solution to Exercise 7.12: The odds ratio is non-collapsible and attenuates toward the null

This Chapter 7 exercise proves that marginalizing a constant conditional odds ratio over a covariate strictly shrinks it toward \(1\), so the marginal odds ratio lies between the null and the conditional value. This is non-collapsibility, and it is the reason odds ratios do not transport across populations with different covariate mixes.

Strategy. Rewrite each marginal odds as a single scalar function \(A(a)\) of the odds-multiplier \(a\) (equal to \(1\) for control, \(\omega\) for treatment), and prove two monotonicity facts about \(A\): it increases (giving \(\omega^{\mathrm M}>1\)) but grows sub-linearly, \(A(a)/a\) decreasing (giving \(\omega^{\mathrm M}<\omega\)).

Uses. Jensen’s inequality Theorem 4.6 and the covariance inequality for like-monotone functions (two functions that both increase in the same variable have nonnegative covariance).

Common mistake (non-collapsibility versus effect modification). The attenuation proved here happens even when the conditional odds ratio \(\omega\) is identical in every stratum, that is, with no effect modification at all. Non-collapsibility is a property of the nonlinear odds scale, not evidence that the treatment works differently in different subgroups. Do not read a marginal-conditional gap as effect modification; a constant conditional odds ratio still produces a population-dependent marginal one.

Let \(X\in\{0,1\}\) with \(P(X=1)=\pi\), constant conditional odds ratio \(\omega>1\) on the logit scale, and baseline risks \(\mu_0(0)\neq\mu_0(1)\). Under the no-confounding identity of Section 7.7, \(\mathbb E[Y(t)]=\mathbb E_X[\mu_t(X)]\), the marginal odds ratio is \(\omega^{\mathrm M}=\dfrac{\mathbb E[\mu_1(X)]/(1-\mathbb E[\mu_1(X)])}{\mathbb E[\mu_0(X)]/(1-\mathbb E[\mu_0(X)])}\).

(a) A numerical example. Take \(\pi=\tfrac12\) and baseline risks \(\mu_0(0)=0.1\), \(\mu_0(1)=0.5\), with \(\omega=4\). Solving \(\mu_1(x)/(1-\mu_1(x))=4\,\mu_0(x)/(1-\mu_0(x))\) gives the treated risks \(\mu_1(0)=\dfrac{4(1/9)}{1+4(1/9)}=\dfrac{4/9}{13/9}=\tfrac{4}{13}=0.30769\) and \(\mu_1(1)=\dfrac{4(1)}{1+4(1)}=\tfrac45=0.8\). The marginal risks are \(\mathbb E[\mu_0]=\tfrac12(0.1+0.5)=0.3\) and \(\mathbb E[\mu_1]=\tfrac12(0.30769+0.8)=0.55385\), so \[ \omega^{\mathrm M}=\frac{0.55385/0.44615}{0.3/0.7}=\frac{1.24138}{0.42857}=2.8966, \] which lies strictly between \(1\) and \(\omega=4\). The conditional log odds ratio \(\log 4=1.3863\) is attenuated to the marginal \(\log 2.8966=1.0635\).

To make the attenuation visible as the covariate prevalence \(\pi=P(X=1)\) moves, hold the conditional risks (and hence the conditional odds ratio \(\omega=4\)) fixed and recompute the marginal odds ratio at several prevalences:

\(\pi=P(X=1)\)	\(\mathbb E[\mu_0]\)	\(\mathbb E[\mu_1]\)	marginal OR \(\omega^{\mathrm M}\)
\(0.00\)	\(0.100\)	\(0.30769\)	\(4.0000\)
\(0.25\)	\(0.200\)	\(0.43077\)	\(3.0270\)
\(0.50\)	\(0.300\)	\(0.55385\)	\(2.8966\)
\(0.75\)	\(0.400\)	\(0.67692\)	\(3.1429\)
\(1.00\)	\(0.500\)	\(0.80000\)	\(4.0000\)

The marginal odds ratio equals the conditional value \(4\) only at the two degenerate endpoints (\(\pi=0\) or \(\pi=1\), where the covariate carries no variation), and dips below it in between, reaching its strongest attenuation near \(\pi=0.5\) where the baseline-odds spread is largest. Non-collapsibility is therefore a distributional effect: the same two treatments carry a different marginal odds ratio in every population, with no change in any conditional parameter.

(b) The general bound. Write the baseline odds as the random variable \(W=\mu_0(X)/(1-\mu_0(X))\), which is non-degenerate because \(\mu_0(0)\neq\mu_0(1)\). The constant-odds-ratio relation gives \(\mu_0(X)=W/(1+W)\) and \(\mu_1(X)=\omega W/(1+\omega W)\). The right object to introduce is the marginal odds that result when every stratum’s baseline odds \(W\) is multiplied by a common factor \(a\). Under that tilt the stratum risk becomes \(aW/(1+aW)\), so the marginal odds (marginal risk over one minus marginal risk) is the ratio below. For \(a>0\) set \[ A(a)=\frac{\mathbb E[aW/(1+aW)]}{\mathbb E[1/(1+aW)]}, \qquad\text{so}\qquad \frac{\mathbb E[\mu_0]}{1-\mathbb E[\mu_0]}=A(1),\quad\frac{\mathbb E[\mu_1]}{1-\mathbb E[\mu_1]}=A(\omega), \] using \(1-aW/(1+aW)=1/(1+aW)\). In words, \(A(a)\) is the marginal odds when every conditional odds is scaled by \(a\): at \(a=1\) it is the control marginal odds and at \(a=\omega\) the treated marginal odds, so \(\omega^{\mathrm M}=A(\omega)/A(1)\). The whole problem has now been reduced to understanding how the single scalar function \(A\) grows.

First, \(A\) is strictly increasing. Writing \(D(a)=\mathbb E[1/(1+aW)]\) and noting \(aW/(1+aW)=1-1/(1+aW)\), we get \(A(a)=(1-D(a))/D(a)=1/D(a)-1\). The integrand \(1/(1+aW)\) is strictly decreasing in \(a\) on the event \(\{W>0\}\), which has probability one (risks lie in \((0,1)\)), so \(D\) is strictly decreasing, \(1/D\) strictly increasing, and \(A\) strictly increasing. Since \(\omega>1\) and \(A(1)>0\), this yields \(\omega^{\mathrm M}=A(\omega)/A(1)>1\).

Second, \(a\mapsto A(a)/a\) is strictly decreasing. Indeed \(A(a)/a=\mathbb E[W/(1+aW)]/\mathbb E[1/(1+aW)]=\mathbb E_{Q_a}[W]\), the mean of \(W\) under the tilted law \(\mathrm dQ_a/\mathrm dP\propto 1/(1+aW)\). By the standard exponential-tilt identity, \[ \frac{\mathrm d}{\mathrm da}\mathbb E_{Q_a}[W] =\operatorname{Cov}_{Q_a}\!\Bigl(W,\ \frac{\mathrm d}{\mathrm da}\log\frac{1}{1+aW}\Bigr) =-\operatorname{Cov}_{Q_a}\!\Bigl(W,\ \frac{W}{1+aW}\Bigr)<0, \] because \(W\) and \(W/(1+aW)\) are both strictly increasing in \(W\), so their covariance is positive (strictly, since \(W\) is non-degenerate). Hence \(A(a)/a\) is strictly decreasing, and with \(\omega>1\), \(\dfrac{\omega^{\mathrm M}}{\omega}=\dfrac{A(\omega)/\omega}{A(1)}=\dfrac{\mathbb E_{Q_\omega}[W]}{\mathbb E_{Q_1}[W]}<1\), that is \(\omega^{\mathrm M}<\omega\).

Together \(1<\omega^{\mathrm M}<\omega\): the marginal odds ratio lies strictly between the null and the conditional value, so the odds ratio is attenuated toward the null and is non-collapsible. The engine of the argument is the covariance inequality for like-monotone functions (both \(w\mapsto w\) and \(w\mapsto w/(1+aw)\) increase in the baseline odds \(w\)), not any convexity of \(a\mapsto a/(1+a)\); that map is in fact strictly concave for \(a>0\), and it is precisely this concavity, fed through Jensen’s inequality Theorem 4.6 to the risk written in odds coordinates, that delivers the same attenuation from the complementary marginal-risk viewpoint. Equality \(\omega^{\mathrm M}=\omega\) would require \(W\) degenerate, that is equal baseline risks, which is the agreement-iff-degenerate clause of Corollary 11.1.

For a health technology assessment the message is concrete: if a trial reports a conditional (adjusted) odds ratio and a competitor reports a marginal (unadjusted) one, the two are not on the same footing even with no bias and no effect modification, because marginalizing over a different patient mix moves the odds ratio on its own. An indirect comparison that mixes conditional and marginal odds ratios, or that transports a marginal odds ratio to a population with a different covariate distribution, will be biased for this reason alone. \(\square\)

F.8 Chapter 8: Hierarchical Models, Shrinkage, and Bayesian Inference

F.8.1 Solution to Exercise 8.10: Stein’s unbiased risk estimate and James-Stein domination

This Chapter 8 exercise derives Stein’s unbiased risk estimate (SURE), uses it to show that the James-Stein shrinkage estimator beats the raw estimator in \(J\ge 3\) dimensions, and reads the result as the frequentist justification for partial pooling in meta-analysis.

Strategy. SURE turns the unknowable risk into a computable expectation by trading the cross term \(\mathbb E[(\mathbf Y-\boldsymbol\theta)^\top\mathbf g]\) for a divergence \(\mathbb E[\nabla\cdot\mathbf g]\) via Stein’s lemma; then the specific James-Stein \(\mathbf g\) makes the divergence and the squared shrinkage combine into a single negative term.

Uses. Stein’s lemma Lemma 8.2 (for \(Y\sim\mathcal N(\theta,\sigma^2)\), \(\mathbb E[(Y-\theta)h(Y)]=\sigma^2\mathbb E[h'(Y)]\)) and the risk identity Theorem 8.6.

Let \(\mathbf Y\sim\mathcal N_J(\boldsymbol\theta,\sigma^2 I_J)\) with \(J\ge 3\) and \(\hat{\boldsymbol\theta}=\mathbf Y+\mathbf g(\mathbf Y)\).

(a) Stein’s unbiased risk estimate. Expand the loss: \[ \|\hat{\boldsymbol\theta}-\boldsymbol\theta\|^2=\|\mathbf Y-\boldsymbol\theta+\mathbf g(\mathbf Y)\|^2 =\|\mathbf Y-\boldsymbol\theta\|^2+2(\mathbf Y-\boldsymbol\theta)^\top\mathbf g(\mathbf Y)+\|\mathbf g(\mathbf Y)\|^2 . \] Take expectations term by term. First \(\mathbb E\|\mathbf Y-\boldsymbol\theta\|^2=J\sigma^2\). For the cross term, apply Stein’s lemma (Lemma 8.2) coordinatewise: with \(Y_j\sim\mathcal N(\theta_j,\sigma^2)\) and the other coordinates held fixed, \(\mathbb E[(Y_j-\theta_j)g_j(\mathbf Y)]=\sigma^2\,\mathbb E[\partial g_j/\partial Y_j]\). Summing over \(j\), \(\mathbb E[(\mathbf Y-\boldsymbol\theta)^\top\mathbf g(\mathbf Y)]=\sigma^2\,\mathbb E[\nabla\cdot\mathbf g(\mathbf Y)]\). Hence \[ \mathbb E\|\hat{\boldsymbol\theta}-\boldsymbol\theta\|^2 =\mathbb E\bigl[J\sigma^2+2\sigma^2\nabla\cdot\mathbf g(\mathbf Y)+\|\mathbf g(\mathbf Y)\|^2\bigr]. \]

(b) The James-Stein choice. Put \(\mathbf g(\mathbf y)=-(J-2)\sigma^2\mathbf y/\|\mathbf y\|^2\). Then \(\|\mathbf g(\mathbf y)\|^2=(J-2)^2\sigma^4/\|\mathbf y\|^2\). For the divergence, differentiate the \(j\)-th component \(g_j(\mathbf y)=-(J-2)\sigma^2\,y_j/\|\mathbf y\|^2\) by the quotient rule, using \(\partial\|\mathbf y\|^2/\partial y_j=2y_j\): \[ \frac{\partial}{\partial y_j}\frac{y_j}{\|\mathbf y\|^2} =\frac{1\cdot\|\mathbf y\|^2-y_j\cdot 2y_j}{\|\mathbf y\|^4} =\frac{\|\mathbf y\|^2-2y_j^2}{\|\mathbf y\|^4}, \qquad\text{so}\qquad \frac{\partial g_j}{\partial y_j}=-(J-2)\sigma^2\,\frac{\|\mathbf y\|^2-2y_j^2}{\|\mathbf y\|^4}. \] Summing over \(j=1,\dots,J\) and using \(\sum_j\|\mathbf y\|^2=J\|\mathbf y\|^2\) together with \(\sum_j y_j^2=\|\mathbf y\|^2\), \[ \nabla\cdot\mathbf g=-(J-2)\sigma^2\,\frac{\sum_j\bigl(\|\mathbf y\|^2-2y_j^2\bigr)}{\|\mathbf y\|^4} =-(J-2)\sigma^2\,\frac{J\|\mathbf y\|^2-2\|\mathbf y\|^2}{\|\mathbf y\|^4} =-(J-2)\sigma^2\,\frac{(J-2)\|\mathbf y\|^2}{\|\mathbf y\|^4} =-\frac{(J-2)^2\sigma^2}{\|\mathbf y\|^2}. \] Substituting, \[ \mathbb E\|\hat{\boldsymbol\theta}^{\mathrm{JS}}-\boldsymbol\theta\|^2 =J\sigma^2+2\sigma^2\Bigl(-(J-2)^2\sigma^2\,\mathbb E\tfrac{1}{\|\mathbf Y\|^2}\Bigr) +(J-2)^2\sigma^4\,\mathbb E\tfrac{1}{\|\mathbf Y\|^2} =J\sigma^2-(J-2)^2\sigma^4\,\mathbb E\!\left[\tfrac{1}{\|\mathbf Y\|^2}\right], \] which is the risk identity of Theorem 8.6, strictly below \(J\sigma^2\) because the subtracted term is positive when \(J\ge 3\). The positive-part estimator \(\hat{\boldsymbol\theta}^{\mathrm{JS+}}=(1-(J-2)\sigma^2/\|\mathbf Y\|^2)_{+}\mathbf Y\) replaces the shrinkage factor by its nonnegative part. When the raw factor \(1-(J-2)\sigma^2/\|\mathbf Y\|^2\) is negative, \(\hat{\boldsymbol\theta}^{\mathrm{JS}}\) points opposite to \(\mathbf Y\) and overshoots the origin, which is never a sensible estimate; truncating the factor at \(0\) replaces it by the origin on that event and leaves it unchanged elsewhere. This truncation dominates in risk: the positive-part estimator has weakly smaller risk than \(\hat{\boldsymbol\theta}^{\mathrm{JS}}\) for every \(\boldsymbol\theta\), and strictly smaller whenever the truncation event has positive probability. The claim is about expected loss, not realized loss: it is not true that the truncated estimate is closer to \(\boldsymbol\theta\) for every sample and every \(\boldsymbol\theta\) (for a \(\boldsymbol\theta\) lying far along the \(-\mathbf Y\) direction the untruncated, sign-flipped estimate can occasionally be closer on a given draw), so the domination is a statement about risk, the average over samples, not a pointwise-in-outcome guarantee.

Why the threshold is \(J\ge 3\). The subtracted term \((J-2)^2\sigma^4\,\mathbb E[1/\|\mathbf Y\|^2]\) strictly improves the risk exactly when two conditions both hold: the coefficient \((J-2)^2\) is nonzero, and the expectation \(\mathbb E[1/\|\mathbf Y\|^2]\) is finite. For \(J=2\) the coefficient is \((2-2)^2=0\), so \(\mathbf g\equiv\mathbf 0\) and James-Stein is just the raw estimator, with no gain. For \(J=1\) the coefficient is nonzero but \(\mathbb E[1/Y^2]=\infty\) (the reciprocal of a one-degree-of-freedom chi-square has an infinite mean), so the SURE identity is vacuous; and indeed in one or two dimensions the raw estimator is admissible (Stein’s theorem), so no shrinkage estimator of this form can dominate it. Only from \(J=3\) on are both \((J-2)^2>0\) and \(\mathbb E[1/\|\mathbf Y\|^2]<\infty\), which is why “three or more” is the exact dividing line.

(c) Meta-analytic reading. Treating the \(J\ge 3\) study-specific effect estimates as \(\mathbf Y\sim\mathcal N_J(\boldsymbol\theta,\sigma^2 I)\), the James-Stein shrinkage pulls each study toward the common center and, by (b), strictly reduces total mean squared error relative to reporting the separate unpooled estimates. This is exactly the partial pooling of random-effects meta-analysis of Chapter 13: the random-effects model shrinks each study’s estimate toward the pooled mean by an amount set by the estimated between-study variance, and the James-Stein result says that such shrinkage cannot hurt and typically helps the total mean squared error once there are at least three studies. The justification is frequentist and prior-free: dominance holds for every \(\boldsymbol\theta\), so the gain from borrowing strength across studies does not depend on adopting any subjective prior distribution for the effects. For an evidence synthesis this is reassurance that pooling three or more related estimates (three treatment contrasts in a network, three subgroup effects) is not merely a Bayesian convenience but improves accuracy under the sampling distribution alone. \(\square\)

F.8.2 Solution to Exercise 8.11: REML variance components in the balanced one-way model

This Chapter 8 exercise computes the restricted maximum likelihood (REML) estimators of the within- and between-group variances in a balanced one-way random-effects model, and shows they equal the analysis-of-variance estimators, with the \(I-1\) (not \(I\)) divisor that ordinary maximum likelihood misses.

Strategy. REML estimates variances from error contrasts, the part of the data that is free of the unknown mean. In the balanced design these contrasts split cleanly into a within-group block and a between-group block, each a scaled chi-square, so matching each mean square to its expectation solves the problem.

Uses. The one-sample REML variance correction Theorem 8.8 and the chi-square distribution of a sample sum of squares of independent normals.

A word on the vocabulary, since it recurs. An error contrast is a linear combination \(\mathbf c^\top \mathbf y\) of the data whose coefficient vector is orthogonal to the design column of the fixed effect, here \(\mathbf 1_{In}\) (the all-ones vector of length \(In\), since there are \(I\) groups of \(n\) observations). Such a combination has \(\mathbb E[\mathbf c^\top\mathbf y]=\mu\,\mathbf c^\top\mathbf 1_{In}=0\), so its distribution is free of the unknown grand mean \(\mu\); REML forms the likelihood from a maximal set of these mean-free combinations, which is exactly what removes the downward bias of ordinary maximum likelihood.

Consider \(y_{ij}=\mu+\theta_i+\varepsilon_{ij}\), \(i=1,\dots,I\), \(j=1,\dots,n\), with \(\theta_i\sim\mathcal N(0,\tau^2)\), \(\varepsilon_{ij}\sim\mathcal N(0,\sigma^2)\), all independent. Let \(\bar y_{i\cdot}=\tfrac1n\sum_j y_{ij}\), \(\bar y_{\cdot\cdot}=\tfrac1{In}\sum_{ij}y_{ij}\), and define the balanced mean squares \[ \mathrm{MS}_{\mathrm W}=\frac{1}{I(n-1)}\sum_{i,j}(y_{ij}-\bar y_{i\cdot})^2, \qquad \mathrm{MS}_{\mathrm B}=\frac{n}{I-1}\sum_{i}(\bar y_{i\cdot}-\bar y_{\cdot\cdot})^2 . \]

(a) REML estimators. REML maximizes the likelihood of a maximal set of error contrasts, the linear combinations orthogonal to the design column \(\mathbf 1_{In}\) (which carries the only fixed effect \(\mu\)). Two chi-square facts make the balanced computation explicit, and each is a short calculation rather than an article of faith.

Within-group block. The within-group contrasts \(y_{ij}-\bar y_{i\cdot}\) subtract off each group mean, so the common \(\mu\) and the group effect \(\theta_i\) cancel and only the errors remain: \(y_{ij}-\bar y_{i\cdot}=\varepsilon_{ij}-\bar\varepsilon_{i\cdot}\). For each group \(i\) these are the centered residuals of \(n\) independent \(\mathcal N(0,\sigma^2)\) errors, so \(\sum_j(y_{ij}-\bar y_{i\cdot})^2/\sigma^2 \sim\chi^2_{n-1}\); the \(I\) groups are independent, and adding \(I\) independent \(\chi^2_{n-1}\) variables gives \(\sum_{i,j}(y_{ij}-\bar y_{i\cdot})^2/\sigma^2\sim\chi^2_{I(n-1)}\), that is \(I(n-1)\mathrm{MS}_{\mathrm W}/\sigma^2\sim\chi^2_{I(n-1)}\), with \(\mathbb E[\mathrm{MS}_{\mathrm W}]=\sigma^2\).

Between-group block. The group means are \(\bar y_{i\cdot}=\mu+\theta_i+\bar\varepsilon_{i\cdot}\) with \(\theta_i\sim\mathcal N(0,\tau^2)\) and \(\bar\varepsilon_{i\cdot}\sim\mathcal N(0,\sigma^2/n)\) independent, so \(\bar y_{i\cdot}\sim\mathcal N(\mu,\tau^2+\sigma^2/n)\) independently across \(i\). The between-group contrasts \(\bar y_{i\cdot}-\bar y_{\cdot\cdot}\) are orthogonal to \(\mathbf 1_{In}\) and to the within-group block, and \(\sum_i(\bar y_{i\cdot}-\bar y_{\cdot\cdot})^2/(\tau^2+\sigma^2/n)\sim\chi^2_{I-1}\) as the centered sum of squares of \(I\) independent normals. Multiplying through by \(n\) turns \(\tau^2+\sigma^2/n\) into \(n\tau^2+\sigma^2\) and \(\sum_i(\bar y_{i\cdot}-\bar y_{\cdot\cdot})^2\) into \((I-1)\mathrm{MS}_{\mathrm B}/n\cdot n\), giving \((I-1)\mathrm{MS}_{\mathrm B}/(n\tau^2+\sigma^2)\sim\chi^2_{I-1}\) with \(\mathbb E[\mathrm{MS}_{\mathrm B}]=n\tau^2+\sigma^2\). Because these two chi-square blocks are independent and together span the error-contrast space, the restricted log-likelihood separates and is maximized by matching each mean square to its expectation: \[ \hat\sigma^2_{\mathrm{REML}}=\mathrm{MS}_{\mathrm W}, \qquad n\hat\tau^2_{\mathrm{REML}}+\hat\sigma^2_{\mathrm{REML}}=\mathrm{MS}_{\mathrm B}, \quad\text{so}\quad \hat\tau^2_{\mathrm{REML}}=\frac{\mathrm{MS}_{\mathrm B}-\mathrm{MS}_{\mathrm W}}{n}. \] These are exactly the analysis-of-variance estimators.

(b) The divisor \(I-1\) versus \(I\). The between-group sum of squares is divided by \(I-1\) in \(\mathrm{MS}_{\mathrm B}\) because REML works with the \(I-1\) between-group contrasts orthogonal to \(\mathbf 1_{In}\), spending one degree of freedom on the estimated grand mean \(\bar y_{\cdot\cdot}\) (equivalently on \(\mu\)). Ordinary maximum likelihood does not remove the fixed effect before estimating the variance components; its profile treats \(\mu\) as known at \(\bar y_{\cdot\cdot}\) and divides the same between-group sum of squares by \(I\), giving \(\hat\tau^2_{\mathrm{ML}}=\bigl(\tfrac{I-1}{I}\mathrm{MS}_{\mathrm B}-\mathrm{MS}_{\mathrm W}\bigr)/n\), which is biased downward. REML’s \(I-1\) corrects precisely for the degree of freedom consumed by estimating \(\mu\), reproducing the analysis-of-variance estimator.

