Execution and conditional error guarantees of the fluid UQ budget algorithm

In fixed-size mode the index set is $\{1,\dots ,N\}$. By [F1] the callback is invoked once at each index and failures do not shorten or restart this loop. Each requested draw is delivered in finite time by hypothesis, and each callback invocation has its supplied finite budget and returns one of its two allowed outcomes. Thus the loop finishes with exactly $N$ outcomes, and comparing their failure flags gives exactly the failed-index set. If it is empty, the definitions use only finite sums, $N>0$, $N-1>0$, and the nonnegative square root of a finite sum of squares, so all reported diagnostics are defined. Otherwise the prescribed output is incomplete. There are a bounded number of arithmetic and flag operations per index; the reductions have the same linear count.

The certified numeric guards ensure that $a=V_h\mathrm{/}(\delta {ϵ}_s^2)$ is a finite nonnegative real. Its ceiling is finite, so $N\ge 1$ is finite and $N\ge a$. Under the total-return IID oracle contract, [F1]'s loop requests exactly $N$ finite samples and computes their finite mean. If an invocation instead reports a failure, its explicit branch returns incomplete; no assertion above requires an actual nonreturning oracle to terminate. The input-guard branches perform no sampling.

Let $C_i=\{{\theta }_i\in A\}$. These are measurable and independent by [F4], each with probability $p$. Step 1.1 makes $C={\bigcap }_{i=1}^N{C}_i$, so $\mathbb{P}(C)=p^N$. The identity $1-p^N=(1-p){\sum }_{j=0}^{N-1}{p}^j$ and $0\le p\le 1$ give $1-p^N\le N(1-p)$. For $p>0$ and measurable $B_1,\dots ,B_N\subseteq \mathrm{\Theta }$, independence gives $\mathbb{P}({\bigcap }_{i=1}^N\{{\theta }_i\in B_i\}|C)=\frac{{\prod }_{i=1}^N\mu (B_i\cap A)}{p^N}={\prod }_{i=1}^N\mu (B_i\mid A)\mathrm{.}$ This rectangle identity says precisely that the conditional coordinates are independent with the asserted marginals. If $0<p<1$, the conditional probability of $A$ in one coordinate is $1$ rather than $p$, so its law differs from $\mu$. For $p=0$ the conditioning event is null; for $p=1$ it has probability one.

For $p=1$, zero extensions of the measurable successful maps are measurable, and modifications on $A^c$ affect no law or expectation. Their values on independent inputs are independent: the preimages of output events belong to the independent input sigma-algebras [F4]. Their common law is the law of $q_h$ under $\mu$ by [F3]. Equal-law integration and finite linearity [F3] give $\mathbb{E}{\bar{q}}_h=N^{-1}{\sum }_{i=1}^N\mathbb{E}{q}_i={\theta }_h$. All moment hypotheses of [F2] now hold with $n=N$, $Y_h=q_h$, and the supplied integrable target $Y$, so [F2] gives the stated MSE identity. It also gives $\mathrm{\mid }{b}_h\mathrm{\mid }\le \mathbb{E}b$ when the pointwise certificate and its integrability premise hold, while finite linearity and equal-law integration [F3] give $\mathbb{E}\bar{b}=\mathbb{E}b$.

Algebra gives ${\sum }_{i=1}^N(q_i-{\bar{q}}_h{)}^2={\sum }_{i=1}^N(q_i-{\theta }_h{)}^2-N({\bar{q}}_h-{\theta }_h{)}^2$. These terms are integrable by the assumed second moment. Applying [F2] with the identical exact and approximate observable $q_h$ gives $\mathbb{E}[({\bar{q}}_h-{\theta }_h{)}^2]={\sigma }_h^2\mathrm{/}N$. Equal-law integration [F3] gives $\mathbb{E}(q_i-{\theta }_h{)}^2={\sigma }_h^2$. Therefore the expectation of the right side is $N{\sigma }_h^2-N({\sigma }_h^2\mathrm{/}N)=(N-1){\sigma }_h^2$. Divide by $N-1$ and then by $N$ to obtain the remaining two moment identities. For the diagnostic warnings, a Bernoulli output with parameter $r\in (0,1)$ has probability $(1-r{)}^N>0$ of an all-zero sample, with zero sample variance and mean but positive population variance and mean. Taking a Bernoulli certificate likewise shows its all-zero empirical average cannot bound its positive expectation. Finally, [F2] supplies precisely the stated infinite-IID limits when their extra hypotheses hold; a single finite tuple is not extended implicitly.

A true certificate supplies exactly [F2]'s finite-sample hypotheses with $B_h={ϵ}_b$ and variance bound $V_h$, hence the probability is at most $V_h\mathrm{/}(N{ϵ}_s^2)\le \delta$ by step 1.2. Its MSE identity is at most $V_h\mathrm{/}N+{ϵ}_b^2$; monotonicity of the nonnegative square root gives the RMS bound. If $V_h=0$, the variance is zero and $N=1$. The centered square has expectation zero, so it vanishes almost surely: for each positive integer $m$, its being greater than $1\mathrm{/}m$ has probability at most $m$ times that zero expectation by [F3]; the union of these events is its positivity event. Thus sampling error is zero almost surely, without any division by a variance or by $N-1$. The algorithm does not report sample variance in this branch. For a callback implementation, step 2.1 and its conditional-law distinction explain the separate almost-sure-success premise. These arguments prove only the certified model guarantees and do not validate the certificate or quantify floating-point/model error.