Polyhedral selective inference

The density $\varphi$ in [F1] is positive and continuous by [F2]. Fix $a<b$. By [F3], $\varphi$ attains a minimum $c$ on $[a,b]$; since $\varphi$ is positive at every point, $c>0$. The density formula and additivity over $(-\mathrm{\infty },a]\bigsqcup (a,b]$ give $\mathrm{\Phi }(b)-\mathrm{\Phi }(a)={\int }_{(a,b]}\varphi (x)dx\ge c(b-a)>0$, so $\mathrm{\Phi }$ is strictly increasing.

Fix $x\in \mathbb{R}$. By [F3], $\varphi$ is bounded by some finite $M$ on the compact interval $[x-1,x+1]$. If $\mathrm{\mid }y-x\mathrm{\mid }<1$, then $\mathrm{\mid }\mathrm{\Phi }(y)-\mathrm{\Phi }(x)\mathrm{\mid }\le M\mathrm{\mid }y-x\mathrm{\mid }$ by integrating the density over the interval between $x$ and $y$. Letting $y\to x$ proves continuity at $x$, and $x$ was arbitrary.

The sets $(-\mathrm{\infty },n]$ increase to $\mathbb{R}$, so [F4] and the total mass in [F1] give ${\lim }_{n\to \mathrm{\infty }}\mathrm{\Phi }(n)=1$. The sets $(-\mathrm{\infty },-n]$ decrease to the empty set and have finite measure, so [F5] gives ${\lim }_{n\to \mathrm{\infty }}\mathrm{\Phi }(-n)=0$. Monotonicity from step 1.1 then gives the two stated limits at infinity.

Since $P_{\eta }y=c_{\eta }({\eta }^{T}y)=c_{\eta }T(y)$ and $Z(y)=(I_n-P_{\eta })y$, we have $y=c_{\eta }T(y)+Z(y)$.

For each row $j$, the inequality $a_j^{T}y\le b_j$ becomes $r_{j}T(y)\le d_j(Z(y))$. If $r_j>0$, division gives $T(y)\le d_j(Z(y))\mathrm{/}{r}_j$; if $r_j<0$, division reverses the inequality and gives $T(y)\ge d_j(Z(y))\mathrm{/}{r}_j$; if $r_j=0$, the row is feasible exactly when $d_j(Z(y))\ge 0$. Intersecting the finitely many row conditions is therefore equivalent to $G(Z(y))$, $T(y)\ge L(Z(y))$, and $T(y)\le U(Z(y))$, with the empty-index conventions in [F2].

Each $d_j$ is affine and hence continuous. The finite maximum defining $L$ and finite minimum defining $U$ are Borel, with the constant infinite values in the empty-index cases; $G$ is a finite intersection of closed half-spaces. Thus the fiber set is Borel. The rowwise equivalence in step 2.1 proves both implications, including zero coefficients and empty or degenerate intervals.

Put $P=P_{\eta }$. Since $\eta \mathrm{\ne }0$, $P^T=P$ and $P^2=P$. For every $y$, $Py=\eta ({\eta }^{T}y)\mathrm{/}\mathrm{\parallel }\eta {\mathrm{\parallel }}^2$ belongs to $\mathrm{span}\{\eta \}$, and ${\eta }^T(y-Py)=0$, so $y-Py\in (\mathrm{span}\{\eta \}{)}^{\perp }$. Thus $P$ is the rank-one orthogonal projection onto $\mathrm{span}\{\eta \}$ in [F4]. The conclusion concerns only the joint law of $T,Z$, so by [F1] it suffices to use the realization $Y=\mu +\sigma W$ with $W\sim N_n(0,I_n)$. Apply [F2] to $P$. This use of AC is exactly [A1].

Because $P\eta =\eta$, $T={\eta }^T\mu +\sigma {\eta }^{T}PW$, while $Z=(I_n-P)\mu +\sigma (I_n-P)W$. Thus $T$ and $Z$ are measurable functions of the independent vectors $PW$ and $(I_n-P)W$; [F3] makes them independent.

By the linear-projection clause of [F1], $T={\eta }^{T}Y$ is normal with mean ${\eta }^T\mu$ and variance ${\eta }^T({\sigma }^2{I}_n)\eta ={\sigma }^2\mathrm{\parallel }\eta {\mathrm{\parallel }}^2$. AC is inherited from [A1] through the normal-law supplier [F1].

Under $P_{\mu }$, $T$ and $Z$ are independent by [F2], so their joint law is the product of their marginal laws by [F3]. Write ${\nu }_T=N(\theta ,s^2)$ and let ${\nu }_Z$ be the marginal law of $Z$. Both are probability measures and hence sigma-finite. By [F1] and [F7], the selection event in these coordinates is the Borel set $D=\{(t,z):G(z),L(z)\le t\le U(z)\}$. For every product-measurable set $B$, conditioning the product law on $D$ gives $Q_{\mu }((T,Z)\in B)={\int }_B{\mathbf{1}}_D(t,z)\mathrm{/}{P}_{\mu }(S)d({\nu }_T\times {\nu }_Z)(t,z)$. Thus the selected joint law has this product density.