(c) Truncation. Both estimators are truncated at zero, since a variance cannot be negative. The REML estimate is truncated when \(\mathrm{MS}_{\mathrm B}<\mathrm{MS}_{\mathrm W}\), that is when the observed between-group spread does not exceed what the within-group noise alone would produce; then \(\hat\tau^2_{\mathrm{REML}}=0\). This is the balanced one-way instance of the truncation in the DerSimonian-Laird estimator Equation 8.8, where \(\hat\tau^2\) is set to zero whenever the heterogeneity statistic falls below its null expectation. \(\square\)

F.9 Chapter 9: Potential Outcomes and Identification

F.9.1 Solution to Exercise 9.7: Positivity failure leaves a counterfactual mean unidentified

This Chapter 9 exercise shows that when one covariate stratum receives a treatment with probability zero, that stratum’s counterfactual mean, and hence the average treatment effect, is not identified from the data at all.

Uses. The adjustment (standardization) formula Equation 9.12 and the positivity requirement of Section 9.9.

Let \(X\in\{0,1\}\) with \(P(X=1)=\tfrac12\), \(e(0)=P(T=1\mid X=0)=0\), \(e(1)=0.5\), under SUTVA and strong ignorability.

(a) The adjustment formula Equation 9.12 expresses \(\mathbb E[Y(1)]\) through the observed regression \(\mathbb E[Y\mid T=1,X=x]\) averaged over \(x\). In the stratum \(X=0\) the conditioning event \(\{T=1,X=0\}\) has probability \(P(T=1,X=0)=P(X=0)e(0)=\tfrac12\cdot 0=0\), so \(\mathbb E[Y\mid T=1,X=0]\) conditions on a null event and is undefined. The treated branch of the adjustment formula therefore cannot be evaluated: there are no treated units in stratum \(X=0\) to estimate \(\mu_1(0)\).

(b) Strong ignorability identifies \(\mu_1(0)=\mathbb E[Y(1)\mid X=0]\) only through treated units in that stratum, of which there are none. Hence the observed-data law places no constraint on \(\mu_1(0)\): two data-generating processes that agree on everything observable can differ in \(\mu_1(0)\). Concretely, keep \(\mu_1(1)=\mathbb E[Y(1)\mid X=1]\), \(\mu_0(0)\), \(\mu_0(1)\) fixed at any values consistent with the observed conditional means, and consider \(\mu_1(0)\in\{0,10\}\). Both choices reproduce the identical observed law (since stratum \(X=0\) contributes no treated outcomes), yet \[ \mathbb E[Y(1)]=\tfrac12\mu_1(0)+\tfrac12\mu_1(1) \] differs by \(\tfrac12(10-0)=5\) between them, so the average treatment effect \(\tau=\mathbb E[Y(1)] -\mathbb E[Y(0)]\) is not identified. Positivity failure in one stratum leaves a whole counterfactual mean free.

(c) This is the covariate-overlap requirement of Section 9.9 in microcosm. An indirect comparison transports an effect from a source population to a target only on the region of covariate space where both treatments are represented. A stratum that receives one treatment with probability zero is an extrapolation region: the comparison there rests entirely on an uncheckable modeling assumption, exactly as \(\mu_1(0)\) is here unconstrained by data. Concretely, in a MAIC or STC (Chapters 18 and 19) suppose the individual-level trial enrolled no patients over age 65 but the aggregate target population contains many such patients. Then the matching weights would have to place weight where the source has no data, and the outcome regression would have to predict outcomes for a covariate value it never saw; the reweighted or standardized comparison in the elderly subgroup is then pure extrapolation, supported by the model’s functional form rather than by observed outcomes. Overlap of the covariate distributions is what makes the population-adjusted estimand identifiable rather than assumption-driven, and its failure is exactly the sort of gap a weight-diagnostic (a near-zero effective sample size) or an out-of-range covariate mean is meant to catch before an indirect comparison is trusted. \(\square\)

F.9.2 Solution to Exercise 9.8: A symmetric decomposition of naive selection bias

This Chapter 9 exercise decomposes the naive (unadjusted) difference in group means into an average treatment effect plus a symmetric selection-bias term, and shows the bias vanishes under randomization.

Uses. The consistency identity Theorem 9.2 (\(\mathbb E[Y\mid T=t]=\mathbb E[Y(t)\mid T=t]\)) and the ATT-side decomposition Equation 9.13.

Control-side decomposition. By the consistency identity (Theorem 9.2), the observed group means are potential-outcome means, \(\mathbb E[Y\mid T=1]=\mathbb E[Y(1)\mid T=1]\) and \(\mathbb E[Y\mid T=0]=\mathbb E[Y(0)\mid T=0]\), so \(\Delta_{\mathrm{naive}}=\mathbb E[Y(1)\mid T=1]-\mathbb E[Y(0)\mid T=0]\). Now add and subtract the counterfactual treated mean among the controls, \(\mathbb E[Y(1)\mid T=0]\): \[ \Delta_{\mathrm{naive}} =\underbrace{\bigl(\mathbb E[Y(1)\mid T=0]-\mathbb E[Y(0)\mid T=0]\bigr)}_{\textstyle \tau_{\mathrm C}} +\underbrace{\bigl(\mathbb E[Y(1)\mid T=1]-\mathbb E[Y(1)\mid T=0]\bigr)}_{\text{treated-outcome selection}} . \] The first bracket is \(\mathbb E[Y(1)-Y(0)\mid T=0]=\tau_{\mathrm C}\), the average effect among the controls (ATC). This is the control-side analogue of Theorem 9.4.

The symmetric average. Averaging this decomposition with the ATT-side decomposition Equation 9.13, \(\Delta_{\mathrm{naive}}=\tau_{\mathrm T}+(\mathbb E[Y(0)\mid T=1]-\mathbb E[Y(0)\mid T=0])\), gives \[ \Delta_{\mathrm{naive}}=\tfrac12(\tau_{\mathrm T}+\tau_{\mathrm C}) +\tfrac12\Bigl[\bigl(\mathbb E[Y(0)\mid T=1]-\mathbb E[Y(0)\mid T=0]\bigr) +\bigl(\mathbb E[Y(1)\mid T=1]-\mathbb E[Y(1)\mid T=0]\bigr)\Bigr]. \] The leading term is the average of ATT and ATC. The bracketed symmetric bias term averages two between-group differences: the gap in the untreated potential outcome \(Y(0)\) and the gap in the treated potential outcome \(Y(1)\) between those who took treatment and those who did not. It compares the two groups on both potential outcomes rather than one, and so is symmetric in the roles of treatment and control.

Vanishing under marginal ignorability. If \((Y(0),Y(1))\perp T\), then \(\mathbb E[Y(0)\mid T=1]=\mathbb E[Y(0)\mid T=0]\) and \(\mathbb E[Y(1)\mid T=1]=\mathbb E[Y(1)\mid T=0]\), so each of the two between-group differences is zero and the symmetric bias term vanishes. Independence also gives \(\tau_{\mathrm T}=\tau_{\mathrm C}=\tau\), so \(\Delta_{\mathrm{naive}}=\tfrac12(\tau+\tau)=\tau\), recovering Corollary 9.2. \(\square\)

F.10 Chapter 10: Adjustment Methods

F.10.1 Solution to Exercise 10.5: Stabilized inverse-probability weights have mean one

This Chapter 10 exercise verifies that stabilized inverse-probability weights average to one and explains why stabilizing lowers the variance of the weighted estimator without changing its target.

Uses. The propensity definition \(\mathbb E[T\mid X]=e(X)\) and the worked example Example 10.3.

Let \(sw=T\,p_1/e(X)+(1-T)\,p_0/(1-e(X))\) with \(p_1=P(T=1)\), \(p_0=P(T=0)=1-p_1\).

(a) Condition on \(X\) and use \(\mathbb E[T\mid X]=e(X)\), \(\mathbb E[1-T\mid X]=1-e(X)\): \[ \mathbb E[sw\mid X]=\frac{p_1}{e(X)}\,\mathbb E[T\mid X]+\frac{p_0}{1-e(X)}\,\mathbb E[1-T\mid X] =\frac{p_1}{e(X)}e(X)+\frac{p_0}{1-e(X)}(1-e(X))=p_1+p_0=1 \] pointwise in \(X\). Taking expectations, \(\mathbb E[sw]=1\).

(b) In Example 10.3, \(e(0)=0.2\), \(e(1)=0.6\), \(P(X=0)=P(X=1)=\tfrac12\), and \(p_1=P(T=1)=\tfrac12(0.2)+\tfrac12(0.6)=0.4\), \(p_0=0.6\). The four cell values of \(sw\) are \[ \begin{aligned} (T=1,x=0):\ &\tfrac{0.4}{0.2}=2.0, & (T=1,x=1):\ &\tfrac{0.4}{0.6}=0.6\overline{6},\\ (T=0,x=0):\ &\tfrac{0.6}{0.8}=0.75, & (T=0,x=1):\ &\tfrac{0.6}{0.4}=1.5 . \end{aligned} \] The cell probabilities are \(P(T=1,x=0)=\tfrac12(0.2)=0.1\), \(P(T=1,x=1)=\tfrac12(0.6)=0.3\), \(P(T=0,x=0)=\tfrac12(0.8)=0.4\), \(P(T=0,x=1)=\tfrac12(0.4)=0.2\). The \(P\)-weighted average is \[ 0.1(2.0)+0.3(0.6\overline 6)+0.4(0.75)+0.2(1.5)=0.2+0.2+0.3+0.3=1.0, \] confirming \(\mathbb E[sw]=1\).

(c) The stabilized weight multiplies the raw weight \(T/e(X)\) by the constant \(p_1\) (and the control weight by \(p_0\)); rescaling by the marginal treatment probability brings the weights nearer to one, so they vary less around their common mean of \(1\). Their reduced dispersion lowers the variance of the inverse-probability estimator while leaving its target unchanged, because the numerator and denominator of the Hajek-form estimand are scaled by the same constant. Here the Hajek-form estimator is the one that divides the weighted sum of outcomes by the weighted sum of weights (a self-normalizing ratio), rather than by the fixed sample size \(n\); the latter, unnormalized version is the Horvitz-Thompson estimator. The two differ only by whether the weights are renormalized to sum to one, and it is precisely because the Hajek form renormalizes that scaling every weight by a common constant \(p_1\) cancels top and bottom and leaves the point estimate untouched while shrinking the weight dispersion. \(\square\)

F.10.2 Solution to Exercise 10.9: Efficient influence function and efficiency bound for the control mean

This Chapter 10 exercise writes down the efficient influence function and the semiparametric efficiency bound for the control-arm mean \(\psi_0=\mathbb E[Y(0)]\), mirroring the treated-arm result, and evaluates the bound on the worked example.

Uses. The treated-arm efficient influence function and bound Theorem 10.6, mirrored by the symmetry \(T\mapsto 1-T\).

An efficiency bound is the smallest asymptotic variance any regular estimator of the target can achieve; it is the semiparametric analogue of the Cramer-Rao lower bound, and the variance of the efficient influence function is that bound. The efficient influence function itself is the mean-zero random variable whose variance an efficient estimator’s error tracks, so reading off its form and squaring it gives the bound. Mirroring Theorem 10.6, which gives the treated-arm influence function \(\varphi_1=T(Y-m_1(X))/e(X)+m_1(X)-\psi_1\), the control-arm version swaps the treatment indicator and the propensity for their complements.

By the symmetry \(T\mapsto 1-T\), the efficient influence function for \(\psi_0=\mathbb E[Y(0)]\) is \[ \varphi_0=\frac{(1-T)\bigl(Y-m_0(X)\bigr)}{1-e(X)}+m_0(X)-\psi_0, \] and the semiparametric efficiency bound (the mirror of Theorem 10.6) is its variance, \[ V_0^\ast=\mathbb E\!\left[\frac{\operatorname{Var}(Y\mid T=0,X)}{1-e(X)}\right]+\operatorname{Var}\bigl(\mu_0(X)\bigr). \]

In Example 10.3, \(\operatorname{Var}(Y\mid T=0,X=x)=4\), \(1-e(0)=0.8\), \(1-e(1)=0.4\), \(\mu_0(0)=0\), \(\mu_0(1)=2\), \(P(X=0)=P(X=1)=\tfrac12\). The first term is \[ \mathbb E\!\left[\frac{4}{1-e(X)}\right]=\tfrac12\frac{4}{0.8}+\tfrac12\frac{4}{0.4}=\tfrac12(5)+\tfrac12(10)=7.5 . \] The regression variance is \(\operatorname{Var}(\mu_0(X))=\tfrac12(0-1)^2+\tfrac12(2-1)^2=1\) (mean \(\mathbb E[\mu_0(X)]=1\)). Hence \[ V_0^\ast=7.5+1=8.5 . \] For the treated mean the analogous computation gives \(V^\ast=\mathbb E[4/e(X)]+\operatorname{Var}(\mu_1(X))=(\tfrac12\cdot 20+\tfrac12\cdot 6.6\overline 6)+1 =13.3\overline 3+1=14.333\), as in the text. The control bound is the smaller of the two, \(8.5<14.333\). The reason is the inverse-propensity inflation factor: the treated term weights each stratum’s conditional variance by \(1/e(x)\), which is large (\(1/0.2=5\)) in stratum \(x=0\) where treatment is scarce, whereas the control term weights by \(1/(1-e(x))\), which is moderate there (\(1/0.8=1.25\)) because controls are plentiful in \(x=0\). Better control-arm overlap in the low-propensity stratum makes \(V_0^\ast<V^\ast\). \(\square\)

F.11 Chapter 11: Effect Modification, Interaction, and Collapsibility

F.11.1 Solution to Exercise 11.6: Odds-ratio attenuation via a tilted-measure argument

This Chapter 11 exercise re-derives the odds-ratio non-collapsibility bound \(1<\omega^{\mathrm M}<\omega\) using the same tilting machinery as the Chapter 7 solution to Exercise 7.12, now framed through an explicit change of measure.

Strategy. Express the marginal odds at tilt level \(a\) as a mean \(A(a)/a=\mathbb E_{Q_a}[W]\) under a tilted probability \(Q_a\), then differentiate that tilted mean; the derivative is a covariance that is negative because two increasing functions are involved.

Uses. The change-of-measure identity Equation 11.10 and the covariance inequality for like-monotone functions.

The identity we lean on is the standard rule for differentiating an expectation under a parametric tilt: if \(\mathrm dQ_a/\mathrm dP\propto h(a,W)\), then \[ \frac{\mathrm d}{\mathrm da}\,\mathbb E_{Q_a}[\varphi(W)] =\operatorname{Cov}_{Q_a}\!\Bigl(\varphi(W),\ \partial_a\log h(a,W)\Bigr), \] the derivative of a tilted mean equals the covariance of the integrand with the tilt’s score. This is what Equation 11.10 records, and below we apply it with \(h(a,W)=1/(1+aW)\), whose score is \(\partial_a\log h=-W/(1+aW)\).

Under the hypotheses of Theorem 11.2(a), write the baseline odds as the random variable \(W=\mu_0(X)/(1-\mu_0(X))\ge 0\), non-degenerate, so that \(\mu_0(X)=W/(1+W)\) and \(\mu_1(X)=\omega W/(1+\omega W)\) under a constant conditional odds ratio \(\omega\). With \(A(a)=\mathbb E[aW/(1+aW)]/\mathbb E[1/(1+aW)]\) the marginal odds ratio is \(\omega^{\mathrm M}=A(\omega)/A(1)\).

(a) \(A\) is strictly increasing, so \(\omega^{\mathrm M}>1\) when \(\omega>1\). Using \(aW/(1+aW)=1-1/(1+aW)\) and writing \(D(a)=\mathbb E[1/(1+aW)]\), \[ A(a)=\frac{1-D(a)}{D(a)}=\frac{1}{D(a)}-1 . \] For each fixed outcome the map \(a\mapsto 1/(1+aW)\) is strictly decreasing on \(\{W>0\}\), an event of probability one, so \(D\) is strictly decreasing, \(1/D\) is strictly increasing, and \(A\) is strictly increasing. Since \(A(1)>0\) and \(\omega>1\), \(\omega^{\mathrm M}=A(\omega)/A(1)>1\).

(b) \(a\mapsto A(a)/a\) is strictly decreasing, so \(\omega^{\mathrm M}<\omega\) when \(\omega>1\). Note \[ \frac{A(a)}{a}=\frac{\mathbb E[W/(1+aW)]}{\mathbb E[1/(1+aW)]}=\mathbb E_{Q_a}[W], \] the mean of \(W\) under the tilt \(\mathrm dQ_a/\mathrm dP\propto 1/(1+aW)\). The change-of-measure identity Equation 11.10 is exactly \(\omega^{\mathrm M}=\omega\,\mathbb E_{Q_\omega}[W]/\mathbb E_{Q_1}[W]\). Differentiating the tilt mean, \[ \frac{\mathrm d}{\mathrm da}\mathbb E_{Q_a}[W] =\operatorname{Cov}_{Q_a}\!\Bigl(W,\ \tfrac{\mathrm d}{\mathrm da}\log\tfrac{1}{1+aW}\Bigr) =-\operatorname{Cov}_{Q_a}\!\Bigl(W,\ \tfrac{W}{1+aW}\Bigr). \] The functions \(w\mapsto w\) and \(w\mapsto w/(1+aw)\) are both strictly increasing, so by the covariance inequality for like-monotone functions their covariance is strictly positive (strict because \(W\) is non-degenerate), and the derivative is strictly negative. Hence \(A(a)/a\) is strictly decreasing, and with \(\omega>1\), \[ \omega^{\mathrm M}=\omega\cdot\frac{\mathbb E_{Q_\omega}[W]}{\mathbb E_{Q_1}[W]}<\omega . \]

Combining (a) and (b), \(1<\omega^{\mathrm M}<\omega\) for \(\omega>1\). The covariance vanishes if and only if \(W\) is degenerate, which is the agreement-iff-degenerate clause of Corollary 11.1: \(\omega^{\mathrm M}=\omega\) exactly when there is a single baseline stratum and no prognostic variation to average over. \(\square\)

F.11.2 Solution to Exercise 11.8: Prognostic variables cancel when anchored, bias when not

This Chapter 11 exercise shows that a purely prognostic covariate cancels from an anchored (comparator-based) contrast but biases an unanchored (single-arm) estimand, which is why unanchored comparisons need a far larger adjustment set.

Uses. The anchored-cancellation lemma Lemma 11.3 and the effect-modifier / prognostic-variable distinction Theorem 11.1.

Split the covariate vector as \((X,W)\) with \(X\) measured and \(W\) possibly unmeasured.

(a) Anchored cancellation of a purely prognostic \(W\). Suppose \(W\) is purely prognostic for the contrast, \(\mu_A(x,w)-\mu_C(x,w)\) free of \(w\) on the identity scale (so \(W\) is not an effect modifier of \(A\) versus \(C\)). By Lemma 11.3 the conditional contrast \(\tau^{\mathrm{id}}_{AC}(x,w)\) does not depend on \(w\), hence equals a function \(\tau^{\mathrm{id}}_{AC}(x)\) of \(x\) alone. For any two populations \(\mathcal P_1,\mathcal P_2\) with the same marginal law of \(X\) but different conditional laws of \(W\mid X\), \[ \Delta^{\mathrm{Cond,id}}_{AC}(\mathcal P_\ell) =\int\!\!\int\tau^{\mathrm{id}}_{AC}(x)\,f_\ell(w\mid x)\,f(x)\,\mathrm dw\,\mathrm dx =\int\tau^{\mathrm{id}}_{AC}(x)\,f(x)\,\mathrm dx,\qquad \ell=1,2, \] the inner integral over \(w\) giving one because \(\tau^{\mathrm{id}}_{AC}(x)\) is constant in \(w\) and \(f_\ell(\cdot\mid x)\) integrates to one. Thus the anchored contrast is identical in the two populations and \(W\) need not be balanced.

(b) An imbalanced prognostic \(W\) biases the unanchored estimand. Take scalar \(X\in\{0,1\}\) with \(P(X=1)=\tfrac12\) common to source and target, \(W\in\{0,1\}\), and the purely prognostic arm mean \(\mu_A(x,w)=w\) (so \(\mu_A\) depends on \(w\) but the \(A\)-versus-\(C\) contrast does not, provided \(\mu_C(x,w)=w-c\) for a constant \(c\)). Let the conditional law of \(W\mid X\) differ between source and target while \(f_X\) agrees: in the source \(\Pr_{\mathrm{src}}(W=1\mid X=x)=0.2\), in the target \(\Pr_{*}(W=1\mid X=x)=0.8\), for both \(x\). The unanchored target is the absolute arm mean \[ \mathbb E_{f_*}[\mu_A(X,W)]=\Pr_{*}(W=1)=0.8, \qquad \mathbb E_{f_{\mathrm{src}}}[\mu_A(X,W)]=\Pr_{\mathrm{src}}(W=1)=0.2, \] which differ by \(0.6\) even though \(f_X\) is the same and \(W\) is purely prognostic. The anchored contrast cancels this difference, but the unanchored arm mean does not, because there is no comparator term to subtract the common \(W\)-effect. Hence an imbalanced purely prognostic variable biases the unanchored estimand while leaving the anchored one untouched. This is precisely the logical distinctness of effect modifiers and prognostic variables in Theorem 11.1(1): purely prognostic covariates are real influences on each arm whose effect on the link-scale contrast is nil, so they matter for unanchored absolute means but cancel in anchored contrasts. \(\square\)

F.12 Chapter 12: Transportability: The Estimand-Based Theory

F.12.1 Solution to Exercise 12.7: Marginal log odds ratios attenuate and are population-dependent

This Chapter 12 exercise shows that a constant conditional log odds ratio \(c\), marginalized over the covariate, produces a marginal log odds ratio strictly between \(0\) and \(c\), with a gap that grows as the covariate spreads out. The upshot is that the marginal odds ratio is a property of the population, not of the treatments.

Strategy. The map \(\varphi\) that sends the control risk to the treated risk (add \(c\) on the logit scale) is strictly increasing and strictly concave; Jensen’s inequality applied to \(\varphi\) then forces the marginal risk below \(\varphi\) of the mean, which is exactly the attenuation.

Uses. Jensen’s inequality Theorem 4.6, and the chain rule for \(\operatorname{expit}\) and \(\operatorname{logit}\).

Let the conditional log odds ratio be the constant \(c>0\), so \(\mu_B(x)=\varphi(\mu_C(x))\) with \(\varphi(p)=\operatorname{expit}(\operatorname{logit}p+c)\).

(a) Monotonicity and concavity of \(\varphi\). Write \(\ell=\operatorname{logit}p\), so \(\varphi(p)=\operatorname{expit}(\ell+c)\) and \(p=\operatorname{expit}(\ell)\). Since \(\mathrm d\ell/\mathrm dp=1/(p(1-p))>0\) and \(\operatorname{expit}'=\operatorname{expit}(1-\operatorname{expit})>0\), \[ \varphi'(p)=\frac{\varphi(p)(1-\varphi(p))}{p(1-p)}>0, \] so \(\varphi\) is strictly increasing. For the second derivative it is cleanest to differentiate the logarithm of \(\varphi'\) (logarithmic differentiation turns the product/quotient in \(\varphi'\) into a sum). Taking logs of \(\varphi'=\varphi(1-\varphi)/[p(1-p)]\), \[ \log\varphi'=\log\varphi+\log(1-\varphi)-\log p-\log(1-p), \] and differentiating each term in \(p\) (using \(\tfrac{\mathrm d}{\mathrm dp}\log\varphi=\varphi'/\varphi\) and \(\tfrac{\mathrm d}{\mathrm dp}\log(1-\varphi)=-\varphi'/(1-\varphi)\) by the chain rule), \[ \frac{\varphi''}{\varphi'} =\frac{\varphi'}{\varphi}-\frac{\varphi'}{1-\varphi}-\frac1p+\frac{1}{1-p} =\varphi'\Bigl(\tfrac1\varphi-\tfrac{1}{1-\varphi}\Bigr)-\Bigl(\tfrac1p-\tfrac{1}{1-p}\Bigr). \] Combine each pair of reciprocals over a common denominator, \(\tfrac1\varphi-\tfrac1{1-\varphi} =(1-2\varphi)/[\varphi(1-\varphi)]\) and \(\tfrac1p-\tfrac1{1-p}=(1-2p)/[p(1-p)]\), so \[ \frac{\varphi''}{\varphi'}=(1-2\varphi)\frac{\varphi'}{\varphi(1-\varphi)}-(1-2p)\frac{1}{p(1-p)} . \] Now substitute \(\varphi'/[\varphi(1-\varphi)]=1/[p(1-p)]\) (which is just \(\varphi'=\varphi(1-\varphi)/[p(1-p)]\) rearranged), so both terms share the denominator \(p(1-p)\): \[ \frac{\varphi''}{\varphi'}=\frac{(1-2\varphi)-(1-2p)}{p(1-p)}=\frac{2(p-\varphi)}{p(1-p)} . \] Since \(c>0\) gives \(\varphi(p)>p\) on \((0,1)\), the numerator \(2(p-\varphi)\) is negative while \(\varphi'>0\) and \(p(1-p)>0\), so \(\varphi''<0\): \(\varphi\) is strictly concave on \((0,1)\).

