Exact sampling variance and a numerical-bias error bound

This argument uses only the given finite $n$-tuple. By [A1], every $Y_{h,j}$ has the law of $Y_h$. The Borel functions $\mathrm{\mid }y\mathrm{\mid }$, $y^2$ and $(y-{\theta }_h{)}^2$, and the identity, have the same law integrals for these variables by [F2]. Since $\mathrm{\mid }y\mathrm{\mid }\le 1+y^2$, all are integrable when needed, so ${\mathbb{E}}_s{Y}_{h,j}={\theta }_h$ and ${\mathrm{Var}}_s(Y_{h,j})={\sigma }_h^2$. Finite linearity [F3] gives ${\mathbb{E}}_s{\bar{Y}}_{h,n}=n^{-1}{\sum }_{j=1}^n{\theta }_h={\theta }_h$. Mutual independence implies pairwise independence; finite variance additivity [F2] gives ${\mathrm{Var}}_s({\sum }_{j=1}^n{Y}_{h,j})=n{\sigma }_h^2$. By the centered-second-moment definition and finite linearity, scaling this sum by $n^{-1}$ scales its variance by $n^{-2}$, hence ${\mathrm{Var}}_s({\bar{Y}}_{h,n})={\sigma }_h^2\mathrm{/}n$. This includes $n=1$ and ${\sigma }_h^2=0$, without constructing an infinite extension.

For the probability-bound clause, assume its supplied bounds and put $Z:={\bar{Y}}_{h,n}-{\theta }_h$ and let $E=\{\mathrm{\mid }{\bar{Y}}_{h,n}-\theta \mathrm{\mid }>{ϵ}_b+{ϵ}_s\}$. On $E$, the reverse triangle inequality and $\mathrm{\mid }{b}_h\mathrm{\mid }\le B_h\le {ϵ}_b$ imply $\mathrm{\mid }Z\mathrm{\mid }\ge \mathrm{\mid }{\bar{Y}}_{h,n}-\theta \mathrm{\mid }-\mathrm{\mid }{b}_h\mathrm{\mid }>{ϵ}_s$. Thus, pointwise, ${ϵ}_s^2{\mathbf{1}}_E\le Z^2$.

By [F3], $\mathrm{\mid }{b}_h\mathrm{\mid }=\mathrm{\mid }\mathbb{E}(Y_h-Y)\mathrm{\mid }\le \mathbb{E}\mathrm{\mid }{Y}_h-Y\mathrm{\mid }\le \mathbb{E}b$ under the pointwise certificate. For an $L$-Lipschitz $J$, substitute $\mathrm{\mid }J(S_h)-J(S)\mathrm{\mid }\le L\mathrm{\parallel }{S}_h-S\mathrm{\parallel }$ into the same expectation inequality. A deterministic envelope $b\le B$ gives $\mathbb{E}b\le B$. An empirical average cannot in general certify this upper bound: if $b$ is Bernoulli with parameter $p\in (0,1)$, an IID sample of $n$ zero values has probability $(1-p{)}^n>0$, while $\mathbb{E}b=p>0$.

Let $A\subseteq \mathrm{\Theta }$ be the measurable callback success set under input law $\mu$. If $\mu (A)=1$, all members of any finite IID sample lie in $A$ almost surely, so their output law and the preceding inequalities are unchanged by any measurable definition on the failure-null set. If $0<\mu (A)<1$, the conditional joint input law given all $n$ calls succeed is $\mu (\cdot \mid A{)}^{\otimes n}$: on rectangles it follows by dividing ${\prod }_i\mu (B_i\cap A)$ by $\mu (A{)}^n$. This is not the original ${\mu }^{\otimes n}$, as already seen by the probability of $A$ in one coordinate. If $\mu (A)=0$, complete execution has probability zero and supplies no conditional certificate. Thus success-only sampling does not justify the original-law guarantee.

By [F4], $s_n^2\to {\sigma }_h^2$ almost surely. Under [A2] and ${\sigma }_h>0$, continuity of the square root gives $s_n\mathrm{/}{\sigma }_h\to 1$ almost surely and hence in probability; the latter implication follows since the indicators of $\mathrm{\mid }{s}_n\mathrm{/}{\sigma }_h-1\mathrm{\mid }>\epsilon$ tend to zero almost surely and are bounded by one. The imported IID CLT, rescaled by the constant ${\sigma }_h$, gives $\sqrt{n}({\bar{Y}}_{h,n}-{\theta }_h)\mathrm{/}{\sigma }_h\Rightarrow N(0,1)$. Slutsky [F4] with denominator $s_n\mathrm{/}{\sigma }_h$ proves the displayed studentized limit under its zero-denominator convention.

By step 1.1, ${\mathbb{E}}_s[Z]=0$ and ${\mathbb{E}}_s[Z^2]={\sigma }_h^2\mathrm{/}n$. Since ${\bar{Y}}_{h,n}-\theta =Z+b_h$, expanding the square and taking expectations makes the cross term $2b_h{\mathbb{E}}_s[Z]$ vanish, so ${\mathbb{E}}_s[({\bar{Y}}_{h,n}-\theta {)}^2]={\sigma }_h^2\mathrm{/}n+b_h^2$.

Taking expectations in step 1.2 gives ${\mathbb{P}}_s(E)\le {\mathbb{E}}_s[Z^2]\mathrm{/}{ϵ}_s^2$. Step 2.1 identifies this second moment as ${\sigma }_h^2\mathrm{/}n$, and ${\sigma }_h^2\le V_h$ gives the stated probability bound.

Almost-sure convergence of $s_n$ to ${\sigma }_h>0$ also gives ${\mathbb{P}}_s(s_n=0)\to 0$ by the same bounded-indicator argument. The standard normal has a continuous density, so CDF convergence in [F4] gives ${\mathbb{P}}_s(\mathrm{\mid }\sqrt{n}({\bar{Y}}_{h,n}-{\theta }_h)\mathrm{/}{s}_n\mathrm{\mid }\le z)\to 1-\alpha$; values at zero denominators alter this probability by at most ${\mathbb{P}}_s(s_n=0)$. On $s_n>0$, that event implies $\mathrm{\mid }{\bar{Y}}_{h,n}-\theta \mathrm{\mid }\le z{s}_n\mathrm{/}\sqrt{n}+\mathrm{\mid }{b}_h\mathrm{\mid }\le z{s}_n\mathrm{/}\sqrt{n}+B$. This proves the interval liminf bound. It adds no finite-sample guarantee, and step 1.3 explains why an empirical bias diagnostic alone cannot provide deterministic $B$.