Put $H(z)=G(z)\wedge (L(z)\le U(z))$. If $H(z)$ fails, the fiber is empty and its section integral is zero. If $H(z)$ holds, the section is $[L(z),U(z)]$; by [F5], [F6], and the endpoint convention in [F7], its ${\nu }_T$-probability is ${\mathrm{\Delta }}_{\theta }(z)$, including a singleton interval and infinite endpoints. Thus the section integral of the selected density is $m(z)={\mathbf{1}}_{H(z)}{\mathrm{\Delta }}_{\theta }(z)\mathrm{/}{P}_{\mu }(S)$. It is finite because $0\le {\mathrm{\Delta }}_{\theta }(z)\le 1$ on feasible fibers and $P_{\mu }(S)>0$. Apply [F4] with filler probability ${\nu }_T$: for almost every selected $z$, $m(z)>0$, so $H(z)$ holds, $L(z)<U(z)$, and ${\mathrm{\Delta }}_{\theta }(z)>0$; the conditional law of $T$ is $K_z(B)={\nu }_T(B\cap [L(z),U(z)])\mathrm{/}{\mathrm{\Delta }}_{\theta }(z)$. Fibers with zero section integral have zero selected marginal probability, so their arbitrary extension value in the pivot definition is immaterial.

For $L(z)\le t\le U(z)$, [F6] gives $K_z((-\mathrm{\infty },t])=[\mathrm{\Phi }((t-\theta )\mathrm{/}s)-\mathrm{\Phi }((L(z)-\theta )\mathrm{/}s)]\mathrm{/}[\mathrm{\Phi }((U(z)-\theta )\mathrm{/}s)-\mathrm{\Phi }((L(z)-\theta )\mathrm{/}s)]=q_{\theta }(t,z)$. The denominator is positive exactly on the selected fibers under consideration. The finite max/min formulas make $L,U,G$ Borel, and [F5] with [F7] makes ${\mathrm{\Delta }}_{\theta }$ Borel. On the Borel set of $y\in S$ for which $G(Z(y))$, $L(Z(y))<U(Z(y))$, and ${\mathrm{\Delta }}_{\theta }(Z(y))>0$, the pivot's numerator and reciprocal denominator are Borel, so the ratio is Borel; the definition extends it by zero on the complement.

Fix such a $z$. The conditional density in step 2.1 is positive throughout the interval and has no atoms; [F5] makes its CDF continuous and strictly increasing from zero to one, including when an endpoint is infinite. For any $u\in (0,1)$ there is a unique $t_u$ in the interval with $q_{\theta }(t_u,z)=u$, so $K_z\{t:q_{\theta }(t,z)\le u\}=K_z((-\mathrm{\infty },t_u])=q_{\theta }(t_u,z)=u$. At $u=0$ the probability is zero and at $u=1$ it is one because the conditional law has a density. Thus the pivot is uniform given $Z=z,S$.

Integrating the conditional identity in step 4.1 against the selected marginal law of $Z$ supplied by [F4] gives $P_{\mu }(q_{\theta }(Y)\le u\mid S)=u$ for every $u\in [0,1]$. Hence the pivot is uniform given $S$ as well.

For any $\mu$ in the conditional experiment, [F1] gives $q_{{\eta }^T\mu }(Y)\sim \mathrm{Uniform}(0,1)$ under $Q_{\mu }$. For $0\le u\le 1$, the event $2\min \{q,1-q\}\le u$ is the union of the two tails $q\le u\mathrm{/}2$ and $q\ge 1-u\mathrm{/}2$, whose uniform probabilities sum to $u$ (their intersection has measure zero). Thus, if $H_0(\vartheta )\mathrm{\ne }\mathrm{\varnothing }$, $p_{\vartheta }$ is uniform under every null law and is measurable by the pivot formula; [F2] makes it a valid p-value.

If $H_0(\vartheta )\mathrm{\ne }\mathrm{\varnothing }$ and the alternative is nonempty, [F3] applied to $p_{\vartheta }$ gives conditional size at most $\alpha$. By [F4] size is the supremum of the null rejection probabilities; since every null law has the exact uniform p-value distribution from step 1.1, $Q_{\mu }(p_{\vartheta }\le \alpha )=\alpha$ for every null $\mu$, so the size is exactly $\alpha$.

By [F1], [F6], and [F8], $q_{\mu }(Y):=q_{{\eta }^T\mu }(Y)$ is a pivot in the conditional experiment. Its full-parameter inversion is $C_{\mathrm{f}\mathrm{u}\mathrm{l}\mathrm{l}}(Y)=\{{\mu }^{\mathrm{\prime }}\in {\mathrm{\Theta }}_S:q_{{\eta }^T{\mu }^{\mathrm{\prime }}}(Y)\in [\alpha \mathrm{/}2,1-\alpha \mathrm{/}2]\}$. The acceptance interval has uniform probability $1-\alpha$, so [F7] gives exact coverage for $C_{\mathrm{f}\mathrm{u}\mathrm{l}\mathrm{l}}$. Its image under ${\mu }^{\mathrm{\prime }}\mapsto {\eta }^T{\mu }^{\mathrm{\prime }}$ is exactly $C_{\eta }(Y)$ by the definition of ${\mathrm{\Theta }}_{\eta }$. For each fixed true $\mu$, membership ${\eta }^T\mu \in C_{\eta }(Y)$ is precisely $\alpha \mathrm{/}2\le q_{{\eta }^T\mu }(Y)\le 1-\alpha \mathrm{/}2$, which has $Q_{\mu }$-probability $1-\alpha$ by [F1]. The membership section is measurable by the Borel pivot formula, so [F5] applies.