(b) Strict attenuation. The marginal risks are \(m_C=\mathbb E[\mu_C(X)]\) and \(m_B=\mathbb E[\varphi(\mu_C(X))]\). By Jensen’s inequality (Theorem 4.6) applied to the strictly concave \(\varphi\) with the non-degenerate \(\mu_C(X)\), \[ m_B=\mathbb E[\varphi(\mu_C(X))]<\varphi(\mathbb E[\mu_C(X)])=\varphi(m_C), \] strictly. Since \(\operatorname{logit}\) is increasing and \(\operatorname{logit}\varphi(m_C)=\operatorname{logit}m_C+c\), \[ \operatorname{logit}m_B<\operatorname{logit}\varphi(m_C)=\operatorname{logit}m_C+c, \qquad\text{so}\qquad \operatorname{logit}m_B-\operatorname{logit}m_C<c . \] The lower bound is positive: \(\varphi(p)>p\) pointwise gives \(m_B=\mathbb E[\varphi(\mu_C(X))]>\mathbb E[\mu_C(X)]=m_C\), so \(\operatorname{logit}m_B>\operatorname{logit}m_C\). Hence \(0<\operatorname{logit}m_B-\operatorname{logit}m_C<c\).

(c) The gap grows with spread. For a point mass at \(p_0\), Jensen holds with equality and the gap \(c-(\operatorname{logit}m_B-\operatorname{logit}m_C)\) is zero; for any non-degenerate law of \(\mu_C(X)\) the strict concavity makes the Jensen inequality strict and the gap positive. The more dispersed \(\mu_C(X)\) is, the larger the Jensen deficit \(\varphi(\mathbb E[\mu_C])-\mathbb E[\varphi(\mu_C)]\), and so the larger the attenuation gap. Because this deficit depends on the distribution of \(\mu_C(X)\), the marginal log odds ratio \(\operatorname{logit}m_B-\operatorname{logit}m_C\) is population-dependent: it is not a property the treatments carry from one population to another.

(d) Numerical check. In Section 12.8.2 the conditional log odds ratio is \(c=2\). In \(\mathcal P_1\) (\(P(X=1)=0.1\)), \(m_C=0.137941\) and \(m_B=0.341378\), so the marginal log odds ratio is \(\operatorname{logit}(0.341378)-\operatorname{logit}(0.137941)=1.17534\), strictly between \(0\) and \(2\). In \(\mathcal P_2\) (\(P(X=1)=0.9\)), \(m_C=0.862060\), \(m_B=0.920871\), giving \(0.62173\), again in \((0,2)\). Both lie strictly between \(0\) and \(c=2\), and they differ (\(1.18\) versus \(0.62\)) because the spread of \(\mu_C(X)\) differs between the two populations, confirming (b) and (c). \(\square\)

F.12.2 Solution to Exercise 12.8: Scale misalignment manufactures effect modification

This Chapter 12 exercise shows that reporting a logit-scale model on the risk-difference scale turns a covariate-free conditional structure into a covariate-varying, population-dependent risk difference: the effect modification is an artifact of the mismatched scales, not of the biology.

Uses. The scale-alignment theorem Theorem 12.4 and the collapsibility of the risk difference.

Take a logistic model with no interactions, \(\eta_k(x)=\alpha+x^\top\boldsymbol\beta_1+\gamma_k\), so SEMA holds trivially (\(\boldsymbol\beta_{2,B}=\boldsymbol\beta_{2,C}=\mathbf 0\)), and report the effect as a risk difference, \(h=\mathrm{id}\) with \(g=\operatorname{logit}\).

(a) Here \(\psi=h\circ g^{-1}=\operatorname{id}\circ\operatorname{expit}=\operatorname{expit}\), which is not affine. The conditional reported contrast is \[ \operatorname{expit}(\eta_B(x))-\operatorname{expit}(\eta_C(x)) =\operatorname{expit}(\alpha+x^\top\boldsymbol\beta_1+\gamma_B)-\operatorname{expit}(\alpha+x^\top\boldsymbol\beta_1+\gamma_C). \] Both arms share the same covariate term \(x^\top\boldsymbol\beta_1\), but they enter \(\operatorname{expit}\) at different levels \(\gamma_B\neq\gamma_C\). Writing \(u(x)=\alpha+x^\top\boldsymbol\beta_1\), the contrast is \(\operatorname{expit}(u+\gamma_B)-\operatorname{expit}(u+\gamma_C)\), whose derivative in \(u\) is \(\operatorname{expit}'(u+\gamma_B)-\operatorname{expit}'(u+\gamma_C)\), generically nonzero because \(\operatorname{expit}'\) is non-constant. So whenever \(\boldsymbol\beta_1\neq\mathbf 0\), the term \(u(x)\) varies with \(x\) and the risk difference is a non-constant function of \(x\), instantiating Theorem 12.4(ii): a shared effect-modifier structure on the logit scale does not make the risk-difference contrast constant in the covariates.

(b) With scalar \(x\in\{0,1\}\), \(\alpha=-2\), \(\beta_1=3\), \(\gamma_C=0\), \(\gamma_B=1.5\): \[ \begin{aligned} x=0:\ &\eta_C=-2,\ \eta_B=-0.5;\quad \operatorname{expit}(-2)=0.119203,\ \operatorname{expit}(-0.5)=0.377541;\\ &\text{risk difference }=0.377541-0.119203=0.258338.\\ x=1:\ &\eta_C=1,\ \eta_B=2.5;\quad \operatorname{expit}(1)=0.731059,\ \operatorname{expit}(2.5)=0.924142;\\ &\text{risk difference }=0.924142-0.731059=0.193083 . \end{aligned} \] The conditional risk difference is \(0.2583\) at \(x=0\) and \(0.1931\) at \(x=1\): non-constant, as predicted. The marginal risk differences are population-weighted averages of these two values. With \(P(X=1)=0.2\), \[ 0.8(0.258338)+0.2(0.193083)=0.245287; \] with \(P(X=1)=0.8\), \[ 0.2(0.258338)+0.8(0.193083)=0.206134 . \] The marginal risk difference depends on the covariate distribution, \(0.2453\) versus \(0.2061\).

(c) The risk difference is a collapsible measure (the marginal risk difference equals the average of the conditional risk differences within any one population, by linearity). Collapsibility, however, concerns averaging across strata at fixed conditional contrasts; it does not make those conditional contrasts constant in \(x\). Here the failure is upstream of collapsibility: the misaligned map \(\psi=\operatorname{expit}\) turns a constant logit-scale contrast into a covariate-varying risk-difference contrast, so the very conditional effect being averaged depends on \(x\). A population shift then reweights those differing conditional values and changes the marginal risk difference. Scale misalignment, not non-collapsibility, is what manufactures the effect modification and the population dependence here. \(\square\)

F.13 Chapter 13: Pairwise Meta-Analysis

F.13.1 Solution to Exercise 13.3: A full fixed- and random-effects meta-analysis by hand

This Chapter 13 exercise runs a three-trial meta-analysis end to end: inverse-variance fixed-effect pooling, Cochran’s \(Q\) and \(I^2\), the DerSimonian-Laird between-study variance, and the random-effects recomputation. Every arithmetic step is shown so the calculation can be checked line by line.

Uses. Inverse-variance weighting Equation 15.17, the expected-\(Q\) constant Equation 13.5, and the DerSimonian-Laird truncation Equation 13.11.

The three trials are \(\hat\theta_1=1.2\ (s_1^2=0.16)\), \(\hat\theta_2=0.8\ (s_2^2=0.09)\), \(\hat\theta_3=1.5\ (s_3^2=0.25)\).

(a) Fixed-effect estimate. The inverse-variance weights are \(w_i=1/s_i^2\): \(w_1=6.25\), \(w_2=11.1111\), \(w_3=4\), with total \(W=\sum_i w_i=21.3611\). Then \[ \hat\theta_{\mathrm{FE}}=\frac{\sum_i w_i\hat\theta_i}{W} =\frac{6.25(1.2)+11.1111(0.8)+4(1.5)}{21.3611}=\frac{22.3889}{21.3611}=1.04812, \] with standard error \(\operatorname{SE}(\hat\theta_{\mathrm{FE}})=1/\sqrt W=1/\sqrt{21.3611}=0.21636\).

(b) Cochran’s \(Q\) and \(I^2\). With deviations \(\hat\theta_i-\hat\theta_{\mathrm{FE}}\), \[ Q=\sum_i w_i(\hat\theta_i-\hat\theta_{\mathrm{FE}})^2 =6.25(0.15188)^2+11.1111(-0.24812)^2+4(0.45188)^2=0.1442+0.6840+0.8168=1.6450 . \] Under homogeneity \(Q\sim\chi^2_{k-1}=\chi^2_2\), with mean \(k-1=2\). Since the observed \(Q=1.645\) is below its null expectation, \(I^2=\max\{0,(Q-(k-1))/Q\}=\max\{0,-0.216\}=0\).

(c) DerSimonian-Laird. The constant of Equation 13.5 is \[ c=W-\frac{\sum_i w_i^2}{W}=21.3611-\frac{6.25^2+11.1111^2+4^2}{21.3611} =21.3611-\frac{178.5193}{21.3611}=21.3611-8.3563=13.0048 . \] Then \(\hat\tau^2_{\mathrm{DL}}=\max\{0,(Q-(k-1))/c\}=\max\{0,(1.645-2)/13.0048\}=\max\{0,-0.0273\}=0\).

(d) Random-effects recomputation. Because \(\hat\tau^2_{\mathrm{DL}}=0\), the random-effects weights \(\tilde w_i=1/(s_i^2+\hat\tau^2_{\mathrm{DL}})\) equal the fixed-effect weights, so the random-effects pooled estimate and standard error coincide with the fixed-effect values, \(\hat\theta_{\mathrm{RE}}=1.04812\) and \(\operatorname{SE}=0.21636\). The data show no detectable heterogeneity: the between-study spread is smaller than sampling noise alone would produce, so the truncation in Equation 13.11 returns \(\hat\tau^2=0\) and the two analyses agree. \(\square\)

F.13.2 Solution to Exercise 13.6: REML heterogeneity variance with equal within-study variances

This Chapter 13 exercise solves the REML estimating equation for the between-study variance \(\tau^2\) in the special case of equal within-study variances, recovering the \(k-1\) divisor that maximum likelihood misses.

Uses. The REML estimating equation Equation 13.13 and the one-sample REML correction Theorem 8.8.

Suppose all within-study variances are equal, \(s_i^2\equiv s^2\). Then the random-effects weights are equal, \(\tilde w_i\equiv\tilde w=1/(s^2+\tau^2)\), and the weighted mean \(\hat\theta_R=\sum_i\tilde w_i\hat\theta_i/\sum_i\tilde w_i\) reduces to the unweighted mean \(\bar{\hat\theta}=\tfrac1k\sum_i\hat\theta_i\). The REML estimating equation Equation 13.13 is \[ \sum_{i=1}^k\tilde w_i^2(\hat\theta_i-\hat\theta_R)^2=\sum_{i=1}^k\tilde w_i-\frac{\sum_i\tilde w_i^2}{\sum_i\tilde w_i}. \] With \(\tilde w_i\equiv\tilde w\) the left side is \(\tilde w^2\sum_i(\hat\theta_i-\bar{\hat\theta})^2\), and the right side is \(k\tilde w-(k\tilde w^2)/(k\tilde w)=k\tilde w-\tilde w=(k-1)\tilde w\). Equating and dividing by \(\tilde w\), \[ \tilde w\sum_{i=1}^k(\hat\theta_i-\bar{\hat\theta})^2=k-1, \qquad\text{that is}\qquad \frac{1}{s^2+\tau^2}\sum_{i=1}^k(\hat\theta_i-\bar{\hat\theta})^2=k-1 . \] Solving, \(s^2+\tau^2=\tfrac{1}{k-1}\sum_i(\hat\theta_i-\bar{\hat\theta})^2\), hence \[ \hat\tau^2_{\mathrm{REML}}=\frac{1}{k-1}\sum_{i=1}^k(\hat\theta_i-\bar{\hat\theta})^2-s^2 \] (truncated at \(0\)). The maximum likelihood counterpart divides the same sum of squares by \(k\) rather than \(k-1\), giving \(\tilde\tau^2_{\mathrm{ML}}=\tfrac1k\sum_i(\hat\theta_i-\bar{\hat\theta})^2-s^2\). The discrepancy is the single degree of freedom spent on estimating the common mean \(\theta\): REML forms the likelihood from the \(k-1\) contrasts orthogonal to \(\mathbf 1_k\), dividing by \(k-1\), whereas maximum likelihood treats \(\bar{\hat\theta}\) as if known and divides by \(k\), underestimating \(\tau^2\). This is the balanced-data instance of the one-sample variance correction of Theorem 8.8. \(\square\)

F.14 Chapter 14: Indirect Comparison and Bucher’s Theorem

F.14.1 Solution to Exercise 14.8: Bucher variance with correlated within-trial contrasts

This Chapter 14 exercise works out the Bucher indirect-comparison variance when the two contrasts share a trial arm, and shows the shared-comparator correlation is exactly \(\tfrac12\), recovering the direct within-trial contrast.

Uses. The Bucher variance formula Theorem 14.2 and the homoskedastic random-effects correlation Theorem 15.4.

The Bucher estimator is \(\hat d_{AB}^{\,\mathrm{Bucher}}=\hat d_{AC}-\hat d_{BC}\), and by the general variance formula of Theorem 14.2, \[ \operatorname{Var}\bigl(\hat d_{AB}^{\,\mathrm{Bucher}}\bigr)=v_{AC}+v_{BC}-2\operatorname{Cov}(\hat d_{AC},\hat d_{BC}) =v_{AC}+v_{BC}-2\rho\sqrt{v_{AC}v_{BC}} . \] For \(\rho>0\) the subtracted term is positive, so the variance is strictly smaller than \(v_{AC}+v_{BC}\): treating correlated contrasts from one trial as independent overstates the variance and is conservative for this comparison.

Homoskedastic three-arm structure. Let arms \(A,B,C\) of a single trial have independent arm-level estimates \(\hat y_A,\hat y_B,\hat y_C\), each of variance \(\sigma^2\), with \(C\) the shared comparator. Then \(\hat d_{AC}=\hat y_A-\hat y_C\) and \(\hat d_{BC}=\hat y_B-\hat y_C\), so \[ v_{AC}=v_{BC}=2\sigma^2, \qquad \operatorname{Cov}(\hat d_{AC},\hat d_{BC})=\operatorname{Var}(\hat y_C)=\sigma^2, \] the covariance coming only from the shared \(C\) arm. Hence \(\rho=\sigma^2/\sqrt{(2\sigma^2)(2\sigma^2)}=\sigma^2/(2\sigma^2)=\tfrac12\). The Bucher variance is \[ \operatorname{Var}\bigl(\hat d_{AB}^{\,\mathrm{Bucher}}\bigr)=2\sigma^2+2\sigma^2-2\sigma^2=2\sigma^2, \] which equals \(\operatorname{Var}(\hat y_A-\hat y_B)=\operatorname{Var}(\hat y_A)+\operatorname{Var}(\hat y_B)=2\sigma^2\), the variance of the direct within-trial \(A\)-versus-\(B\) contrast. The shared-comparator correlation \(\rho=\tfrac12\) is exactly the value derived for the homoskedastic random-effects network in Theorem 15.4, and it makes the Bucher subtraction recover the direct within-study contrast with no loss. \(\square\)

F.14.2 Solution to Exercise 14.9: Effect-modifier imbalance breaks Bucher’s transitivity

This Chapter 14 exercise puts numbers to the transitivity failure of an unadjusted indirect comparison: an effect modifier distributed differently in the \(BC\) trial and the target population biases the Bucher estimand, and the bias is exactly the reweighting that population adjustment later corrects.

Uses. Network additivity of contrasts and the effect-modifier reweighting introduced from Chapter 17.

The binary effect modifier \(Z\) acts on the \(B\)-versus-\(C\) contrast with conditional values \(d_{BC}(Z=0)=0.2\), \(d_{BC}(Z=1)=0.8\) (treated as collapsible). Study \(BC\) has \(\Pr(Z=1)=0.7\); the target population \(\mathcal P_{AB}\) has \(\Pr(Z=1)=0.3\). The \(A\)-versus-\(C\) contrast is unmodified, \(d_{AC}=0.6\) in every population.

Marginal \(B\)-versus-\(C\) contrasts. Averaging over \(Z\), \[ d_{BC(\mathcal P_{BC})}=0.3(0.2)+0.7(0.8)=0.62, \qquad d_{BC(\mathcal P_{AB})}=0.7(0.2)+0.3(0.8)=0.38 . \]

True target contrast. In the target population, by network additivity, \[ d_{AB}=d_{AC}-d_{BC(\mathcal P_{AB})}=0.6-0.38=0.22 . \]

Bucher estimand and its bias. Bucher subtracts the contrast as reported by trial \(BC\): \[ d_{AC}-d_{BC(\mathcal P_{BC})}=0.6-0.62=-0.02, \] so the bias of the unadjusted indirect comparison is \[ \bigl(d_{AC}-d_{BC(\mathcal P_{BC})}\bigr)-d_{AB}=-0.02-0.22=-0.24 . \] The bias arises entirely because the \(B\)-versus-\(C\) contrast must be reweighted from the \(BC\)-study modifier distribution \(\Pr(Z=1)=0.7\) to the target distribution \(\Pr(Z=1)=0.3\); the effect-modifier adjustment of Chapter 17 onward is precisely the reweighting of \(d_{BC}\) to the target marginal. A purely prognostic variable, which shifts the baseline of both arms equally and leaves the conditional contrast \(d_{BC}(z)\) constant in \(z\), would make \(d_{BC}\) population-free and produce no bias at all. \(\square\)

F.15 Chapter 15: Network Meta-Analysis: Models and Identifiability

F.15.1 Solution to Exercise 15.9: Network pooling is inverse-variance blending of direct and indirect

This Chapter 15 exercise shows that combining a direct and an indirect estimate of the same contrast through generalized least squares reduces to the familiar inverse-variance-weighted average, the algebra behind node-splitting.

Uses. The network generalized-least-squares estimator Equation 15.13 and the definition of node-splitting Definition 15.4.

Take a contrast \(d_{AB}\) informed by two independent unbiased sources, a direct estimate \(\hat d^{\mathrm{dir}}_{AB}\) with variance \(v_{\mathrm{dir}}\) and an indirect estimate \(\hat d^{\mathrm{ind}}_{AB}\) with variance \(v_{\mathrm{ind}}\). Stack them as a one-parameter generalized-least-squares problem with design \(\mathbf X=\binom{1}{1}\), response \(\mathbf y=(\hat d^{\mathrm{dir}}_{AB},\hat d^{\mathrm{ind}}_{AB})^\top\), and covariance \(\mathbf V=\operatorname{diag}(v_{\mathrm{dir}},v_{\mathrm{ind}})\). The GLS estimator Equation 15.13 is \(\hat d^{\mathrm{net}}_{AB}=(\mathbf X^\top\mathbf V^{-1}\mathbf X)^{-1}\mathbf X^\top\mathbf V^{-1}\mathbf y\). Here, with \(w_{\mathrm{dir}}=1/v_{\mathrm{dir}}\) and \(w_{\mathrm{ind}}=1/v_{\mathrm{ind}}\), \[ \mathbf X^\top\mathbf V^{-1}\mathbf X=w_{\mathrm{dir}}+w_{\mathrm{ind}}, \qquad \mathbf X^\top\mathbf V^{-1}\mathbf y=w_{\mathrm{dir}}\hat d^{\mathrm{dir}}_{AB}+w_{\mathrm{ind}}\hat d^{\mathrm{ind}}_{AB}, \] so \[ \hat d^{\mathrm{net}}_{AB}=\frac{w_{\mathrm{dir}}\hat d^{\mathrm{dir}}_{AB}+w_{\mathrm{ind}}\hat d^{\mathrm{ind}}_{AB}}{w_{\mathrm{dir}}+w_{\mathrm{ind}}}, \] the inverse-variance-weighted mean Equation 15.17, with variance \((\mathbf X^\top\mathbf V^{-1}\mathbf X)^{-1}=1/(w_{\mathrm{dir}}+w_{\mathrm{ind}})\). This is the best linear unbiased combination of the two sources (the \(k=2\) fixed-effect meta-analysis). In a larger network, any contrast whose only evidence separates into one direct stream (studies comparing \(A\) and \(B\) directly) and one indirect stream (the rest of the network with the direct \(AB\) evidence removed), the two streams being independent, reduces to exactly this two-source blend; this is the construction behind node-splitting (Definition 15.4), where the network estimate of the contrast is the inverse-variance blend of its direct and indirect parts. \(\square\)

F.15.2 Solution to Exercise 15.10: Between-study-only covariates are confounded with baselines

This Chapter 15 exercise shows that when a covariate varies only between studies (constant within each study), its prognostic effect cannot be separated from the study baselines, the individual-level seed of ecological bias.

Uses. The IPD network meta-regression identifiability theorem Theorem 15.5 and the definition of ecological bias Definition 16.2.

In the IPD network meta-regression Equation 15.20 with a single scalar covariate \(x\), \[ g\bigl(\mathbb E[Y_{ijk}]\bigr)=\mu_j+x_{ijk}\beta_1+d_k+x_{ijk}\beta_{2,k}, \] suppose \(x\) is constant within each study, \(x_{ijk}=x_j\) for all individuals and arms of study \(j\).

Identifiability condition (iii) fails. Condition (iii) of Theorem 15.5 requires the within-study covariate covariance to be nonsingular. When \(x\) takes a single value \(x_j\) in study \(j\), the within-study variance of \(x\) is zero, so the covariance is singular and (iii) is violated.

Confounding of \(\beta_1\) with the baselines. On arm \(k\) of study \(j\) the linear predictor is \(\eta_{jk}=\mu_j+x_j\beta_1+d_k+x_j\beta_{2,k}\). For any scalar \(t\), define the alternative parameters \[ \mu_j'=\mu_j+t\,x_j,\qquad \beta_1'=\beta_1-t, \] leaving \(d_k\) and \(\beta_{2,k}\) unchanged. Then \[ \mu_j'+x_j\beta_1'+d_k+x_j\beta_{2,k}=\mu_j+t x_j+x_j(\beta_1-t)+d_k+x_j\beta_{2,k}=\mu_j+x_j\beta_1+d_k+x_j\beta_{2,k}=\eta_{jk}, \] so every arm of every study has the identical conditional mean. The one-parameter family \((\boldsymbol\mu+t\,\mathbf x_{\mathrm{study}},\,\beta_1-t,\,\mathbf d,\,\{\beta_{2,k}\})\), \(t\in\mathbb R\), gives indistinguishable conditional-mean functions, so the prognostic main effect \(\beta_1\) is confounded with the study baselines \(\mu_j\) and is not identified.

This is the individual-level seed of ecological bias (Definition 16.2): with only between-study variation in \(x\), a study-level association between \(x_j\) and the outcome can be explained equally by a prognostic effect \(\beta_1\) or by differences in the baselines \(\mu_j\), and the two cannot be separated. Within-study covariate variation rescues identifiability because then \(x_{ijk}\beta_1\) varies across individuals within a study and can no longer be absorbed into a single per-study intercept, so \(\beta_1\) is pinned down by the within-study covariate-outcome relationship. \(\square\)

F.16 Chapter 16: The Limits of Standard Network Meta-Analysis

F.16.1 Solution to Exercise 16.8: Anchored bias depends only on effect-modifier marginals

This Chapter 16 exercise shows the Bucher bias of an anchored comparison depends only on the marginal distributions of the effect modifiers, so balancing those marginals alone removes it, while an unanchored comparison reinstates the prognostic variables.

Uses. The failure-of-Bucher bias formula Theorem 16.1.

