The truncated normal selective pivot is conditionally uniform

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.