Split \(X=(V,W)\) with \(V\) an effect modifier and \(W\) purely prognostic for the anchored contrast on scale \(g\), so \(\tau^g_{tC}(v,w)=\tau^g_{tC}(v)\) for \(t\in\{A,B\}\), depending on \(v\) but not \(w\). By Theorem 16.1 the Bucher bias is \(b_{AC}-b_{BC}\) with \(b_{tC}=\int\tau^g_{tC}(x)[f_{tC}(x)-f_{*}(x)]\,\mathrm dx\). Substituting \(\tau^g_{tC}(v,w)=\tau^g_{tC}(v)\) and integrating out \(w\), \[ b_{tC}=\int\!\!\int\tau^g_{tC}(v)\,[f_{tC}(v,w)-f_{*}(v,w)]\,\mathrm dw\,\mathrm dv =\int\tau^g_{tC}(v)\,[f^V_{tC}(v)-f^V_{*}(v)]\,\mathrm dv, \] where \(f^V_{tC}\) and \(f^V_{*}\) are the \(V\)-marginals of \(f_{tC}\) and \(f_{*}\). Thus each \(b_{tC}\), and hence the total Bucher bias \(b_{AC}-b_{BC}\), depends only on the marginal distributions of \(V\) in \(\mathcal P_{AC}\), \(\mathcal P_{BC}\), and \(\mathcal P^{*}\). It is invariant to every feature of the distribution of \(W\), including the conditional law of \(W\) given \(V\) and any difference in the \(V\)-\(W\) dependence across the three populations, since \(W\) has been integrated out against densities that each integrate to one in \(w\).

Consequently an anchored adjustment that balances the marginal distribution of \(V\) alone removes the bias: once \(f^V_{AC}=f^V_{BC}=f^V_{*}\), every \(b_{tC}=0\). The argument breaks for an unanchored comparison built from absolute arm means \(\mathbb E_{f_*}[\mu_A(V,W)]\): there \(\mu_A\) depends on the prognostic \(W\), so the integral over \(w\) does not collapse to a \(V\)-marginal statement, and an imbalance in the conditional law of \(W\) given \(V\) shifts the absolute mean with no comparator term to cancel it. Anchoring is what confines the adjustment set to effect modifiers; unanchoring reinstates prognostic variables. \(\square\)

F.16.2 Solution to Exercise 16.9: Ecological attenuation of the effect-modifier slope

This Chapter 16 exercise quantifies how within-study covariate variation attenuates the effect-modification slope seen at the aggregate (ecological) level, gives the exact probit answer, and explains why global convexity of the inverse link is not enough to guarantee attenuation toward zero.

Strategy. Expand the inverse link to second order in the within-study spread \(\sigma^2\); the leading correction to the aggregate slope is a curvature term. For the probit the integral is exact, so the approximation becomes an identity; for a general link the sign of the correction is not fixed, which is the point of the last part.

Uses. The delta-method-style Taylor expansion of Chapter 4 and the Gaussian integral identity for \(\mathbb E[\Phi(a+\rho X)]\) derived in part (b).

We organize the solution as (a) the general-link expansion, (b) the exact probit specialization, and (c) why convexity alone does not force attenuation.

(a) General-link Taylor expansion. Let each trial have \(X\sim\mathcal N(\bar x,\sigma^2)\) and arm linear predictors \(\eta_t(x)=a_t+\rho_t x\). Write \(m_t=a_t+\rho_t\bar x\) for the linear predictor at the trial mean and let \(\psi=g^{-1}\). Expand \(\psi\) to second order about \(m_t\): \[ \psi(a_t+\rho_t X)=\psi(m_t)+\psi'(m_t)\rho_t(X-\bar x)+\tfrac12\psi''(m_t)\rho_t^2(X-\bar x)^2+R_t, \] with \(R_t\) collecting the remainder. Taking expectations and using \(\mathbb E[X-\bar x]=0\), \(\mathbb E[(X-\bar x)^2]=\sigma^2\), \[ \bar\mu_t(\bar x)=\mathbb E[\psi(a_t+\rho_t X)]=\psi(m_t)+\tfrac12\psi''(m_t)\rho_t^2\sigma^2+O(\sigma^4). \] Apply \(g\) and expand again, using \(g'(\psi(m))=1/\psi'(m)\): \[ g(\bar\mu_t(\bar x))=m_t+g'(\psi(m_t))\cdot\tfrac12\psi''(m_t)\rho_t^2\sigma^2+O(\sigma^4) =m_t+\tfrac12\rho_t^2\sigma^2\frac{\psi''(m_t)}{\psi'(m_t)}+O(\sigma^4). \] Therefore the aggregate link-scale contrast is \[ D(\bar x)=g(\bar\mu_1(\bar x))-g(\bar\mu_0(\bar x)) =(m_1-m_0)+\tfrac12\sigma^2\!\left[\rho_1^2\frac{\psi''(m_1)}{\psi'(m_1)}-\rho_0^2\frac{\psi''(m_0)}{\psi'(m_0)}\right]+O(\sigma^4), \] and differentiating in \(\bar x\) (note \(\mathrm dm_t/\mathrm d\bar x=\rho_t\)), \[ D'(\bar x)=(\rho_1-\rho_0)+\tfrac12\sigma^2\frac{\mathrm d}{\mathrm d\bar x}\!\left[\rho_1^2\frac{\psi''(m_1)}{\psi'(m_1)}-\rho_0^2\frac{\psi''(m_0)}{\psi'(m_0)}\right]+O(\sigma^4). \] To zeroth order in \(\sigma^2\), \(D'(\bar x)=\rho_1-\rho_0=\beta_2\), the individual effect-modification coefficient. The \(O(\sigma^2)\) correction is governed by the curvature \(\psi''=(g^{-1})''\) and, through the \(\bar x\)-derivative, the third derivative \(\psi'''=(g^{-1})'''\).

(b) Probit specialization (exact). For \(g=\Phi^{-1}\), \(\psi=\Phi\) and the integral is exact (no remainder). The identity we need is \(\mathbb E[\Phi(a+\rho X)]=\Phi\bigl((a+\rho\bar x)/\sqrt{1+\rho^2\sigma^2}\bigr)\) for \(X\sim\mathcal N(\bar x,\sigma^2)\), and it is worth deriving rather than quoting, because a beginner cannot check it otherwise. Introduce an auxiliary standard normal \(Z\sim\mathcal N(0,1)\) independent of \(X\). Then \(\Phi(a+\rho X)=P(Z\le a+\rho X\mid X)\), so taking the expectation over \(X\), \[ \mathbb E[\Phi(a+\rho X)]=P(Z-\rho X\le a). \] Now \(Z-\rho X\) is a linear combination of independent normals, hence normal, with mean \(\mathbb E[Z-\rho X]=-\rho\bar x\) and variance \(\operatorname{Var}(Z)+\rho^2\operatorname{Var}(X)=1+\rho^2\sigma^2\). Standardizing, \[ P(Z-\rho X\le a)=\Phi\!\left(\frac{a-(-\rho\bar x)}{\sqrt{1+\rho^2\sigma^2}}\right) =\Phi\!\left(\frac{a+\rho\bar x}{\sqrt{1+\rho^2\sigma^2}}\right), \] the stated identity (this is the standard Gaussian convolution of Chapter 4, obtained here by the auxiliary normal rather than by completing the square). Applying it to each arm, \(\bar\mu_t(\bar x)=\mathbb E[\Phi(a_t+\rho_t X)]=\Phi\bigl((a_t+\rho_t\bar x)/\sqrt{1+\rho_t^2\sigma^2}\bigr)\), so \(g(\bar\mu_t)=(a_t+\rho_t\bar x)/\sqrt{1+\rho_t^2\sigma^2}\) and \[ D'(\bar x)=\frac{\rho_1}{\sqrt{1+\rho_1^2\sigma^2}}-\frac{\rho_0}{\sqrt{1+\rho_0^2\sigma^2}}=b_2, \] the exact slope Equation 16.12. Expanding \(1/\sqrt{1+\rho^2\sigma^2}=1-\tfrac12\rho^2\sigma^2+O(\sigma^4)\) gives \[ b_2=(\rho_1-\rho_0)-\tfrac12\sigma^2(\rho_1^3-\rho_0^3)+O(\sigma^4), \] the leading-order term: the slope equals \(\beta_2\) corrected at order \(\sigma^2\).

(c) Why global convexity is not enough. The probit attenuation \(0<b_2<\beta_2\) holds for every parameter configuration because the exact factor \(\rho/\sqrt{1+\rho^2\sigma^2}\) is a strictly increasing odd contraction of \(\rho\). For a general link the \(O(\sigma^2)\) correction depends on the signed combination \(\rho_1^2\psi''(m_1)/\psi'(m_1)-\rho_0^2\psi''(m_0)/\psi'(m_0)\) and its \(\bar x\)-derivative, which involve the levels \(m_0,m_1\) and the slopes \(\rho_0,\rho_1\) with both signs. A globally convex \(g^{-1}\) fixes the sign of \(\psi''\) but not the sign of the difference, so the correction can push the ecological slope either toward or away from zero depending on the configuration. Global convexity therefore does not guarantee attenuation toward zero, in contrast to the exact, configuration-free probit result. \(\square\)

F.17 Chapter 17: Population Adjustment Theory

F.17.1 Solution to Exercise 17.8: The unanchored adjustment set must include prognostic variables

This Chapter 17 exercise derives the omitted-covariate bias of an unanchored (single-arm) population adjustment and shows it is nonzero for imbalanced prognostic variables, which cancel in the anchored case. The moral is that unanchored comparisons need every imbalanced prognostic variable in the adjustment set, not just effect modifiers.

Uses. The unanchored bias formula Equation 17.17 and the prognostic-cancellation lemma Lemma 17.1.

The bias formula. The unanchored estimand standardizes each arm’s conditional mean to the target, \(m_t(\mathcal P^{*})=\int\mu_t(x)f_{*}(x)\,\mathrm dx\). Suppose the true conditional mean depends on \((X,W)\) with \(W\) omitted from the adjustment set, \(\mu_A=\mu_A(x,w)\). Standardizing only over \(X\) uses the source conditional law of \(W\) given \(X\), \(f_{\mathrm{src},A}(w\mid x)\), whereas the target uses \(f^{*}(w\mid x)\). The \(A\)-arm bias is the difference between the source-plugged and target arm means, \[ \int\!\!\int\mu_A(x,w)\,f_{\mathrm{src},A}(w\mid x)\,f_{*}(x)\,\mathrm dw\,\mathrm dx -\int\!\!\int\mu_A(x,w)\,f^{*}(w\mid x)\,f_{*}(x)\,\mathrm dw\,\mathrm dx =\int\!\!\int\mu_A(x,w)\,[f_{\mathrm{src},A}(w\mid x)-f^{*}(w\mid x)]\,f_{*}(x)\,\mathrm dw\,\mathrm dx, \] which is Equation 17.17. It is nonzero precisely when \(\mu_A\) depends on the omitted \(w\) (prognostic) and the conditional law of \(W\) given \(X\) differs between source and target (imbalance), and there is no comparator arm to cancel it.

Binary specialization. Take \(\mu_A(x,w)=\alpha(x)+\rho w\) with binary \(W\), \(\Pr_{\mathrm{src}}(W=1)=p\), \(\Pr_{*}(W=1)=q\), with \(W\) independent of \(X\) for transparency. The bias of the \(A\) marginal mean is \[ \int\!\!\int(\alpha(x)+\rho w)\,[f_{\mathrm{src}}(w)-f^{*}(w)]\,f_{*}(x)\,\mathrm dw\,\mathrm dx =\rho\int w\,[f_{\mathrm{src}}(w)-f^{*}(w)]\,\mathrm dw =\rho\,(p-q), \] the \(\alpha(x)\) term integrating to zero because \(f_{\mathrm{src}}(\cdot)\) and \(f^{*}(\cdot)\) each integrate to one in \(w\). So an imbalanced purely prognostic \(W\) biases the unanchored \(A\) mean by \(\rho(p-q)\).

Why the anchored comparison is multiplied by zero. In an anchored comparison the same \(W\) contributes to both the \(A\) arm and the comparator arm through the contrast \(\mu_A(x,w)-\mu_C(x,w)\). If \(W\) is purely prognostic, this contrast is free of \(w\), so the \(\rho w\) term cancels and the corresponding bias coefficient is zero: the imbalance \(p-q\) is multiplied by the contrast’s \(w\)-derivative, which vanishes. This is Lemma 17.1. Consequently the unanchored adjustment set of Theorem 17.2 must contain every imbalanced prognostic variable and every imbalanced effect modifier, whereas the anchored set needs only effect modifiers. The unanchored requirement is therefore far harder to satisfy in practice, since prognostic variables are typically far more numerous than effect modifiers and need not even be suspected in advance. \(\square\)

F.17.2 Solution to Exercise 17.9: The marginal odds ratio depends on the target population

This Chapter 17 exercise computes how the marginal odds ratio of \(A\) versus \(B\) changes across target populations that differ only in covariate prevalence, isolating the non-collapsibility (Jensen) gap even though every conditional risk is held fixed.

Uses. The anchored consistency theorem Theorem 17.1 and the SEMA-insufficiency obstruction Theorem 12.6.

Use the conditional risks of Example 17.2: \(\mu_A(0)=0.20\), \(\mu_A(1)=0.50\), \(\mu_B(0)=\mu_B(1)=0.30\), the same functions in every population. We compute the marginal odds ratio of \(A\) versus \(B\) as the covariate prevalence \(\Pr(X=1)=p\) varies over \(\{0.25,0.50,0.75\}\).

Marginal risks and log odds ratios. Since \(\mu_B\) is constant, \(m_B(p)=0.30\) for all \(p\), while \(m_A(p)=(1-p)(0.20)+p(0.50)=0.20+0.30p\). The marginal log odds ratio is \(\operatorname{logit}m_A(p)-\operatorname{logit}(0.30)\) with \(\operatorname{logit}(0.30)=-0.847298\): \[ \begin{aligned} p=0.25:\ & m_A=0.275,\ \operatorname{logit}=-0.969401,\ \text{log OR}=-0.122103;\\ p=0.50:\ & m_A=0.350,\ \operatorname{logit}=-0.619039,\ \text{log OR}=\phantom{-}0.228259;\\ p=0.75:\ & m_A=0.425,\ \operatorname{logit}=-0.302280,\ \text{log OR}=\phantom{-}0.545018 . \end{aligned} \] The marginal log odds ratio changes across the three populations, from \(-0.122\) to \(+0.545\), even though the conditional risks (and therefore the conditional odds ratios at each \(x\)) are identical in all three: no conditional parameter has changed, only \(\Pr(X=1)\).

Isolating the non-collapsibility component. The population-standardized conditional log odds ratio is \(\Delta^{\mathrm{Cond}}_{AB}(p)=(1-p)\,\delta(0)+p\,\delta(1)\) with \(\delta(x)=\operatorname{logit}\mu_A(x)-\operatorname{logit}\mu_B(x)\), that is \(\delta(0)=-0.538997\) and \(\delta(1)=0.847298\): \[ \Delta^{\mathrm{Cond}}_{AB}(0.25)=-0.192423,\quad \Delta^{\mathrm{Cond}}_{AB}(0.50)=0.154151,\quad \Delta^{\mathrm{Cond}}_{AB}(0.75)=0.500724 . \] The difference between the marginal log odds ratio and this population-standardized conditional value is the non-collapsibility (Jensen) gap, “logit of the average minus average of the logit.” Showing the subtraction at each prevalence, \[ \begin{aligned} p=0.25:\ &-0.122103-(-0.192423)=0.070320,\\ p=0.50:\ &\phantom{-}0.228259-\phantom{-}0.154151=0.074108,\\ p=0.75:\ &\phantom{-}0.545018-\phantom{-}0.500724=0.044294. \end{aligned} \] This gap is a pure non-collapsibility effect: it is nonzero, and population-dependent, even though every conditional risk is held fixed across the three populations. (In this particular table the \(A\)-versus-\(B\) contrast is itself modified by \(X\), so the conditional standardized value also moves; the clean case where the conditional log odds ratio is held constant across \(x\) and only the baseline varies is the SEMA worked example of Section 12.8.2, which exhibits the non-collapsibility attenuation in isolation.)

Conclusion. On the odds-ratio scale the marginal effect is a property of the population, not of the two treatments: even a fully adjusted anchored comparison, which recovers the conditional contrast exactly (part (a) of Theorem 17.1), does not deliver a population-free marginal odds ratio (part (c)). This is exactly the obstruction of Theorem 12.6: SEMA and scale alignment pin down the conditional effect but cannot make the non-collapsible marginal odds ratio transport. \(\square\)

F.18 Chapter 18: Matching-Adjusted Indirect Comparison

F.18.1 Solution to Exercise 18.8: Why MAIC is exact on the identity scale but not the logit scale

This Chapter 18 exercise shows that on the identity scale, with a benefit linear in the covariates, any first-moment-balancing weight gives the same (correct) MAIC point estimate, so the choice of balancing objective affects only the variance; and that this exactness fails on the noncollapsible logit scale.

Uses. First-moment balance of the MAIC weights and the non-collapsibility gap Definition 7.7.

Generalize Example 18.3 by replacing the conditional benefit \(4+x_1\) with \(b_0+b_1x_1+b_2x_2\) and the comparator mean by a general target \((\bar x_1,\bar x_2)\). The MAIC-adjusted benefit is the weighted index mean of the conditional benefit, \[ \hat d_{AC}^{\mathrm{MAIC}}=\sum_i w_i\,(b_0+b_1x_{i1}+b_2x_{i2})\Big/\sum_i w_i =b_0+b_1\frac{\sum_i w_i x_{i1}}{\sum_i w_i}+b_2\frac{\sum_i w_i x_{i2}}{\sum_i w_i}. \] Any weights achieving first-moment balance satisfy \(\sum_i w_i x_{i\ell}/\sum_i w_i=\bar x_\ell\) for \(\ell=1,2\), by definition of moment balance. Substituting, \[ \hat d_{AC}^{\mathrm{MAIC}}=b_0+b_1\bar x_1+b_2\bar x_2, \] which is exactly the true comparator-population benefit. The derivation used only that the weights match the two covariate means; it did not use that the weights are the entropy (exponential-tilting) weights. Hence on the identity scale with a benefit linear in the covariates, every moment-balancing weight vector gives the same point estimate, and the choice of balancing objective (entropy, empirical likelihood, or any other) affects only the effective sample size, and through it the variance, not the point estimate.

This exactness fails on the logit scale. There the adjusted contrast is \(\operatorname{logit}(\sum_i w_i\mu_A(x_i)/\sum_i w_i)-\operatorname{logit}(\sum_i w_i\mu_C(x_i)/\sum_i w_i)\), and because \(\operatorname{logit}\) is nonlinear, balancing the first moments of the covariates does not make the weighted average risk equal the target-population risk. The discrepancy is governed by higher moments of the covariate distribution under the weights, which is why MAIC on a noncollapsible scale must consider second-moment balancing (Section 18.4), and why even exact moment balance leaves a residual non-collapsibility gap (Definition 7.7). \(\square\)

F.18.2 Solution to Exercise 18.9: Sandwich versus naive variance for the MAIC weighted mean

This Chapter 18 exercise compares the honest (sandwich) variance of a MAIC weighted mean with the naive fixed-weight variance \(\sigma^2/\mathrm{ESS}\), and identifies exactly the extra term the naive analysis drops by pretending the balancing weights are known.

Strategy. Stack the balancing equation and the weighted mean into one estimating function, read its asymptotic variance off the M-estimation sandwich, and isolate the correction that estimating \(\beta\) adds.

Uses. The MAIC estimating function Equation 18.24 and the sandwich variance Definition 5.11.

Two definitions first, since this is the densest notation in the appendix. For an M-estimator solving \(\sum_i\boldsymbol\psi(\cdot;\boldsymbol\theta)=\mathbf 0\), the bread is the expected Jacobian \(A=\mathbb E[\nabla_{\boldsymbol\theta}\boldsymbol\psi]\) and the meat is the outer-product matrix \(B=\mathbb E[\boldsymbol\psi\boldsymbol\psi^\top]\); the influence function of the estimator is the random vector \(-A^{-1}\boldsymbol\psi\) whose (co)variance gives the estimator’s asymptotic variance, because the first-order expansion of an M-estimator is \[ \sqrt n\,(\hat{\boldsymbol\theta}-\boldsymbol\theta)=-A^{-1}\cdot\frac{1}{\sqrt n}\sum_{i=1}^n\boldsymbol\psi(\cdot;\boldsymbol\theta)+o_p(1), \] so \(\sqrt n(\hat{\boldsymbol\theta}-\boldsymbol\theta)\xrightarrow{d}\mathcal N(\mathbf 0,A^{-1}BA^{-\top})\). A matrix is block lower-triangular when its upper-right block is zero, which here means the balance equation does not involve \(\mu_A\); that zero block is what lets us extract one component’s variance cleanly.

Drop the comparator block of Equation 18.24 and consider the single-arm weighted mean with scalar covariate. Stack \(\boldsymbol\theta=(\beta,\mu_A)\) and per-unit estimating function \[ \boldsymbol\psi(x,Y;\boldsymbol\theta)=\begin{pmatrix}e^{\beta\Delta}\,\Delta\\ e^{\beta\Delta}\,(Y-\mu_A)\end{pmatrix}, \qquad \Delta=x-\bar x_{\mathrm{target}} . \]

The Jacobian and its inverse. With \(A=\mathbb E[\nabla_{\boldsymbol\theta}\boldsymbol\psi]\), \[ A=\begin{pmatrix}\mathbb E[e^{\beta\Delta}\Delta^2] & 0\\[2pt] \mathbb E[e^{\beta\Delta}\Delta(Y-\mu_A)] & -\mathbb E[e^{\beta\Delta}]\end{pmatrix} =\begin{pmatrix}a_{11}&0\\ a_{21}&a_{22}\end{pmatrix}, \] block lower-triangular because the balance equation does not involve \(\mu_A\). Its inverse is \[ A^{-1}=\begin{pmatrix}1/a_{11}&0\\ -a_{21}/(a_{11}a_{22})&1/a_{22}\end{pmatrix}. \] Now apply the M-estimator expansion. The influence function of the pair \(\hat{\boldsymbol\theta}=(\hat\beta,\hat\mu_A)\) is \(-A^{-1}\boldsymbol\psi\), and we want its second component, the one for \(\hat\mu_A\). Multiplying out, \[ -A^{-1}\boldsymbol\psi =-\begin{pmatrix}1/a_{11}&0\\ -a_{21}/(a_{11}a_{22})&1/a_{22}\end{pmatrix} \begin{pmatrix}\psi_1\\\psi_2\end{pmatrix} =\begin{pmatrix}-\psi_1/a_{11}\\[2pt] \dfrac{a_{21}\psi_1}{a_{11}a_{22}}-\dfrac{\psi_2}{a_{22}}\end{pmatrix}, \] so the second row (the influence function of \(\hat\mu_A\)) is \[ \mathrm{IF}_{\mu_A}=\frac{a_{21}\psi_1}{a_{11}a_{22}}-\frac{\psi_2}{a_{22}} =\frac{1}{-a_{22}}\Bigl(\psi_2-\frac{a_{21}}{a_{11}}\psi_1\Bigr). \] The block lower-triangular structure is what makes this clean: because the upper-right entry of \(A\) (hence of \(A^{-1}\)) is zero, the \(\hat\mu_A\) row picks up only \(\psi_2\) and a single correction proportional to \(\psi_1\), not a full mixing of both. Its asymptotic variance is \[ \operatorname{Var}(\hat\mu_A)=\frac{1}{n\,a_{22}^2}\operatorname{Var}\!\Bigl(\psi_2-\frac{a_{21}}{a_{11}}\psi_1\Bigr) =\frac{1}{n\,a_{22}^2}\Bigl[B_{22}-2\frac{a_{21}}{a_{11}}B_{12}+\frac{a_{21}^2}{a_{11}^2}B_{11}\Bigr], \] with \(B=\mathbb E[\boldsymbol\psi\boldsymbol\psi^\top]\).

Fixed-weight term plus a correction. Had \(\beta\) been known, only \(\psi_2\) would enter, giving the fixed-weight variance \(B_{22}/(n\,a_{22}^2)\). With \(\operatorname{Var}(Y\mid x)=\sigma^2\) this is \(\sigma^2\mathbb E[e^{2\beta\Delta}]/(n\,\mathbb E[e^{\beta\Delta}]^2)=\sigma^2/\mathrm{ESS}\), the variance of a weighted mean with effective sample size \(\mathrm{ESS}=(\sum_i w_i)^2/\sum_i w_i^2\). Estimating \(\beta\) adds the extra terms \[ \frac{1}{n\,a_{22}^2}\Bigl[\frac{a_{21}^2}{a_{11}^2}B_{11}-2\frac{a_{21}}{a_{11}}B_{12}\Bigr], \] the correction attributable to fitting the balancing parameter.

When the correction vanishes. The correction is zero precisely when \(a_{21}=\mathbb E[e^{\beta\Delta} \Delta(Y-\mu_A)]=0\). Taking the conditional expectation over \(Y\) given \(x\), \(a_{21}=\mathbb E[w(x)\Delta(x) (\mu(x)-\mu_A)]\), the weighted covariance between the balancing covariate \(\Delta\) and the conditional outcome mean. It vanishes exactly when, within the weighting, the outcome is uncorrelated with the balancing covariate, that is when the covariate carries no prognostic information about the outcome under the weights. In that case the naive fixed-weight variance \(\sigma^2/\mathrm{ESS}\) equals the sandwich variance. Generically \(a_{21}\neq 0\), so the two differ; as the remark following Theorem 18.9 notes, the off-diagonal coupling that the naive analysis drops is usually positively correlated noise, so fixed-weight standard errors are typically anticonservative, though the sign is not universal, which is why the full sandwich is required. \(\square\)

Common mistake (naive versus sandwich MAIC variance). It is tempting to report the standard error of a MAIC estimate as if the matching weights were fixed constants, giving the tidy formula \(\sigma^2/\mathrm{ESS}\). That treats the estimated balancing parameter \(\beta\) as known, dropping the correction term derived above. Because that term is usually of the anticonservative sign, the fixed-weight standard error tends to be too small, understating the uncertainty of an indirect comparison. For a health technology assessment the safe default is the sandwich (robust) standard error, or a bootstrap that resamples and re-estimates the weights; reporting \(\sigma^2/\mathrm{ESS}\) alone can make a MAIC look more precise than it is.

F.19 Chapter 19: Simulated Treatment Comparison

F.19.1 Solution to Exercise 19.8: The log link makes the plug-in STC contrast collapsible

This Chapter 19 exercise shows that on the log link the marginal and conditional (plug-in) STC contrasts differ by a term that cancels when there is no effect modification, so the log-scale rate ratio is collapsible even though each arm separately carries a nonzero plug-in gap; the logit scale is different.

Uses. The Gaussian moment generating function (Chapter 4) and the logit non-collapsibility counterexample Theorem 19.4.

Work under Example 19.1: \(g=\log\), \(g^{-1}(\eta)=e^{\eta}\), with \(X\sim\mathcal N(\bar{\mathbf x}_{BC}, \boldsymbol\Sigma_{BC})\) in the comparator population, control-arm slope \(\boldsymbol\beta_1\) and active-arm slope \(\boldsymbol\beta_1+\boldsymbol\beta_2\). The conditional plug-in STC effect is the link contrast at the mean, \(d^{\mathrm{cond}}=\eta_A(\bar{\mathbf x}_{BC})-\eta_C(\bar{\mathbf x}_{BC})\). The marginal effect is \(d^{\mathrm{marg}}=\log\mathbb E[e^{\eta_A(X)}]-\log\mathbb E[e^{\eta_C(X)}]\). By the Gaussian moment generating function (Chapter 4), \[ \mathbb E[e^{\eta_k(X)}]=\exp\!\Bigl(\eta_k(\bar{\mathbf x}_{BC})+\tfrac12\boldsymbol\beta_{(k)}^\top\boldsymbol\Sigma_{BC}\boldsymbol\beta_{(k)}\Bigr), \] with \(\boldsymbol\beta_{(C)}=\boldsymbol\beta_1\) and \(\boldsymbol\beta_{(A)}=\boldsymbol\beta_1+\boldsymbol\beta_2\). Hence \[ d^{\mathrm{marg}}-d^{\mathrm{cond}} =\tfrac12\boldsymbol\beta_{(A)}^\top\boldsymbol\Sigma_{BC}\boldsymbol\beta_{(A)}-\tfrac12\boldsymbol\beta_{(C)}^\top\boldsymbol\Sigma_{BC}\boldsymbol\beta_{(C)} . \] Expand the difference of quadratic forms with \(\boldsymbol\beta_{(A)}=\boldsymbol\beta_1+\boldsymbol\beta_2\) and \(\boldsymbol\beta_{(C)}=\boldsymbol\beta_1\), using the symmetry of \(\boldsymbol\Sigma_{BC}\) (so \(\boldsymbol\beta_1^\top\boldsymbol\Sigma_{BC}\boldsymbol\beta_2=\boldsymbol\beta_2^\top\boldsymbol\Sigma_{BC}\boldsymbol\beta_1\)): \[ (\boldsymbol\beta_1+\boldsymbol\beta_2)^\top\boldsymbol\Sigma_{BC}(\boldsymbol\beta_1+\boldsymbol\beta_2)-\boldsymbol\beta_1^\top\boldsymbol\Sigma_{BC}\boldsymbol\beta_1 =2\,\boldsymbol\beta_1^\top\boldsymbol\Sigma_{BC}\boldsymbol\beta_2+\boldsymbol\beta_2^\top\boldsymbol\Sigma_{BC}\boldsymbol\beta_2 . \] Halving, \[ d^{\mathrm{marg}}-d^{\mathrm{cond}} =\boldsymbol\beta_1^\top\boldsymbol\Sigma_{BC}\boldsymbol\beta_2+\tfrac12\boldsymbol\beta_2^\top\boldsymbol\Sigma_{BC}\boldsymbol\beta_2 . \] Therefore the conditional plug-in is unbiased for the marginal effect (\(d^{\mathrm{cond}}=d^{\mathrm{marg}}\)) if and only if \[ \boldsymbol\beta_1^\top\boldsymbol\Sigma_{BC}\boldsymbol\beta_2+\tfrac12\boldsymbol\beta_2^\top\boldsymbol\Sigma_{BC}\boldsymbol\beta_2=0 . \] In particular this holds whenever \(\boldsymbol\beta_2=\mathbf 0\) (no effect modification): then the two arm slopes are equal, the two Gaussian variance corrections coincide, and they cancel in the contrast, so the log-link contrast is collapsible even though each arm separately suffers a nonzero plug-in gap. Contrast the logit link, where Theorem 19.4 shows the plug-in is biased for the marginal log odds ratio even with \(\boldsymbol\beta_2=\mathbf 0\). The cancellation is therefore a property of the link: the log scale makes the rate ratio collapsible (as the risk ratio is in Chapter 11), while the logit scale makes the odds ratio non-collapsible. \(\square\)

F.19.2 Solution to Exercise 19.9: Delta-method variance of the marginal STC contrast

This Chapter 19 exercise derives the asymptotic variance of the marginal STC contrast by the delta method, connects it to the Bayesian g-computation posterior, and states what changes under model misspecification.

Uses. The STC functional Equation 19.19, the outcome-MLE limit law Theorem 19.1, the delta method Theorem 5.3, and the sandwich variance Definition 5.11.

Let \(\Psi(\boldsymbol\xi)\) be the STC functional Equation 19.19, \(\Psi(\boldsymbol\xi)=g(\int\mu_C(\mathbf x;\boldsymbol\xi)f_{BC})-g(\int\mu_A(\mathbf x;\boldsymbol\xi)f_{BC})\) with \(\mu_k(\mathbf x;\boldsymbol\xi)=g^{-1}(\eta_k(\mathbf x;\boldsymbol\xi))\). By Theorem 19.1 the outcome MLE satisfies \(\sqrt n(\hat{\boldsymbol\xi}-\boldsymbol\xi_0)\xrightarrow{d}\mathcal N(\mathbf 0, \mathcal I(\boldsymbol\xi_0)^{-1})\). Since \(\Psi\) is continuously differentiable in \(\boldsymbol\xi\), the delta method (Theorem 5.3) gives \[ \sqrt n\bigl(\hat d_{AC}^{\,\mathrm{marg}}-d_{AC(\mathcal P_{BC})}^{\,\mathrm{marg}}\bigr) =\sqrt n\bigl(\Psi(\hat{\boldsymbol\xi})-\Psi(\boldsymbol\xi_0)\bigr) \xrightarrow{d}\mathcal N\!\bigl(0,\ \nabla\Psi(\boldsymbol\xi_0)^\top\mathcal I(\boldsymbol\xi_0)^{-1}\nabla\Psi(\boldsymbol\xi_0)\bigr). \]

The gradient. Writing \(\bar\mu_k=\int\mu_k(\mathbf x;\boldsymbol\xi_0)f_{BC}(\mathbf x)\,\mathrm d\mathbf x\) and differentiating under the integral, \[ \frac{\partial\Psi}{\partial\xi_j} =g'(\bar\mu_C)\!\int (g^{-1})'(\eta_C(\mathbf x))\frac{\partial\eta_C}{\partial\xi_j}(\mathbf x)\,f_{BC}(\mathbf x)\,\mathrm d\mathbf x -g'(\bar\mu_A)\!\int (g^{-1})'(\eta_A(\mathbf x))\frac{\partial\eta_A}{\partial\xi_j}(\mathbf x)\,f_{BC}(\mathbf x)\,\mathrm d\mathbf x, \] so each coordinate of \(\nabla\Psi(\boldsymbol\xi_0)\) is an integral of \((g^{-1})'\), weighted by the corresponding linear-predictor derivative, against the comparator density \(f_{BC}\), scaled by the outer \(g'\) at the marginal mean.

Bayesian g-computation. By the Bernstein-von Mises theorem (part 2 of Theorem 19.5) the posterior of \(\boldsymbol\xi\) is asymptotically \(\mathcal N(\hat{\boldsymbol\xi},n^{-1}\mathcal I^{-1})\), and pushing it through the smooth map \(\Psi\) gives a posterior for the target effect asymptotically \(\mathcal N(\Psi(\hat{\boldsymbol\xi}),n^{-1}\nabla\Psi^\top\mathcal I^{-1}\nabla\Psi)\), the same variance as the delta method. The Bayesian route avoids computing \(\nabla\Psi\) because it draws posterior samples \(\boldsymbol\xi^{(s)}\) and forms \(\Psi(\boldsymbol\xi^{(s)})\) directly; the spread of the pushed-forward draws estimates the variance without any explicit derivative.

Under misspecification. If the outcome model is misspecified, \(\hat{\boldsymbol\xi}\) converges to the pseudo-true value \(\boldsymbol\xi^\ast\) minimizing the Kullback-Leibler divergence, and the M-estimation limit law replaces \(\mathcal I^{-1}\) by the sandwich \(A^{-1}BA^{-\top}\) of Definition 5.11, so the asymptotic variance becomes \(\nabla\Psi(\boldsymbol\xi^\ast)^\top A^{-1}BA^{-\top}\nabla\Psi(\boldsymbol\xi^\ast)\). The point estimand is then \(\Psi(\boldsymbol\xi^\ast)\), the marginal STC contrast induced by the best-fitting (Kullback-Leibler-projected) parameter, which need not equal the true marginal effect when the working outcome surface is wrong. \(\square\)

F.20 Chapter 20: Multilevel Network Meta-Regression I: The Aggregation Principle

F.20.1 Solution to Exercise 20.3: Aggregation bias of the logit contrast, computed by hand

This Chapter 20 exercise recomputes the integrated (ML-NMR) aggregate risks and the naive plug-in risks for a two-point covariate, and reads off the ecological (aggregation) bias arm by arm. Every number is shown so the aggregation identity can be checked with a calculator.

Uses. The no-closed-form logit aggregation theorem Theorem 20.5 and the ecological-bias definition Definition 16.2.

Repeat Example 20.1 with effect-modifier prevalence \(q=P(x=1)=0.3\), keeping \(\mu_j=-1\), \(\beta_1=0.5\), \(\beta_{2,B}=1.0\), \(\gamma_B=0.8\), so \(\eta_C(x)=-1+0.5x\), \(\eta_B(x)=-0.2+1.5x\). The conditional risks do not depend on \(q\): \(\operatorname{expit}(\eta_C(0))=0.26894\), \(\operatorname{expit}(\eta_C(1))=0.37754\), \(\operatorname{expit}(\eta_B(0))=0.45017\), \(\operatorname{expit}(\eta_B(1))=0.78583\).

(a) Integrated aggregate risks. With weights \((1-q,q)=(0.7,0.3)\), \[ \theta_{\bullet C}=0.7(0.26894)+0.3(0.37754)=0.30152, \qquad \theta_{\bullet B}=0.7(0.45017)+0.3(0.78583)=0.55087 . \]

(b) Plug-in risks at \(\bar x_j=q=0.3\). Here \(\eta_C(0.3)=-0.85\), \(\eta_B(0.3)=0.25\), so \[ \theta^{\mathrm{naive}}_{\bullet C}=\operatorname{expit}(-0.85)=0.29943, \qquad \theta^{\mathrm{naive}}_{\bullet B}=\operatorname{expit}(0.25)=0.56218 . \]

(c) Plug-in marginal log odds ratio. Because \(\operatorname{logit}\circ\operatorname{expit}\) is the identity, the plug-in contrast telescopes to the conditional contrast at the mean, \[ \operatorname{logit}\theta^{\mathrm{naive}}_{\bullet B}-\operatorname{logit}\theta^{\mathrm{naive}}_{\bullet C} =\eta_B(0.3)-\eta_C(0.3)=0.25-(-0.85)=1.10, \] which equals \(\gamma_B+\beta_{2,B}\bar x_j=0.8+1.0(0.3)=1.10\) exactly. The integrated (ML-NMR) marginal log odds ratio is \[ \operatorname{logit}(0.55087)-\operatorname{logit}(0.30152)=0.20420-(-0.84002)=1.04421, \] so the ecological bias of the contrast is \(1.10-1.04421=+0.05579\).

(d) Effect of decreasing \(q\). Decreasing \(q\) from \(0.5\) to \(0.3\) shifts both arms’ mean linear predictors downward, since \(\eta_k\) increases in \(x\) and less weight falls on \(x=1\): the reference arm has \(m_C=\eta_C(0.3)=-0.85<0\) and the active arm \(m_B=\eta_B(0.3)=0.25>0\). By Theorem 20.5(c) the sign of each arm’s bias follows the curvature of \(\operatorname{expit}\) at its mean linear predictor: \(\operatorname{expit}\) is convex where the argument is negative, so the reference arm is underestimated (\(\mathrm{EB}_C=0.29943-0.30152=-0.00209<0\)), and concave where positive, so the active arm is overestimated (\(\mathrm{EB}_B=0.56218-0.55087=+0.01131>0\)). As \(q\) decreases further, \(m_B\) falls toward and through zero, at which point the active-arm bias would change sign; the biases do not cancel across arms but combine in the contrast. \(\square\)

F.20.2 Solution to Exercise 20.4: Exact aggregation characterizes the identity link

This Chapter 20 exercise proves that the plug-in-at-the-mean shortcut is exact for every covariate distribution if and only if the link is the identity (up to affine reparameterization); on any nonlinear scale, aggregation bias is unavoidable.

Uses. The identity-link exactness result Theorem 20.3 and the fact that a continuous midpoint-affine function is affine.

We show a twice-differentiable link \(g\) satisfies \(\theta_{\bullet jk}=\theta^{\mathrm{naive}}_{\bullet jk}\) for every nondegenerate covariate distribution and every total slope \(\boldsymbol\beta_{\mathrm{tot},k}\) if and only if \(g^{-1}(z)=a+bz\) for constants \(a,b\) with \(b\neq 0\) (the identity link up to affine reparameterization).

Sufficiency. If \(g^{-1}(z)=a+bz\), then for the linear predictor \(\eta(\mathbf x)\), \[ \theta_{\bullet jk}=\int g^{-1}(\eta(\mathbf x))f_j(\mathbf x)\,\mathrm d\mathbf x =a+b\,\mathbb E[\eta(X)]=a+b\,\eta(\bar{\mathbf x}_j)=g^{-1}(\eta(\bar{\mathbf x}_j))=\theta^{\mathrm{naive}}_{\bullet jk}, \] using linearity of \(g^{-1}\) and of \(\eta\) together with linearity of expectation (this is Theorem 20.3).

Necessity. Suppose \(\theta_{\bullet jk}=\theta^{\mathrm{naive}}_{\bullet jk}\) for every nondegenerate covariate law and every total slope. Fix any open interval \(I\subseteq\mathbb R\). Choosing a slope and a covariate law so that the linear predictor \(W=\eta(X)\) takes its values in \(I\), the equality reads \(g^{-1}(\mathbb E[W])=\mathbb E[g^{-1}(W)]\) for every distribution of \(W\) supported in \(I\). Apply it to the two-point law placing mass \(\tfrac12\) at each of \(w_1,w_2\in I\): with \(\psi=g^{-1}\), \[ \psi\!\Bigl(\frac{w_1+w_2}{2}\Bigr)=\tfrac12\psi(w_1)+\tfrac12\psi(w_2), \] the midpoint-affine (Jensen) equation. A continuous function satisfying the midpoint-affine identity for every pair in an interval is affine on that interval; since \(g\) is twice differentiable, so is \(\psi\), and twice differentiating the midpoint identity (or invoking the classical result for measurable midpoint-affine functions) gives \(\psi''\equiv 0\) on \(I\), hence \(\psi(z)=a_I+b_I z\) on \(I\). Because \(I\) was an arbitrary open interval and the affine pieces must agree on overlaps, \(\psi=g^{-1}\) is globally affine, \(g^{-1}(z)=a+bz\). The slope \(b\neq 0\) because \(g^{-1}\) is a strictly monotone inverse link. Translating to \(g\), the relation \(\mu=a+b\,g(\mu)\) gives \(g(\mu)=(\mu-a)/b\), also affine: \(g\) is the identity link up to an affine reparameterization.

Conclusion. Exact aggregation for all covariate distributions characterizes the identity link. Equivalently, on any genuinely nonlinear scale, including the logit, log, probit, and complementary log-log, there exist covariate distributions for which \(\theta_{\bullet jk}\neq\theta^{\mathrm{naive}}_{\bullet jk}\), so aggregation bias is unavoidable. \(\square\)

F.21 Chapter 21: Multilevel Network Meta-Regression II: Discrete Outcomes and Approximation

F.21.1 Solution to Exercise 21.6: Le Cam’s Poisson approximation by coupling

This Chapter 21 exercise proves Le Cam’s total-variation bound between a Poisson-binomial sum and a Poisson, using an explicit coupling of a Bernoulli with a Poisson, and recovers the classical Poisson limit theorem as a corollary.

Uses. The coupling inequality Equation 21.13 (\(\|\mu-\nu\|_{\mathrm{TV}}\le P(X\ne Z)\) for any coupling) and a union bound.

(a) Marginals and disagreement of the single pair. The coupling Equation 21.15 assigns \(P(B_i=0,P_i=0)=1-p\), \(P(B_i=1,P_i=1)=pe^{-p}\), \(P(B_i=1,P_i=0)=e^{-p}-(1-p)\), and \(P(B_i=1,P_i=k)=e^{-p}p^k/k!\) for \(k\ge 2\). The \(B_i\)-marginal is \(P(B_i=0)=1-p\) and \(P(B_i=1)=p\) (summing the row), so \(B_i\sim\mathrm{Ber}(p)\); the \(P_i\)-marginal at \(0\) is \((1-p)+(e^{-p}-(1-p))=e^{-p}\), at \(1\) is \(pe^{-p}\), and at \(k\ge 2\) is \(e^{-p}p^k/k!\), so \(P_i\sim\mathrm{Poi}(p)\). The pair agrees only on \((0,0)\) and \((1,1)\), so \[ P(B_i\ne P_i)=1-(1-p)-pe^{-p}=p-pe^{-p}=p(1-e^{-p}). \]

(b) Coupling inequality and the bound. For any coupling \((X,Z)\) of \(\mu,\nu\) and any event \(A\), \(\mu(A)-\nu(A)=P(X\in A,Z\notin A)-P(Z\in A,X\notin A)\le P(X\ne Z)\); by symmetry \(|\mu(A)-\nu(A)|\le P(X\ne Z)\), and taking the supremum over \(A\) gives the coupling inequality Equation 21.13, \(\|\mu-\nu\|_{\mathrm{TV}}\le P(X\ne Z)\). Now take the single-pair couplings independent across \(i=1,\dots,n\), set \(Y_\bullet=\sum_i B_i\sim\mathrm{PoBin}(\mathbf p)\) and \(Z=\sum_i P_i\sim\mathrm{Poi}(\lambda)\) (a sum of independent Poissons, \(\lambda=\sum_i p_i\)). Then \(\{Y_\bullet\ne Z\}\subseteq\bigcup_i\{B_i\ne P_i\}\), so by the coupling inequality and a union bound, \[ \|\mathrm{PoBin}(\mathbf p)-\mathrm{Poi}(\lambda)\|_{\mathrm{TV}}\le P(Y_\bullet\ne Z) \le\sum_{i=1}^n P(B_i\ne P_i)=\sum_{i=1}^n p_i(1-e^{-p_i})\le\sum_{i=1}^n p_i^2, \] the last step by \(1-e^{-p}\le p\). This is Equation 21.14.

(c) Equal-risk case. With \(p_i\equiv p\) and \(n\) terms, \(\lambda=np\) and the bound is \(\sum_i p_i^2=np^2\), so \(\|\mathrm{Bin}(n,p)-\mathrm{Poi}(np)\|_{\mathrm{TV}}\le np^2\). Holding \(np=\lambda\) fixed and letting \(n\to\infty\), \(p=\lambda/n\) gives \(np^2=\lambda^2/n\to 0\), so \(\mathrm{Bin}(n,\lambda/n)\to\mathrm{Poi}(\lambda)\) in total variation, recovering the Poisson limit theorem Theorem 4.13.

(d) Sharpness to leading order. Expanding \(1-e^{-p}=p-\tfrac12p^2+O(p^3)\), the per-pair disagreement is \(p(1-e^{-p})=p^2-\tfrac12p^3+O(p^4)\). The bound replaces this by \(p^2\), an overstatement of order \(\tfrac12p^3\); relative to the leading \(p^2\) term the error is \(O(p)\), so for small risks \(\sum_i p_i(1-e^{-p_i})=\sum_i p_i^2-\tfrac12\sum_i p_i^3+O(\sum_i p_i^4)\) and the bound \(\sum_i p_i^2\) is tight to leading order. \(\square\)

F.21.2 Solution to Exercise 21.8: The adjusted binomial matches two moments but not three

This Chapter 21 exercise computes the third central moment of a Poisson-binomial sum and shows that the moment-matched adjusted binomial, fixed by the first two moments, generically misses the third, so it is a genuinely two-moment surrogate with no total-variation guarantee.

Uses. The Poisson-binomial generating function Equation 21.8 and the adjusted-binomial construction Theorem 21.4.

(a) The Poisson binomial third central moment. Write \(Y_\bullet=\sum_i B_i\) with independent \(B_i\sim\mathrm{Ber}(p_i)\). For a single Bernoulli, \(B_i-p_i\) takes value \(1-p_i\) with probability \(p_i\) and \(-p_i\) with probability \(1-p_i\), so \[ \mathbb E[(B_i-p_i)^3]=p_i(1-p_i)^3+(1-p_i)(-p_i)^3=p_i(1-p_i)\bigl[(1-p_i)^2-p_i^2\bigr]=p_i(1-p_i)(1-2p_i). \] The third central moment is additive over independent summands (the third cumulant of a sum of independent variables is the sum of their third cumulants; equivalently, expanding \(\mathbb E[(\sum_i(B_i-p_i))^3]\) all cross terms vanish by independence and mean-zero centering), so \[ \mathbb E[(Y_\bullet-\lambda)^3]=\sum_i p_i(1-p_i)(1-2p_i), \] as claimed, which also follows from the third derivative of the generating function Equation 21.8 at \(z=1\).

(b) The example. For \(\mathbf p=(0.1,0.3,0.4,0.5,0.7)\) of Example 21.5, \[ \begin{aligned} \sum_i p_i(1-p_i)(1-2p_i)&=0.1(0.9)(0.8)+0.3(0.7)(0.4)+0.4(0.6)(0.2)+0.5(0.5)(0)+0.7(0.3)(-0.4)\\ &=0.072+0.084+0.048+0-0.084=0.120 . \end{aligned} \] The matched adjusted binomial is \(\mathrm{Bin}(N^\ast,\pi^\ast)=\mathrm{Bin}(4,0.5)\) (since \(\lambda=2\), \(\nu=\sum_i p_i^2=1\), so \(\pi^\ast=\nu/\lambda=0.5\) and \(N^\ast=\lambda^2/\nu=4\)). Its third central moment is \(N^\ast\pi^\ast(1-\pi^\ast)(1-2\pi^\ast)=4(0.5)(0.5)(0)=0\): the adjusted binomial is symmetric and misses the Poisson binomial’s right skew \(0.120>0\). Hence no binomial can match three moments of a generic Poisson binomial: once \((N,\pi)\) are fixed to match the mean and variance, the third moment is determined and generically differs (here it is forced to zero by \(\pi^\ast=\tfrac12\)). The adjusted binomial is genuinely a two-moment surrogate, consistent with the remark after Theorem 21.4 that it carries no total-variation guarantee. \(\square\)

F.22 Chapter 22: Multilevel Network Meta-Regression III: Numerical Integration Theory

F.22.1 Solution to Exercise 22.7: The latent correlation is not the covariate correlation

This Chapter 22 exercise shows that the Gaussian-copula parameter \(\boldsymbol\Omega\) is the correlation of the latent normals, which equals the covariate Pearson correlation only when the margins are normal; a nonlinear margin shrinks the observed Pearson correlation while preserving the rank correlation.

Uses. The NORTA map \(X_j=F_j^{-1}(\Phi(Z_j))\) and the maximal-correlation property of the bivariate normal.

Standard-normal margins. With both margins standard normal and a Gaussian copula at latent correlation \(\rho\), the NORTA map \(X_j=\Phi^{-1}(\Phi(Z_j))=Z_j\) is the identity, so \((X_1,X_2)=(Z_1,Z_2)\sim\mathcal N_2\) with correlation \(\rho\). The covariate Pearson correlation therefore equals the latent correlation \(\rho\) exactly: for standard-normal margins the Gaussian-copula parameter and the covariate correlation coincide.

Exponential first margin. Now replace the first margin by an exponential, keeping the Gaussian copula at latent correlation \(\rho\), so \(X_1=F_1^{-1}(\Phi(Z_1))\) with \(F_1\) exponential and \(X_2=Z_2\). The covariate Pearson correlation is \(\operatorname{corr}(X_1,X_2)\). The transform \(u\mapsto F_1^{-1}(\Phi(u))\) is a nonlinear, strictly increasing reshaping of \(Z_1\); Pearson correlation is not invariant under nonlinear monotone transforms, and for a Gaussian vector a strictly increasing nonlinear transform of one coordinate strictly shrinks the magnitude of the Pearson correlation toward zero (this is the content of the maximal-correlation property of the bivariate normal, which is attained only by linear, that is affine, transforms). Hence \(|\operatorname{corr}(X_1,X_2)|<|\rho|\) for \(\rho\neq 0\). By contrast, the Spearman rank correlation depends only on the copula, and applying the strictly increasing map \(F_1^{-1}\circ\Phi\) to a coordinate leaves all ranks unchanged, so the Spearman correlation is exactly the same as in the standard-normal case.

This is precisely the remark in Section 22.9 that \(\boldsymbol\Omega\) is the correlation of the latent normals \(Z\), not of the covariates \(X\): only for normal margins do the two coincide; a nonlinear margin attenuates the covariate Pearson correlation while preserving the rank correlation. \(\square\)

Common mistake (latent versus observed correlation). When specifying the Gaussian copula for ML-NMR integration, it is tempting to plug the observed covariate correlation matrix directly in as \(\boldsymbol\Omega\). That is wrong whenever any margin is non-normal (a skewed biomarker, a binary indicator): the latent correlation that reproduces a target covariate correlation is generally larger in magnitude, because the nonlinear margin shrinks correlation on the way out. Feeding the observed correlation in as the latent one understates the dependence among the integrated covariates and biases the aggregate mean. The fix is to match on the rank (Spearman) correlation, which the copula preserves, or to invert the NORTA relationship numerically to recover the latent \(\boldsymbol\Omega\) that yields the desired covariate correlation.

F.22.2 Solution to Exercise 22.8: A four-point quasi-Monte Carlo aggregation, and its copula dependence

This Chapter 22 exercise traces four Sobol’ points through the full NORTA / Gaussian-copula pipeline to a quasi-Monte Carlo estimate of an aggregate logistic mean, and shows the estimate depends on the copula correlation through the variance of the linear predictor.

Uses. The QMC pipeline Equation 22.17, the Cholesky factor of the copula correlation, and the aggregation/ecological dependence on covariance Theorem 20.2.

Extend Example 22.3 to two standard-normal margins with Gaussian-copula correlation \(\rho=0.5\) and linear predictor \(\eta(\mathbf x)=0.5+x_1+0.5x_2\) on the logit scale. With standard-normal margins the pipeline Equation 22.17 reduces to \(\mathbf x=L\,\Phi^{-1}(\mathbf u)\), where \(L=\begin{pmatrix}1&0\\0.5&0.8660\end{pmatrix}\) is the Cholesky factor of \(\boldsymbol\Omega=\begin{pmatrix}1&0.5\\0.5&1\end{pmatrix}\) (the two normal transforms \(\Phi\) and \(F^{-1}=\Phi^{-1}\) cancel). Using \(\Phi^{-1}(\tfrac12)=0\), \(\Phi^{-1}(\tfrac14)=-0.6745\), \(\Phi^{-1}(\tfrac34)=0.6745\), \(\Phi^{-1}(\tfrac18)=-1.1503\), \(\Phi^{-1}(\tfrac58)=0.3186\), the four Sobol’ points trace to

\(\mathbf u\)	\(\mathbf z=\Phi^{-1}(\mathbf u)\)	\(\mathbf x=L\mathbf z\)	\(\eta(\mathbf x)\)	\(\operatorname{expit}(\eta)\)
\((\tfrac12,\tfrac12)\)	\((0,0)\)	\((0,0)\)	\(0.5000\)	\(0.62246\)
\((\tfrac14,\tfrac34)\)	\((-0.6745,0.6745)\)	\((-0.6745,0.24687)\)	\(-0.05107\)	\(0.48724\)
\((\tfrac34,\tfrac14)\)	\((0.6745,-0.6745)\)	\((0.6745,-0.24687)\)	\(1.05107\)	\(0.74102\)
\((\tfrac18,\tfrac58)\)	\((-1.1503,0.3186)\)	\((-1.1503,-0.29924)\)	\(-0.79992\)	\(0.31005\)

The four-point QMC estimate at \(\rho=0.5\) is \[ \widehat\theta_{\bullet}=\tfrac14(0.62246+0.48724+0.74102+0.31005)=\tfrac14(2.16077)=0.54019 . \] Setting \(\rho=0\) makes \(L=I\), so \(\mathbf x=\mathbf z\) and the same four points give \(\operatorname{expit}(\eta)\) equal to \(0.62246\), \(0.54060\), \(0.69788\), \(0.37966\), with estimate \[ \widehat\theta_{\bullet}=\tfrac14(0.62246+0.54060+0.69788+0.37966)=\tfrac14(2.24060)=0.56015 . \] The estimate changes (\(0.540\) versus \(0.560\)) even though the margins are unchanged, because the aggregate mean \(\mathbb E[\operatorname{expit}(\eta(X))]\) of a nonlinear link depends on the joint distribution through the variance of the linear predictor. Using \(\operatorname{Var}(aX_1+bX_2)=a^2\operatorname{Var}(X_1) +b^2\operatorname{Var}(X_2)+2ab\operatorname{Cov}(X_1,X_2)\) with \(a=1\), \(b=0.5\), standard-normal margins (\(\operatorname{Var}(X_1)=\operatorname{Var}(X_2)=1\)), and \(\operatorname{Cov}(X_1,X_2)=\rho\), \[ \operatorname{Var}(\eta)=\operatorname{Var}(X_1+0.5X_2)=1+0.25+2(1)(0.5)\rho=1.25+\rho, \] which is \(1.75\) at \(\rho=0.5\) and \(1.25\) at \(\rho=0\). The mean linear predictor is \(0.5>0\), where \(\operatorname{expit}\) is concave, so larger variance pulls the aggregate mean further below \(\operatorname{expit}(0.5)=0.622\); the higher-correlation case therefore has the smaller aggregate risk. This is the aggregation/ecological dependence on the covariate covariance of Theorem 20.2 and Definition 16.2: the correlation enters the integrated mean through the spread of \(\eta\), not through the margins. \(\square\)

F.23 Chapter 23: Multilevel Network Meta-Regression IV: Estimands, Identifiability, and Reduction

F.23.1 Solution to Exercise 23.7: SEMA fixes the conditional contrast but not the marginal one

This Chapter 23 exercise writes the non-collapsibility gap between the conditional and marginal \(A\)-versus-\(B\) log odds ratios in closed form under the shared effect-modifier assumption (SEMA), identifies when it vanishes, and checks it numerically. The point is that SEMA pins the conditional contrast to a constant but cannot make the marginal contrast population-free.

Strategy. Compute each arm’s marginal risk as a two-point average, take the logit difference for the marginal contrast, and subtract from the SEMA conditional constant. The gap is “logit of an average minus an average of logits,” the Jensen discrepancy, so it vanishes exactly when there is nothing to average over.

Uses. The aggregation-bias theorem Theorem 20.2 and the two-point marginal-mean construction.

(a) Setup. Single binary effect modifier \(X\in\{0,1\}\) with \(\Pr(X=1)=q\), logit link, SEMA. The arm linear predictors in a population with reference baseline \(\mu_{*}\) are \(\eta_C(x)=\mu_{*}+\boldsymbol\beta_1 x\) and, for \(k\in\{A,B\}\), \(\eta_k(x)=\mu_{*}+(\boldsymbol\beta_1+\boldsymbol\beta_2)x+\gamma_k\) (taking \(\gamma_C=0\)). The conditional \(A\)-versus-\(B\) contrast is the SEMA constant \(\gamma_A-\gamma_B\) (the interaction cancels). Evaluating each arm’s linear predictor at the two covariate values, \(\eta_k(0)=\mu_{*}+\gamma_k\) and \(\eta_k(1)=\mu_{*}+\boldsymbol\beta_1+\boldsymbol\beta_2+\gamma_k\).

(b) The closed-form gap. Rather than write one three-line formula, name the pieces. The marginal risk under each arm is the two-point average \[ \bar\theta_A=(1-q)\operatorname{expit}(\mu_{*}+\gamma_A)+q\operatorname{expit}(\mu_{*}+\boldsymbol\beta_1+\boldsymbol\beta_2+\gamma_A), \] \[ \bar\theta_B=(1-q)\operatorname{expit}(\mu_{*}+\gamma_B)+q\operatorname{expit}(\mu_{*}+\boldsymbol\beta_1+\boldsymbol\beta_2+\gamma_B), \] the marginal contrast is the logit difference \(\Delta_{BA}(\mathcal P^{*})=\operatorname{logit}\bar\theta_A-\operatorname{logit}\bar\theta_B\), and the non-collapsibility gap is the conditional minus the marginal, \[ \text{gap}=(\gamma_A-\gamma_B)-\Delta_{BA}(\mathcal P^{*}) =(\gamma_A-\gamma_B)-\bigl(\operatorname{logit}\bar\theta_A-\operatorname{logit}\bar\theta_B\bigr). \] Substituting \(\bar\theta_A,\bar\theta_B\) gives the single closed-form expression, now assembled from four readable ingredients rather than presented as one wall of symbols.

(c) Degenerate cases. If \(q\in\{0,1\}\) the covariate is degenerate: at \(q=0\), \(\bar\theta_k=\operatorname{expit}(\mu_{*}+\gamma_k)\) and \(\operatorname{logit}\bar\theta_A-\operatorname{logit}\bar\theta_B=\gamma_A-\gamma_B\), so the gap is zero; the same holds at \(q=1\). If \(\boldsymbol\beta_1+\boldsymbol\beta_2=0\), the active-arm linear predictors \(\eta_A,\eta_B\) are constant in \(x\) (no spread), so again \(\bar\theta_k=\operatorname{expit}(\mu_{*}+\gamma_k)\) and the gap is zero. Both vanishing conditions are exactly those of Theorem 20.2: a degenerate covariate or no spread in the active-arm linear predictors leaves nothing to average over.

(d) Numerical check. At the inputs of Example 23.3 (\(\mu_{*}=-0.5\), \(\boldsymbol\beta_1=0.4\), \(\boldsymbol\beta_2=0.5\), \(\gamma_A=0.8\), \(\gamma_B=0.3\), \(q=0.6\)), \(\bar\theta_A=0.4\operatorname{expit}(0.3)+0.6\operatorname{expit}(1.2)=0.690892\) and \(\bar\theta_B=0.4\operatorname{expit}(-0.2)+0.6\operatorname{expit}(0.7)=0.580979\), so \(\Delta_{BA}(\mathcal P^{*})=\operatorname{logit}(0.690892)-\operatorname{logit}(0.580979)=0.804292-0.326794=0.477498\), and the gap is \((\gamma_A-\gamma_B)-0.477498=0.5-0.477498=0.022502\), matching the chapter. SEMA cannot eliminate the gap because it pins the conditional contrast to a constant but the marginal contrast involves logit-of-average minus average-of-logit, the non-collapsibility (Jensen) discrepancy, which persists whenever the covariate has spread and \(q\in(0,1)\). \(\square\)

F.23.2 Solution to Exercise 23.8: Marginal STC is a special case of ML-NMR

This Chapter 23 exercise shows that marginal STC is exactly ML-NMR restricted to fitting on the IPD alone and fixing the target to the comparator population; the extra aggregate-data information and the free choice of target population are what a full ML-NMR fit adds.

Uses. The ML-NMR aggregate arm mean Equation 23.2, the aggregation identity Theorem 20.1, and STC consistency Theorem 19.6.

Fix a logistic outcome family with linear predictor \(\eta_k(\mathbf x;\boldsymbol\xi)\) and inverse link \(g^{-1}=\operatorname{expit}\). Marginal STC (Chapter 19) for an IPD trial on \(\{A,C\}\) standardized to the comparator covariate distribution \(f_{BC}\) predicts the arm means \[ \hat\mu_k^{\mathrm{STC}}=\int \operatorname{expit}\!\bigl(\eta_k(\mathbf x;\hat{\boldsymbol\xi})\bigr)\,f_{BC}(\mathbf x)\,\mathrm d\mathbf x,\qquad k\in\{A,C\}, \] and reports \(g(\hat\mu_C^{\mathrm{STC}})-g(\hat\mu_A^{\mathrm{STC}})\). The ML-NMR aggregate arm mean Equation 23.2 over the same target \(f_{BC}\) using the same fitted linear predictor is \[ \bar\theta_{\bullet k}=\int \operatorname{expit}\!\bigl(\eta_k(\mathbf x;\hat{\boldsymbol\xi})\bigr)\,f_{BC}(\mathbf x)\,\mathrm d\mathbf x . \] These are the identical functional of \(\hat{\boldsymbol\xi}\): both integrate \(\operatorname{expit}(\eta_k)\) against \(f_{BC}\) and contrast on the link scale. Hence marginal STC is ML-NMR restricted to (i) fitting the outcome surface on the IPD alone and (ii) fixing the target population to the comparator \(f_{BC}\).

Two things distinguish a full ML-NMR fit from this restriction. First, ML-NMR’s likelihood incorporates the aggregate-data outcomes through the aggregation identity Theorem 20.1: the AgD arm contributes its observed event summary to the likelihood, so the fitted \(\hat{\boldsymbol\xi}\) uses information that STC (fitting on the IPD only) ignores. Second, ML-NMR’s estimand Theorem 23.1 admits any target population, not only \(f_{BC}\), so it can report the contrast standardized to a decision population different from the comparator trial.

Therefore STC is the special case of ML-NMR obtained by fitting on the IPD alone and fixing the target to \(f_{BC}\). The extra information ML-NMR uses changes the answer when the AgD outcomes are informative about shared parameters, for instance when the comparator arm’s observed event rate sharpens the prognostic or baseline parameters that also enter the IPD arms, shifting \(\hat{\boldsymbol\xi}\) and hence the contrast. It does not change the answer when the IPD already identifies \(\boldsymbol\xi\) and the target is the comparator population with no extrapolation, the no-extrapolation regime of Chapter 18, in which STC is consistent (Theorem 19.6) and the AgD outcomes add precision but not a different limit. \(\square\)

F.24 Chapter 24: Bayesian Computation and Diagnostics

F.24.1 Solution to Exercise 24.4: The algebraic floor of the potential scale reduction factor

This Chapter 24 exercise shows that the split-\(\widehat R\) convergence diagnostic has a hard lower bound \(\sqrt{(N-1)/N}\) set by the number of iterations, is centered at \(1\) at convergence, and is pushed above \(1\) by autocorrelation-inflated between-chain variance.

Uses. The between/within variance definitions Equation 24.6, Equation 24.7 and the expectation lemmas Lemma 24.4, Lemma 24.5.

The algebraic floor. From the definitions Equation 24.6 and Equation 24.7, \(\widehat V=(1-\tfrac1N)W+\tfrac1N B\), so dividing by \(W>0\), \[ \frac{\widehat V}{W}=\Bigl(1-\frac1N\Bigr)+\frac1N\frac{B}{W}=\frac{N-1}{N}+\frac{B}{NW}. \] Since \(B\ge 0\), \(\widehat V/W\ge (N-1)/N\), hence \(\widehat R=\sqrt{\widehat V/W}\ge\sqrt{(N-1)/N}\) always. This floor is a striking practical fact for short chains: at \(N=4\) draws it is \(\sqrt{3/4}=0.866\), at \(N=100\) draws it is \(\sqrt{99/100}=0.995\), and it rises to \(1\) as \(N\to\infty\). So with only four iterations per chain the diagnostic can never read below \(0.866\) no matter how perfectly the chains agree; the floor reflects the estimator’s construction, not convergence. Moreover \(\widehat R\ge 1\) if and only if \(\widehat V/W\ge 1\), that is \((N-1)/N+B/(NW)\ge 1\), that is \(B/(NW)\ge 1/N\), that is \(B\ge W\).

Centering at convergence. In the iid-within-chain idealization a single chain mean of \(N\) independent draws has variance \(v_N=\sigma^2/N\). By Lemma 24.4, \(\mathbb E[W]=\tfrac{N}{N-1}(\sigma^2-v_N)=\tfrac{N}{N-1}\sigma^2\tfrac{N-1}{N}=\sigma^2\), and by Lemma 24.5, \(\mathbb E[B]=Nv_N=\sigma^2\). Thus \(\mathbb E[B]=\mathbb E[W]=\sigma^2\), so \(\widehat V/W\) is centered at \((N-1)/N+\sigma^2/(N\sigma^2)=(N-1)/N+1/N=1\) and \(\widehat R\) is centered at \(1\) at convergence.

What pushes \(\widehat R\) above \(1\). The term that lifts \(\widehat R\) above one is the between-chain term \(B\), through \(B/(NW)\). When the chains are positively autocorrelated but share a common stationary law, the variance of a chain mean is inflated by the integrated autocorrelation time \(\tau_{\mathrm{int}}=1+2\sum_{k\ge 1}\rho_k>1\), namely \(v_N\approx(\sigma^2/N)\tau_{\mathrm{int}}\), so \(\mathbb E[B]=Nv_N\approx\sigma^2\tau_{\mathrm{int}}>\sigma^2\) while \(\mathbb E[W]\) stays near \(\sigma^2\). Then \(\widehat V/W\approx(N-1)/N+\tau_{\mathrm{int}}/N=1+(\tau_{\mathrm{int}}-1)/N\) and \(\widehat R\approx\sqrt{1+(\tau_{\mathrm{int}}-1)/N}>1\). The excess of the autocorrelation-inflated between-chain variance over the within-chain variance, captured by \(B/(NW)\), is what holds \(\widehat R\) above one until the chains mix. \(\square\)

F.24.2 Solution to Exercise 24.10: Generalized Pareto tail moments and the PSIS-LOO shape thresholds

This Chapter 24 exercise finds which moments of a generalized Pareto tail are finite as a function of its shape parameter \(k\), then reads those thresholds as the reliability tiers of the Pareto-smoothed importance sampling (PSIS) leave-one-out diagnostic.

Strategy. A power-law tail \(f(x)\sim Cx^{-1/k-1}\) has a finite \(m\)-th moment exactly when the integrand \(x^m f(x)\) decays fast enough to integrate at infinity; the threshold is a one-line exponent comparison.

Uses. The generalized Pareto moment lemma Lemma 24.7 and the PSIS reliability theorem Theorem 24.4. As a reminder of the vocabulary, PSIS-LOO is leave-one-out cross-validation carried out by importance sampling: each held-out point is scored by reweighting the full-data posterior draws with importance ratios \(r_i\) (the ratio of the leave-one-out to the full posterior density), and the predictive quantity is a self-normalized importance estimator (a weighted average with weights \(r_i\) divided by their sum). PSIS stabilizes the largest of these ratios by fitting a generalized Pareto distribution to their upper tail, whose estimated shape \(\widehat k\) is the diagnostic below.

Moments of the generalized Pareto tail. Let \(\bar F(x)=(1+kx/\sigma)^{-1/k}\) for \(x\ge 0\) with \(k>0\), density \(f(x)=\sigma^{-1}(1+kx/\sigma)^{-1/k-1}\). To read off the tail behavior, factor out the dominant power inside the bracket: \(1+kx/\sigma=(kx/\sigma)\bigl(1+\sigma/(kx)\bigr)\), so \[ f(x)=\sigma^{-1}\Bigl(\tfrac{kx}{\sigma}\Bigr)^{-1/k-1}\Bigl(1+\tfrac{\sigma}{kx}\Bigr)^{-1/k-1} \xrightarrow[x\to\infty]{}\ \sigma^{-1}\Bigl(\tfrac{k}{\sigma}\Bigr)^{-1/k-1}x^{-1/k-1}=C\,x^{-1/k-1}, \] since \((1+\sigma/(kx))^{-1/k-1}\to 1\), with \(C=\sigma^{-1}(k/\sigma)^{-1/k-1}>0\). Hence \(x^m f(x)\sim C x^{m-1/k-1}\). The tail integral \(\int^{\infty}x^{m-1/k-1}\,\mathrm dx\) converges if and only if the exponent is below \(-1\), that is \(m-1/k-1<-1\), that is \(m<1/k\). Hence \(\mathbb E[X^m]<\infty\) if and only if \(m<1/k\) (Lemma 24.7). Setting \(m=1\) gives a finite mean iff \(k<1\); setting \(m=2\) gives a finite variance iff \(k<1/2\).

Relation to PSIS-LOO reliability. The importance ratios \(r_i\) in PSIS-LOO have a generalized Pareto upper tail with estimated shape \(\widehat k_i\). By the moment thresholds: if \(\widehat k_i<\tfrac12\) the ratios have finite variance, so the self-normalized importance estimator obeys a \(\sqrt S\) central limit theorem and the estimate is reliable; if \(\tfrac12\le\widehat k_i<1\) the mean is finite but the variance is infinite, so the law of large numbers still gives convergence to the right value but without a Gaussian error rate; if \(\widehat k_i\ge 1\) even the mean is infinite and the estimate does not converge, so the point must be evaluated by refitting. This is Theorem 24.4.

Why the alarm is at \(0.7\) rather than \(0.5\). The theoretical variance-finiteness threshold is \(k=\tfrac12\), but for \(\tfrac12\le k<1\) the estimator still converges to the correct value by the law of large numbers, and PSIS replaces the offending upper tail of the ratios by a smooth generalized Pareto fit, truncating and stabilizing the largest weights. Empirically this smoothing keeps the bias and variance small enough to be usable up to about \(\widehat k\approx 0.7\). The operational alarm is therefore set at the practical \(0.7\), where the smoothed estimate begins to degrade, rather than at the theoretical variance threshold \(0.5\), below which a central limit theorem holds but above which the estimate is often still acceptable. \(\square\)

F.25 Chapter 25: General Likelihoods: Survival and Ordered Categorical Outcomes

F.25.1 Solution to Exercise 25.7: The marginal hazard ratio drifts away from the conditional one

This Chapter 25 exercise shows that under a mixed (frailty) population the marginal hazard ratio departs from the constant conditional hazard ratio \(\psi\) over time, because the two arms shed high-frailty individuals from their risk sets at different rates.

Strategy. The marginal hazard ratio is \(\psi\) times a ratio of survivor-weighted frailty means; the key identity is that the derivative of such a tilted mean is a negative variance, which makes the mean strictly decreasing and forces the ratio above one.

Uses. The time-varying marginal hazard ratio theorem Theorem 25.3 and the survivor-weighted mean Equation 25.9.

A word on the survival vocabulary first. Among individuals still alive (still at risk) at time \(s\), the frailty distribution is no longer the population one: high-frailty individuals die faster, so the survivors are enriched for low frailty. Because surviving to \(s\) under hazard \(\rho\) has probability \(e^{-\rho s}\), the risk-set or survivor-weighted mean of \(\rho\) at time \(s\) tilts the population law by the weight \(e^{-\rho s}\), giving \(\mathbb E[\rho\mid\text{alive at }s]=\mathbb E[\rho\,e^{-\rho s}]/\mathbb E[e^{-\rho s}]\). This tilted average is what the notation \(\mathbb E_{\tilde f_a(\cdot\mid s)}[\rho]\) denotes.

Binary frailty \(\rho(X)\in\{\rho_0,\rho_1\}\) with \(\rho_0<\rho_1\), \(P(X=1)=q\), constant baseline \(h_{a0}(s)=1\), constant conditional hazard ratio \(\psi\in(0,1)\). The control survival is \(S_a(s\mid x)=e^{-\rho(x)s}\) and the treated survival \(S_b(s\mid x)=e^{-\psi\rho(x)s}\). By Theorem 25.3, \(\mathrm{mHR}(s)=\psi\,R(s)\) with \(R(s)=\mathbb E_{\tilde f_b(\cdot\mid s)}[\rho]/\mathbb E_{\tilde f_a(\cdot\mid s)}[\rho]\), the ratio of survivor-weighted frailty means Equation 25.9.

Closed form. The risk-set (survivor-weighted) mean of \(\rho\) under arm \(a\) at time \(s\) is \(g(s):=\mathbb E_{\tilde f_a(\cdot\mid s)}[\rho]=\dfrac{(1-q)\rho_0 e^{-\rho_0 s}+q\rho_1 e^{-\rho_1 s}}{(1-q)e^{-\rho_0 s}+q e^{-\rho_1 s}}\), and under arm \(b\) the survival weights are \(e^{-\psi\rho(x)s}\), replacing \(s\) by \(\psi s\): \(\mathbb E_{\tilde f_b(\cdot\mid s)}[\rho]=g(\psi s)\). Hence \[ R(s)=\frac{g(\psi s)}{g(s)},\qquad g(c)=\frac{(1-q)\rho_0 e^{-\rho_0 c}+q\rho_1 e^{-\rho_1 c}}{(1-q)e^{-\rho_0 c}+q e^{-\rho_1 c}}, \qquad \mathrm{mHR}(s)=\psi\,\frac{g(\psi s)}{g(s)}. \]

Monotonicity of \(g\) and the bound \(R(s)\ge 1\). The map \(g(c)=\mathbb E[\rho\,e^{-\rho c}]/\mathbb E[e^{-\rho c}]\) is the mean of \(\rho\) under the tilt \(\propto e^{-\rho c}\). Its derivative is the key identity, so we derive it. Write \(N(c)=\mathbb E[\rho\,e^{-\rho c}]\) and \(D(c)=\mathbb E[e^{-\rho c}]\), so \(g=N/D\), and note \(N'(c)=-\mathbb E[\rho^2 e^{-\rho c}]\) and \(D'(c)=-\mathbb E[\rho\,e^{-\rho c}]=-N(c)\). By the quotient rule, \[ g'(c)=\frac{N'D-ND'}{D^2}=\frac{-\mathbb E[\rho^2 e^{-\rho c}]\,D+N^2}{D^2} =-\frac{\mathbb E[\rho^2 e^{-\rho c}]}{D}+\Bigl(\frac{N}{D}\Bigr)^2 =-\bigl(\mathbb E_c[\rho^2]-\mathbb E_c[\rho]^2\bigr)=-\operatorname{Var}_c(\rho), \] where \(\mathbb E_c[\cdot]\) is expectation under the tilt (so \(\mathbb E_c[\rho^2]=\mathbb E[\rho^2 e^{-\rho c}]/D\) and \(\mathbb E_c[\rho]=N/D\)). Thus \(g'(c)=-\operatorname{Var}_c(\rho)\le 0\), strictly negative for \(c\in(0,\infty)\) because \(\rho\) is non-degenerate (\(\rho_0\neq\rho_1\), \(q\in(0,1)\)). So \(g\) is strictly decreasing. Since \(\psi<1\), \(\psi s<s\), hence \(g(\psi s)\ge g(s)\) and \(R(s)\ge 1\), that is \(\mathrm{mHR}(s)\ge\psi\). At \(s=0\) both tilts equal the population law, \(g(0)=(1-q)\rho_0+q\rho_1\), so \(R(0)=1\); as \(s\to\infty\) the risk set in both arms is dominated by the low-frailty stratum, \(g(c)\to\rho_0\), so \(R(\infty)=\rho_0/\rho_0=1\). For every \(s\in(0,\infty)\) the strict monotonicity gives \(g(\psi s)>g(s)\), so \(R(s)>1\) and \(\mathrm{mHR}(s)>\psi\). Thus \(\mathrm{mHR}(s)>\psi\) on \((0,\infty)\) with \(\lim_{s\to 0^+}\mathrm{mHR}(s)=\lim_{s\to\infty}\mathrm{mHR}(s)=\psi\).

Time of maximal attenuation. Setting \(\tfrac{\mathrm d}{\mathrm ds}R(s)=0\) gives, since \(R=g(\psi s)/g(s)\), \[ \psi\,\frac{g'(\psi s)}{g(\psi s)}=\frac{g'(s)}{g(s)}, \] that is, the survivor-selection rate \(\mathrm d\log g/\mathrm dc=-\operatorname{Var}_c(\rho)/g(c)\), scaled by \(\psi\) in the treated arm, matches that of the control arm. At this \(s\) the differential rate at which the two arms shed high-frailty individuals from their risk sets is balanced; it is the moment of maximal survivor selection, where the frailty composition of the two risk sets differs most and the marginal hazard ratio departs furthest from the conditional \(\psi\). With \(\rho_0=1,\rho_1=3,q=\tfrac12, \psi=\tfrac12\) this reproduces the table of Example 25.3, for instance \(\mathrm{mHR}(1)=\tfrac12\,g(0.5)/g(1)=\tfrac12(1.537883/1.238399)=0.62092\). \(\square\)

F.25.2 Solution to Exercise 25.8: RMST is collapsible but not transportable

This Chapter 25 exercise shows that the restricted mean survival time (RMST) contrast is collapsible within each population yet differs across populations with different covariate mixes, so collapsibility does not imply transportability.

Uses. The RMST collapsibility theorem Theorem 25.4 and the shared-hazard (SEMA) assumption.

Two populations \(\mathcal P,\mathcal P^{*}\) share the binary covariate \(X\) but have prevalences \(q\neq q^{*}\), with the exponential proportional-hazards model of Example 25.3: conditional control hazard \(\lambda_x\) and treated hazard \(\psi\lambda_x\), shared across populations (SEMA on the log-hazard scale). The conditional RMST at horizon \(\tau\) is \(\operatorname{RMST}_t(\tau\mid x)=(1-e^{-\lambda_x^{(t)}\tau})/\lambda_x^{(t)}\) with \(\lambda_x^{(C)}=\lambda_x\), \(\lambda_x^{(A)}=\psi\lambda_x\).

Collapsibility within a population. By Theorem 25.4 the marginal RMST contrast is the covariate average of the conditional contrasts, \(\Delta\!\operatorname{RMST}_{AC}(\tau\mid\mathcal P)=\mathbb E_{X\sim\mathcal P}[d(X)]\) with \(d(x)=\operatorname{RMST}_A(\tau\mid x)-\operatorname{RMST}_C(\tau\mid x)\). For binary \(X\) this is \((1-q)d(0)+q\,d(1)\) in \(\mathcal P\).

Difference between populations. Since the conditional contrasts \(d(0),d(1)\) are the same functions in both populations (shared hazards), \[ \Delta\!\operatorname{RMST}_{AC}(\tau\mid\mathcal P)-\Delta\!\operatorname{RMST}_{AC}(\tau\mid\mathcal P^{*}) =\bigl[(1-q)d(0)+q\,d(1)\bigr]-\bigl[(1-q^{*})d(0)+q^{*}d(1)\bigr] =(q-q^{*})\bigl[d(1)-d(0)\bigr]. \] This is nonzero whenever \(d(1)\neq d(0)\), that is whenever the conditional RMST contrast is modified by \(X\). It is, because RMST is a nonlinear function of the hazard: with \(\lambda_0=1,\lambda_1=3,\psi=\tfrac12,\tau=2\), \[ d(0)=\operatorname{RMST}_A(2\mid 0)-\operatorname{RMST}_C(2\mid 0)=1.2642-0.8647=0.3995, \qquad d(1)=0.6335-0.3325=0.3010, \] so \(d(1)-d(0)=-0.0985\) and the two populations’ RMST contrasts differ by \((q-q^{*})(-0.0985)\); for example \(q=\tfrac12,q^{*}=0.3\) gives a difference of \(0.2(-0.0985)=-0.0197\) years.

Conclusion. RMST is collapsible (the within-population marginal equals the average conditional) yet not transportable (the marginal differs across populations with different covariate mixes), because the shift in the covariate distribution acts on the effect-modified conditional RMST contrast. This is unlike the conditional log-hazard model, which is the single shared number \(\log\psi\) in both populations by assumption, hence trivially population-free. The distinction is the scale-misalignment remark following Theorem 25.4: collapsibility is a within-population identity and does not confer the cross-population identity of transportability, exactly the phenomenon carried into Chapter 26 (Theorem 26.4). \(\square\)

Common mistake (collapsibility versus transportability). These two properties are easy to conflate but are different guarantees. Collapsibility is a within-population statement: the marginal effect equals a straight average of the conditional effects, with no Jensen gap (risk difference and RMST are collapsible; the odds ratio and hazard ratio are not). Transportability is a between-population statement: the same effect number holds in a new population with a different covariate distribution. A measure can be collapsible yet not transportable, as RMST is here, because averaging an effect-modified conditional contrast over a different covariate mix changes the marginal value. Do not treat “collapsible” as a license to carry a marginal effect from a trial population to a decision population; only a conditional effect that is genuinely constant across the relevant covariates, or an explicit standardization to the target, transports.

F.26 Chapter 26: Multilevel Unanchored Meta-Regression

F.26.1 Solution to Exercise 26.5: Identifying the comparator intercept in unanchored ML-UMR

This Chapter 26 exercise solves the two-point aggregation equation for the comparator-arm intercept \(\alpha_B\) from an observed aggregate response, then computes the conditional and marginal log odds ratios, exhibiting the non-collapsibility gap.

Uses. The aggregation-equation monotonicity Lemma 23.2 and the shared prognostic-factor assumption (SPFA).

Logit link, SPFA prognostic slope \(\beta_{\mathrm{PF}}=0.8\), single binary covariate with comparator prevalence \(\bar x_B=P(X=1)=0.4\), observed comparator response proportion \(\theta_{\bullet B}^{\mathrm{obs}}=0.30\). The two-point aggregation equation Equation 26.9 is \[ \Theta_B(\alpha_B)=(1-0.4)\operatorname{expit}(\alpha_B)+0.4\operatorname{expit}(\alpha_B+0.8)=0.30 . \]

Solving by bracketing. \(\Theta_B\) is strictly increasing in \(\alpha_B\) (Lemma 23.2), so the root is unique. Evaluating: \[ \Theta_B(-1.3)=0.6(0.21417)+0.4(0.37754)=0.27952,\qquad \Theta_B(-1.2)=0.6(0.23148)+0.4(0.40131)=0.29941 . \] Since \(\Theta_B(-1.3)<0.30<\Theta_B(-1.19)=0.30144\), linear interpolation between \(-1.20\) and \(-1.19\) places the root at \(\alpha_B\approx-1.20\) (to two decimals; more precisely \(\alpha_B\approx-1.197\)).

Comparator-population marginal response. By construction \(\Theta_B(\alpha_B)=0.30\), so the comparator-population marginal response under arm \(B\) is \(0.30\), as required.

Log odds ratios with the index fit \(\alpha_A=-0.2\). Under SPFA the conditional log odds ratio of \(A\) versus \(B\) is the covariate-free constant \[ \Delta^{\mathrm{Cond}}_{AB}=\alpha_A-\alpha_B=-0.2-(-1.197)=0.997\approx 1.00 . \] The comparator-population marginal response under arm \(A\) is \[ \Theta_A=(1-0.4)\operatorname{expit}(\alpha_A)+0.4\operatorname{expit}(\alpha_A+0.8) =0.6\operatorname{expit}(-0.2)+0.4\operatorname{expit}(0.6)=0.6(0.45017)+0.4(0.64566)=0.52836, \] so the comparator-population marginal log odds ratio is \[ \Delta^{\mathrm{Marg}}_{AB}=\operatorname{logit}(0.52836)-\operatorname{logit}(0.30)=0.11352-(-0.84730)=0.96082 . \] The marginal log odds ratio \(0.96\) is smaller in magnitude than the conditional \(1.00\), a non-collapsibility gap of about \(0.04\): averaging the logistic model over the comparator covariate distribution attenuates the contrast toward the null, exactly as in the anchored ML-NMR computation. \(\square\)

F.26.2 Solution to Exercise 26.10: Population-specific RMST from an unanchored ML-UMR survival fit

This Chapter 26 exercise computes the RMST contrast at two target covariate distributions from an exponential proportional-hazards fit, again showing the RMST contrast is collapsible within a population but differs between populations, while the conditional log hazard ratio is a single population-free scalar.

Uses. The RMST collapsibility theorem Theorem 25.4 and the ML-UMR time-to-event result Theorem 26.4.

For the exponential proportional-hazards survival coda, \(h_k(t\mid x)=\lambda_k e^{\beta x}\) with \(\beta=\log 2\), \(\lambda_A=0.10\), \(\lambda_B=0.15\), so \[ \operatorname{RMST}_k(\tau\mid x)=\frac{1-\exp(-\lambda_k e^{\beta x}\tau)}{\lambda_k e^{\beta x}}. \] The population RMST contrast at \(P(X=1)=\pi^{*}\) is, by collapsibility (Theorem 25.4), the two-point average \[ \Delta\!\operatorname{RMST}_{AB}(\tau\mid\mathcal P^{*})=(1-\pi^{*})\,d(0)+\pi^{*}d(1), \qquad d(x)=\operatorname{RMST}_A(\tau\mid x)-\operatorname{RMST}_B(\tau\mid x). \]

Conditional contrasts at \(\tau=10\). With \(e^{\beta\cdot 0}=1\), \(e^{\beta\cdot 1}=2\), \[ \begin{aligned} \operatorname{RMST}_A(10\mid 0)&=\tfrac{1-e^{-1}}{0.10}=6.32121, & \operatorname{RMST}_A(10\mid 1)&=\tfrac{1-e^{-2}}{0.20}=4.32332,\\ \operatorname{RMST}_B(10\mid 0)&=\tfrac{1-e^{-1.5}}{0.15}=5.17913, & \operatorname{RMST}_B(10\mid 1)&=\tfrac{1-e^{-3}}{0.30}=3.16738, \end{aligned} \] so \(d(0)=6.32121-5.17913=1.14208\) and \(d(1)=4.32332-3.16738=1.15594\), with \(d(1)-d(0)=0.01387\).

Collapsible but not transportable. Within each population the marginal contrast is the average of these conditional contrasts (collapsibility). Across populations, \[ \Delta\!\operatorname{RMST}_{AB}(10\mid \pi^{*}=0.5)=0.5(1.14208)+0.5(1.15594)=1.14901, \qquad \Delta\!\operatorname{RMST}_{AB}(10\mid \pi^{*}=0.3)=0.7(1.14208)+0.3(1.15594)=1.14624, \] which differ by \((0.5-0.3)(d(1)-d(0))=0.2(0.01387)=0.00277\) years. So the RMST contrast is collapsible within each population yet differs between the two populations: collapsibility does not imply transportability. By contrast the conditional log hazard ratio of \(A\) versus \(B\) is the single number \(\log(\lambda_B/\lambda_A)=\log(0.15/0.10)=\log 1.5\) in every population, covariate-free and time-invariant. This is Theorem 26.4(e): an unanchored ML-UMR analysis can report the population-specific RMST contrast at any target \(\pi^{*}\), but only the conditional log hazard ratio is a single population-free scalar. \(\square\)

F.27 Chapter 27: Aggregate-Level Covariate-Distance Matching

F.27.1 Solution to Exercise 27.6: Mahalanobis distance is affine-invariant, Euclidean is not

This Chapter 27 exercise proves the Mahalanobis distance between covariate means is unchanged by any invertible affine reparameterization of the covariates, whereas the standardized Euclidean distance can even reverse rankings, because it discards the off-diagonal covariance.

Uses. The Mahalanobis distance definition Definition 27.3 and the whitening interpretation (Mahalanobis is Euclidean distance after standardizing by the covariance).

Let the covariates be reparameterized by an invertible affine map \(\mathbf x\mapsto A\mathbf x+\mathbf c\), with the covariance transformed accordingly to \(A\boldsymbol\Sigma A^\top\). A gap between two means \(\boldsymbol\delta=\bar{\mathbf x}_0-\bar{\mathbf x}_j\) transforms to \(A\boldsymbol\delta\) (the constant \(\mathbf c\) cancels in the difference). The Mahalanobis distance (Definition 27.3) becomes \[ d_M^2=(A\boldsymbol\delta)^\top(A\boldsymbol\Sigma A^\top)^{-1}(A\boldsymbol\delta) =\boldsymbol\delta^\top A^\top A^{-\top}\boldsymbol\Sigma^{-1}A^{-1}A\boldsymbol\delta =\boldsymbol\delta^\top\boldsymbol\Sigma^{-1}\boldsymbol\delta, \] using \((A\boldsymbol\Sigma A^\top)^{-1}=A^{-\top}\boldsymbol\Sigma^{-1}A^{-1}\) and \(A^\top A^{-\top}=I\), \(A^{-1}A=I\). So the Mahalanobis distance is unchanged under every invertible affine reparameterization. The standardized Euclidean distance uses only the diagonal of \(\boldsymbol\Sigma\) as scales; under \(\mathbf x\mapsto A\mathbf x+\mathbf c\) the scales become the diagonal of \(A\boldsymbol\Sigma A^\top\), which for a nondiagonal \(A\) mixes the coordinates and is not the coordinatewise rescaling of the original scales. For diagonal \(A=\operatorname{diag}(a_1,\dots,a_p)\) each coordinate is scaled by \(a_\ell\) and its standard deviation by \(|a_\ell|\), so the ratio \(\delta_\ell/s_\ell\) is unchanged in magnitude and the standardized Euclidean distance is invariant; for nondiagonal \(A\) it generally is not.

A reversing example. Work in \(\mathbb R^2\) with original covariance \(\boldsymbol\Sigma=I\), so initially Mahalanobis and standardized Euclidean both equal ordinary Euclidean distance. Take the two candidate gaps \(\boldsymbol\delta_a=(1.4,0)\) and \(\boldsymbol\delta_b=(0,1)\), so Mahalanobis ranks \(a\) as the more distant (\(1.4>1\)). Apply the nondiagonal shear \(A=\begin{pmatrix}1&1\\0&1\end{pmatrix}\), giving \(\boldsymbol\delta_a'=A\boldsymbol\delta_a=(1.4,0)\), \(\boldsymbol\delta_b'=A\boldsymbol\delta_b=(1,1)\), and covariance \(\boldsymbol\Sigma'=AA^\top=\begin{pmatrix}2&1\\1&1\end{pmatrix}\) with \((\boldsymbol\Sigma')^{-1}=\begin{pmatrix}1&-1\\-1&2\end{pmatrix}\). The Mahalanobis distances are preserved, \(d_M(\boldsymbol\delta_a')^2=(1.4,0)(\boldsymbol\Sigma')^{-1}(1.4,0)^\top=1.96\) and \(d_M(\boldsymbol\delta_b')^2=(1,1)(\boldsymbol\Sigma')^{-1}(1,1)^\top=1\), so \(a\) remains more distant (\(1.4>1\)). But the standardized Euclidean distance now uses the new scales \(\operatorname{diag}(\boldsymbol\Sigma')=(2,1)\): \[ d_{SE}(\boldsymbol\delta_a')^2=\frac{1.4^2}{2}+\frac{0^2}{1}=0.98, \qquad d_{SE}(\boldsymbol\delta_b')^2=\frac{1^2}{2}+\frac{1^2}{1}=1.5, \] so the standardized Euclidean ranking is reversed: it now calls \(b\) the more distant candidate (\(\sqrt{1.5}>\sqrt{0.98}\)), whereas before the reparameterization it agreed with Mahalanobis that \(a\) was farther. The Mahalanobis ranking is preserved throughout. This is the precise sense in which Mahalanobis distance is the coordinate-free member of the family: it depends only on the geometry of the covariate cloud (it is the Euclidean distance after whitening), so any invertible affine relabeling leaves it and its rankings intact, while standardized Euclidean distance, which discards the off-diagonal covariance, is tied to the chosen coordinates. \(\square\)

F.27.2 Solution to Exercise 27.7: A small moment distance does not certify an unbiased comparison

This Chapter 27 exercise builds two covariate distributions with identical first and second moments but different target effects under a nonlinear (quartic) contrast, showing that matching a finite set of moments leaves a residual bias for contrasts whose curvature reaches beyond those moments.

Uses. The diagnostic-not-validity proposition Proposition 27.4 and the second-moment MAIC balancing of Chapter 18.

Construct two scalar covariate distributions with identical first and second moments but different target effects under a nonlinear contrast, strengthening Proposition 27.4. Let \(\mathcal P_0\) put mass \(\tfrac12\) at each of \(\pm 1\), and \(\mathcal P_1\) put mass \(\tfrac34\) at \(0\) and mass \(\tfrac18\) at each of \(\pm 2\).

Equal first two moments. Both are symmetric about \(0\), so each has mean \(0\). Their variances are \[ \operatorname{Var}_{\mathcal P_0}=\tfrac12(1)+\tfrac12(1)=1, \qquad \operatorname{Var}_{\mathcal P_1}=\tfrac34(0)+\tfrac18(4)+\tfrac18(4)=1 . \] So any diagnostic built from the first two moments (mean and variance, hence standardized Euclidean or Mahalanobis distance between the two distributions’ moment summaries) reports zero distance.

Different target effect under a quartic contrast. Take \(\tau(x)=x^4\). The target effects are \[ \mathbb E_{\mathcal P_0}[\tau(X)]=\tfrac12(1)+\tfrac12(1)=1, \qquad \mathbb E_{\mathcal P_1}[\tau(X)]=\tfrac34(0)+\tfrac18(16)+\tfrac18(16)=4 . \] The two target effects are \(1\) and \(4\), so the bias from treating the two populations as exchangeable under this contrast is \(4-1=3\), despite zero distance under any first-two-moment diagnostic.

Implication for MAIC second-moment balancing. Matching through any fixed finite set of moments controls only the corresponding low-order features of the covariate distribution. A sufficiently nonlinear contrast (here a quartic, sensitive to the fourth moment) responds to features the matched moments do not constrain, leaving a residual bias. Balancing means and variances is therefore not enough in general: MAIC weights that equate the first two moments still leave a target-effect discrepancy for contrasts whose curvature reaches beyond those moments, which is why a small moment distance is a diagnostic of similarity, not a certificate of validity. Concretely, a regulator reviewing a submitted MAIC should not accept a small Mahalanobis distance, or a report that “all covariate means and variances are balanced,” as evidence that the indirect comparison is unbiased. The example above has zero first-two-moment distance yet a target-effect bias of \(3\); the missing ingredient is the shape of the outcome-covariate relationship, which no fixed number of matched moments can guarantee. The right posture is to treat moment balance as necessary housekeeping and to argue separately, from subject-matter knowledge, that the outcome model is not so nonlinear in the unbalanced higher moments that a residual bias survives. \(\square\)

F.28 Chapter 28: Quantitative Bias Analysis for Indirect Comparisons

F.28.1 Solution to Exercise 28.6: The E-value for an indirect comparison, and when it is infinite

This Chapter 28 exercise recomputes the sensitivity curve and E-value for an indirect comparison under a stronger unreported covariate, and shows the E-value can be infinite, a certificate that no single-covariate imbalance can explain away the observed effect.

Strategy. The adjusted effect is linear in the unreported covariate’s mean, \(\Delta(m)=K+\beta_U m\); the whole analysis rests on the constant \(K\), which is a Gaussian-MGF integral of the log-linear model. Derive \(K\) once, then read the bias factor and E-value off the standard formulas.

Uses. The indirect-comparison E-value proposition Proposition 28.1 and the lognormal (Gaussian moment-generating-function) mean of a log-linear model.

(a) Recomputed constants. First, where \(K\) comes from. The worked-example model is log-linear, \(\log\mu(R,U)=\beta_0+\beta_R R+\beta_U U\), with independent Gaussian covariates \(R\sim\mathcal N(m_R,\sigma_R^2)\) and \(U\sim\mathcal N(m,\sigma_U^2)\). Taking the marginal mean by the Gaussian moment generating function (\(\mathbb E[e^{tZ}]=e^{tm_Z+t^2\sigma_Z^2/2}\) for \(Z\sim\mathcal N(m_Z,\sigma_Z^2)\)), \[ \log\mathbb E[Y]=\beta_0+\beta_R m_R+\tfrac12\beta_R^2\sigma_R^2+\beta_U m+\tfrac12\beta_U^2\sigma_U^2 . \] The adjusted effect \(\Delta(m)\) is this marginal log-mean minus the observed baseline \(\log\mu_0\), so grouping every term that does not involve \(m\) into a single constant, \[ \Delta(m)=\underbrace{\bigl(\beta_0+\beta_R m_R+\tfrac12\beta_R^2\sigma_R^2+\tfrac12\beta_U^2\sigma_U^2-\log\mu_0\bigr)}_{K}+\beta_U m, \] which is the sensitivity curve and defines \(K\). Now use \(\beta_0=-1.0\), \(\beta_R=0.4\), \(m_R=1.0\), \(\sigma_R=0.5\), \(\sigma_U=1.0\), \(\mu_0=0.47\), \(\bar x_U=2.0\), but now a stronger unreported covariate \(\beta_U=-0.8\): \[ K=\beta_0+\beta_R m_R+\tfrac12\beta_R^2\sigma_R^2+\tfrac12\beta_U^2\sigma_U^2-\log\mu_0 =-1.0+0.4+\tfrac12(0.16)(0.25)+\tfrac12(0.64)(1)-\log(0.47), \] that is \(K=-1.0+0.4+0.02+0.32+0.75502=0.49502\). The sensitivity curve is \(\Delta(m)=K+\beta_U m=0.49502-0.8m\), so the adjusted effect at the IPD value is \(\Delta(2.0)=0.49502-1.6=-1.10498\). The observed bias factor is \(\mathrm{BF}_{\mathrm{obs}}=\exp(|\Delta(2.0)|)=\exp(1.10498)=3.0192\), and the outcome association is \(\mathrm{RR}_{UD}=\exp(|\beta_U|)=\exp(0.8)=2.2255\).

(b) The E-value. By Proposition 28.1 the indirect-comparison E-value is finite only when \(1<\mathrm{BF}_{\mathrm{obs}}<\mathrm{RR}_{UD}\). Here \(\mathrm{RR}_{UD}=2.2255<3.0192=\mathrm{BF}_{\mathrm{obs}}\), so \(\mathrm{RR}_{UD}\le\mathrm{BF}_{\mathrm{obs}}\) and \(E_{\mathrm{ITC}}=+\infty\): no across-arm imbalance of the single covariate, however extreme, can manufacture a bias factor as large as the observed one, given that the outcome association is pinned at \(\mathrm{RR}_{UD}=2.2255\). This is a robustness certificate. As a check of the finite formula on the artificial pair \(\mathrm{BF}_{\mathrm{obs}}=1.5\), \(\mathrm{RR}_{UD}=2.5\) (where \(1<1.5<2.5\)), \[ E_{\mathrm{ITC}}=\frac{\mathrm{BF}_{\mathrm{obs}}(\mathrm{RR}_{UD}-1)}{\mathrm{RR}_{UD}-\mathrm{BF}_{\mathrm{obs}}} =\frac{1.5(2.5-1)}{2.5-1.5}=\frac{2.25}{1}=2.25 . \]

(c) The certificate. For fixed \(\mathrm{RR}_{UD}=y>1\), the bias factor \(\mathrm{BF}_U(x,y)=\dfrac{xy}{x+y-1}=\dfrac{y}{1+(y-1)/x}\) as a function of \(x=\mathrm{RR}_{EU}\) has derivative \[ \frac{\partial}{\partial x}\frac{xy}{x+y-1}=\frac{y(x+y-1)-xy}{(x+y-1)^2}=\frac{y(y-1)}{(x+y-1)^2}>0, \] so it is strictly increasing in \(x\). Moreover \(1+(y-1)/x>1\) for all finite \(x>0\), so \(\mathrm{BF}_U(x,y)<y=\mathrm{RR}_{UD}\), and as \(x\to\infty\), \((y-1)/x\to 0\) and \(\mathrm{BF}_U(x,y)\uparrow y\). Hence the bias factor is strictly increasing in \(\mathrm{RR}_{EU}\) and bounded above by \(\mathrm{RR}_{UD}\), the bound approached but never attained. Consequently, when \(\mathrm{RR}_{UD}\le\mathrm{BF}_{\mathrm{obs}}\), the target is unreachable for every finite imbalance and the result cannot be explained away by imbalance of this one covariate, which is the meaning of \(E_{\mathrm{ITC}}=+\infty\). \(\square\)

F.28.2 Solution to Exercise 28.8: NORTA marginal recovery, copula non-uniqueness, and Pearson drift

This Chapter 28 exercise checks three facts about the NORTA (normal-to-anything) construction: (a) it recovers a discrete Bernoulli margin exactly, (b) for discrete margins the copula is not uniquely determined by the joint law (Sklar non-uniqueness), and (c) for continuous margins the covariate Pearson correlation need not equal the latent normal correlation.

Strategy. Separate the two questions the exercise mixes together: marginal recovery uses only that \(\Phi(Z)\) is uniform, while the dependence questions are about the copula. Keep the identification of the Gaussian parameter distinct from the Sklar non-uniqueness of the copula function.

Uses. The NORTA/Gaussian-copula theorem Theorem 28.2 and the Gaussian moment generating function.

Some copula vocabulary, since the exercise leans on it. The Frechet (Frechet-Hoeffding) bounds say every joint distribution with given margins lies between two extremes: the countermonotone coupling (perfect negative dependence, each variable a decreasing function of the other) below and the comonotone coupling (perfect positive dependence) above. Sklar’s theorem guarantees a copula \(C\) with \(F(x_1,x_2)=C(F_1(x_1),F_2(x_2))\); this \(C\) is unique on the set \(\operatorname{Range}(F_1)\times\operatorname{Range}(F_2)\), which is all of \([0,1]^2\) for continuous margins (so \(C\) is unique) but only a coarse grid for discrete margins (so \(C\) is free between grid points). That grid-only determination is the Sklar non-uniqueness of part (b).

Let \(F\) be the \(\mathrm{Ber}(\pi)\) distribution function and \(Z\sim\mathcal N(0,1)\).

(a) Discrete margin recovery. The generalized inverse of the Bernoulli distribution function is \[ F^{-1}(u)=\inf\{x:F(x)\ge u\}=\begin{cases}0,& 0<u\le 1-\pi,\\ 1,& 1-\pi<u\le 1.\end{cases} \] Since \(\Phi(Z)\sim\mathrm{Unif}(0,1)\) (the probability integral transform), \(X=F^{-1}(\Phi(Z))\) equals \(1\) exactly when \(\Phi(Z)>1-\pi\), an event of probability \(1-(1-\pi)=\pi\), and equals \(0\) otherwise. Hence \(X\sim\mathrm{Ber}(\pi)\), confirming Theorem 28.2(1) for a discrete margin, with no continuity needed.

(b) Non-uniqueness of the copula for discrete margins. It is important to separate two claims that are easy to run together. Take two such Bernoulli variables \(X_1=F^{-1}(\Phi(Z_1))\), \(X_2=F^{-1}(\Phi(Z_2))\) with \((Z_1,Z_2)\) bivariate normal of correlation \(\rho^N\). The joint law of \((X_1,X_2)\) is determined by the single cell probability \[ P(X_1=1,X_2=1)=P\bigl(\Phi(Z_1)>1-\pi,\ \Phi(Z_2)>1-\pi\bigr)=P(Z_1>z_\pi,Z_2>z_\pi),\qquad z_\pi=\Phi^{-1}(1-\pi), \] which is a continuous, strictly increasing function of \(\rho^N\) on \([-1,1]\) for fixed \(\pi\in(0,1)\).

First claim (the Gaussian parameter is identified). Because this cell probability is strictly increasing in \(\rho^N\), it is injective: distinct latent correlations \(\rho^N\) give distinct cell probabilities and hence distinct bivariate Bernoulli laws. So within the Gaussian (NORTA) family the parameter \(\rho^N\) is recovered from the joint law; there is no ambiguity of the Gaussian parameter in the nondegenerate case.

Second claim (the copula function is not unique). The non-uniqueness the exercise is after is a different, Sklar-level statement, and it is not about \(\rho^N\) at all. Sklar’s theorem determines the copula only on \(\operatorname{Range}(F_1)\times\operatorname{Range}(F_2)\); for a Bernoulli margin that range is the finite set \(\{0,\,1-\pi,\,1\}\), so the copula \(C\) is pinned only at those grid points and is free to be interpolated in infinitely many ways between them. Many distinct copula functions, most of them not Gaussian at all, reproduce the same bivariate Bernoulli law, because a discrete margin can only “see” the copula on its coarse grid. That is the sense in which the copula is non-unique for discrete margins: not that two Gaussian parameters collide, but that the copula function is underdetermined once the margins are discrete. The range of attainable cell probabilities is itself limited by the Frechet bounds, with the comonotone and countermonotone couplings as the extremes; a target joint law outside that attainable interval has no Gaussian \(\rho^N\) realizing it. Marginal recovery in (a), by contrast, holds for every \(\rho^N\), because it used only that \(\Phi(Z_j)\) is uniform, irrespective of the dependence.

(c) Pearson correlation need not equal \(\rho^N\) for continuous margins. Take \(F_1=F_2\) a lognormal margin, \(X_j=\exp(\sigma Z_j)\) with \((Z_1,Z_2)\) standard bivariate normal of correlation \(\rho^N\). The degenerate value \(\rho^N=1\) is not informative here: then \(Z_1=Z_2\), so \(X_1=X_2\) and trivially \(\operatorname{corr}(X_1,X_2)=1=\rho^N\), which cannot exhibit strict inequality. So take a genuinely dependent case, \(\rho^N\in(-1,1)\), and compute the covariate Pearson correlation in closed form from the Gaussian moment generating function. With \(\mathbb E[e^{tZ}]=e^{t^2/2}\) for a standard normal and \(Z_1+Z_2\sim\mathcal N(0,\,2+2\rho^N)\), \[ \mathbb E[X_j]=e^{\sigma^2/2},\qquad \mathbb E[X_j^2]=e^{2\sigma^2},\qquad \mathbb E[X_1X_2]=\mathbb E\bigl[e^{\sigma(Z_1+Z_2)}\bigr]=e^{\sigma^2(1+\rho^N)} . \] Hence \(\operatorname{Var}(X_j)=e^{2\sigma^2}-e^{\sigma^2}=e^{\sigma^2}(e^{\sigma^2}-1)\) and \(\operatorname{Cov}(X_1,X_2)=e^{\sigma^2(1+\rho^N)}-e^{\sigma^2}=e^{\sigma^2}(e^{\sigma^2\rho^N}-1)\), so the \(e^{\sigma^2}\) factors cancel and \[ \operatorname{corr}(X_1,X_2)=\frac{e^{\sigma^2\rho^N}-1}{e^{\sigma^2}-1} . \] For \(\sigma=1\) and \(\rho^N=0.5\) this is \((e^{0.5}-1)/(e-1)=0.64872/1.71828=0.37752\neq 0.5\). The Pearson correlation of the lognormal covariates (\(0.3775\)) differs from the latent normal correlation (\(0.5\)), because the nonlinear exponential margin reshapes the dependence; only the rank correlation is preserved. This is exactly why \(\boldsymbol\Omega\) in the NORTA and Gaussian-copula construction is the correlation of the latent normals, not of the covariates. \(\square\)

F.29 Chapter 29: Simulation Study Theory

F.29.1 Solution to Exercise 29.7: Sample size, power, and the test-interval duality

This Chapter 29 exercise computes the standardized signal and the sample size for a target power, checks that the wrong-side rejection term is negligible, and shows the power calculation and the confidence-interval-width calculation are two readings of the same inequality.

Uses. The simulation power formula Theorem 29.3 and its exact two-term version Equation 29.17.

A study targets a true contrast \(\theta=0.4\) against \(\theta_0=0\) at level \(\alpha=0.05\), with standard error \(\sigma_n=c/\sqrt n\), \(c=2.0\).

(a) Standardized signal. By Theorem 29.3 the standardized signal is \[ \Delta_n=\frac{\theta-\theta_0}{\sigma_n}=\frac{0.4}{2.0/\sqrt n}=\frac{0.4}{2.0}\sqrt n=0.2\sqrt n . \]

(b) Sample size for power \(0.80\). Using the one-sided approximation \(\mathrm{Power}\approx\Phi(\Delta_n-z_{\alpha/2})\) with \(z_{\alpha/2}=z_{0.025}=1.95996\) and \(\Phi^{-1}(0.80)=0.84162\), set \(\Delta_n-1.95996=0.84162\), so \(\Delta_n=2.80158\). Then \(0.2\sqrt n=2.80158\), \(\sqrt n=14.0079\), and \(n=196.2\), rounded up to \(n=197\).

(c) The dropped term is negligible. The exact two-term power Equation 29.17 is \(\Phi(\Delta_n-z_{\alpha/2})+\Phi(-\Delta_n-z_{\alpha/2})\). At \(\Delta_n=2.80158\) the dropped term is \[ \Phi(-\Delta_n-z_{\alpha/2})=\Phi(-2.80158-1.95996)=\Phi(-4.7615)\approx 9.6\times 10^{-7}, \] below one part in a million, so the one-term and two-term power formulas agree to far more than the precision of any simulation; the second term is the probability of rejecting on the wrong side and is negligible for a signal this large.

(d) Test-interval duality. The Wald test rejects \(\theta_0\) exactly when \(\theta_0\) lies outside the interval \(\hat\theta_n\pm z_{\alpha/2}\sigma_n\). Power \(0.80\) to reject \(\theta_0=0\) when the truth is \(\theta=0.4\) is therefore the probability \(0.80\) that the \(95\%\) confidence interval of half-width \(z_{\alpha/2}\sigma_n\) excludes \(0\), which holds when \(|\theta-\theta_0|\ge(z_{\alpha/2}+z_{0.20})\sigma_n\), the same standardized-signal condition \(\Delta_n\ge z_{\alpha/2}+\Phi^{-1}(0.80)\) and hence the same \(n\). The power calculation and the confidence-interval-width calculation are two readings of one inequality. \(\square\)

F.29.2 Solution to Exercise 29.8: When balancing fewer covariates lowers mean squared error

This Chapter 29 exercise sets up the bias-variance trade-off of covariate balancing: dropping a covariate saves effective sample size (variance) at the cost of some bias, and balancing the smaller set wins on mean squared error exactly when that bias is below a variance-set threshold.

Uses. The bias-variance (mean-squared-error) decomposition Theorem 29.1 and the MAIC effective sample-size bound Theorem 18.6.

In the geometry of Example 29.1, compare two adjusted estimators: balancing only the effect modifier \(x_1\) leaves a small omitted-covariate bias \(\beta\) on a non-collapsible scale but retains effective sample size \(\mathrm{ESS}_1\), with variance \(\sigma^2 n/\mathrm{ESS}_1\); balancing both \(x_1\) and the prognostic \(x_2\) removes the bias but lowers the effective sample size to \(\mathrm{ESS}_2<\mathrm{ESS}_1\), with variance \(\sigma^2 n/\mathrm{ESS}_2\).

(a) Mean squared errors. By the bias-variance decomposition Theorem 29.1, \(\mathrm{MSE}=\mathrm{bias}^2+\mathrm{variance}\): \[ \mathrm{MSE}_1=\beta^2+\frac{\sigma^2 n}{\mathrm{ESS}_1}, \qquad \mathrm{MSE}_2=0+\frac{\sigma^2 n}{\mathrm{ESS}_2}. \]

(b) When balancing only \(x_1\) wins. \(\mathrm{MSE}_1<\mathrm{MSE}_2\) if and only if \[ \beta^2+\frac{\sigma^2 n}{\mathrm{ESS}_1}<\frac{\sigma^2 n}{\mathrm{ESS}_2}, \qquad\text{that is}\qquad \beta^2<\sigma^2 n\Bigl(\frac{1}{\mathrm{ESS}_2}-\frac{1}{\mathrm{ESS}_1}\Bigr) =\sigma^2 n\,\frac{\mathrm{ESS}_1-\mathrm{ESS}_2}{\mathrm{ESS}_1\,\mathrm{ESS}_2}. \] Equivalently \(|\beta|<\sigma\sqrt{n(\mathrm{ESS}_1-\mathrm{ESS}_2)/(\mathrm{ESS}_1\mathrm{ESS}_2)}\): the smaller-set estimator achieves the lower mean squared error exactly when its bias is below the threshold set by the variance saved through the larger effective sample size.

(c) Relation to ESS and the identity scale. Balancing the additional covariate \(x_2\) adds a moment constraint to the weights, which by Theorem 18.6 can only lower the effective sample size, \(\mathrm{ESS}_2\le\mathrm{ESS}_1\), raising the variance; this is the variance cost on the right-hand side of the condition. The trade-off in (b) is the bias-variance balance of the remark following Lemma 29.4: balancing a covariate is worthwhile only when the bias it removes exceeds the variance it costs. On the identity scale a purely prognostic covariate carries no bias, \(\beta=0\), because it cancels in the contrast (it shifts both arms equally); then the condition \(\beta^2=0<\sigma^2 n(\cdots)\) holds automatically, so balancing \(x_2\) removes no bias while strictly lowering the effective sample size, and it can only hurt. \(\square\)

F.30 Chapter 30: Machine-Checked Foundations: The Coq Formalization

F.30.1 Solution to Exercise 30.8: Why one Coq theorem needs no premise and another carries one

This Chapter 30 exercise contrasts two machine-checked results and explains why the discrete balancing theorem is proved outright while the continuous inverse-probability-weighting identity must carry a named, unproved premise.

A reminder on the tiers (for readers who jumped straight here). The formalization sorts theorems by how much they assume. A tier F (“fully proved”) theorem is proved from first principles inside Coq with no unproved input. A tier P (“proved given a premise”) theorem is a complete Coq proof of an implication whose hypothesis is a named premise: an abstract, explicitly stated proposition that the current libraries cannot yet discharge, made visible in the theorem’s own statement rather than hidden. Schematically the two shapes are

(* tier F: proved outright, no probabilistic hypothesis *)
Theorem propensity_score_is_balancing :
  forall x, balancing_property x.

(* tier P: a complete proof of an implication whose premise Hipw1 is stated, not proved *)
Theorem mu1_IPW_unbiased :
  Hipw1 -> estimator_IPW = mu1_true.

The point of the exercise is to explain why the first needs nothing to the left of its conclusion while the second must display Hipw1.

Contrast propensity_score_is_balancing, a tier-F result with no probabilistic premise, against mu1_IPW_unbiased, a tier-P result carrying the named premise Hipw1.

Why the discrete balancing theorem needs no premise. The Rosenbaum-Rubin balancing property is proved over a finite discrete model, a joint probability mass function over finite types, with conditional independence written in the cross-multiplied form Equation 30.1, \(q_{XYZ}(x,y,z)\,q_Z(z)=q_{XZ}(x,z)\,q_{YZ}(y,z)\). In this multiplicative form the statement needs no positivity hypothesis: when a conditioning mass is zero, both sides vanish by marginal monotonicity (every joint marginal is dominated by the conditioning mass), so the identity holds trivially at zero-mass covariate points rather than requiring a nonzero denominator. There is therefore no division to justify and no probabilistic fact to admit; the proof goes through outright in Coq for every covariate value, including those of measure zero, and the theorem is tier F.

Why the continuous IPW identity defers to Hipw1. The inverse-probability-weighting unbiasedness mu1_IPW_unbiased is a statement in the continuous kernel model, and its proof needs the tower property (law of total expectation) for conditional expectations under a continuous covariate law. As Section 30.9 explains, the development is built on mathcomp-analysis, which does not yet supply the conditional-expectation and disintegration library needed to discharge that step inside Coq. So rather than hide the gap, the tower-property consequence is admitted as the visible named premise Hipw1, and the theorem is tier P: a fully proved Coq object given Hipw1, but resting on a stated, unproved probabilistic input.

What would close the gap, and where the paper proof lives. Discharging Hipw1 inside Coq requires developing, on top of mathcomp-analysis, a measure-theoretic conditional-expectation theory with the tower property and a disintegration (regular conditional distribution) library for the kernel model. Until that exists, the paper proof of the premise is supplied by the potential-outcomes and adjustment chapters: the IPW unbiasedness identity is proved there (Theorem 10.3 in Chapter 10, resting on the identification results of Chapter 9), which is exactly the tower-property argument that Hipw1 abbreviates. \(\square\)

F.30.2 Solution to Exercise 30.9: The placeholder trap and statement faithfulness

This Chapter 30 exercise shows how a formalization can type-check yet certify nothing, if a modeling assumption is replaced by the trivially true proposition, and explains the structural policy that guards against it.

Uses. The statement-faithfulness guarantee Section 30.5 and the visible-premise policy Section 30.9.

Suppose a developer had defined a modeling assumption as the trivially true proposition, Definition conditional_constancy : Prop := True, instead of as an abstract named proposition with real content.

The placeholder trap. A “consistency theorem” of the shape conditional_constancy -> (estimator = truth) then unfolds to True -> (estimator = truth), which is logically equivalent to estimator = truth. If, as is typical of a placeholder development, the estimator and truth terms are wired to be definitionally or provably equal (for instance both reduce to the same standardized integral), the implication is discharged by intros; reflexivity with no mathematical content: the premise contributes nothing because every proof of True is vacuous, and the conclusion was already provable on its own. The theorem type-checks and the kernel certifies it, yet it certifies nothing about the substantive assumption its name advertises.

Which guarantee is violated. This is precisely the statement faithfulness gap of Section 30.5: the kernel checks that the proof proves the stated theorem, but it cannot check that the stated theorem is the one the mathematician meant. A formalization with an unintended hypothesis that renders the proposition vacuous type-checks just as readily as the intended one. Replacing a real modeling restriction by True substitutes a vacuous proposition for the intended conditional-constancy assumption, so the certified statement is no longer a faithful formalization of “under conditional constancy, the estimator is consistent,” even though it compiles.

The structural defense. The development’s stated policy, recorded in MAIC.v and STC.v, is to refuse True-valued assumption definitions and instead make every modeling assumption a visible, abstract named premise that appears in the theorem’s own statement (Section 30.9). Because such a premise is a genuine, non-trivial Prop that the conclusion actually consumes, a vacuous proof is impossible: the theorem cannot be discharged without using the assumption, and a human reading the statement can confirm that the premise says what the book says. Making each assumption a load-bearing visible premise is the structural guard against the placeholder trap, since it converts the unprovable question “is this the intended theorem?” into the checkable question “is this premise the intended assumption?”, which a reader can answer by inspection. \(\square\)