IID Monte Carlo integration and error

Every $Y_i$ has the law of $Y_1$ by [F1]. The identity and $q(y)=(y-\mu {)}^2$ are continuous polynomial maps by the elementary continuity rules for sums and products; inverse images of open sets under continuous maps are open, and inverse images preserve complements and countable unions, so both maps are Borel. Applying [F2] to the identity gives $\mathbb{E}[Y_i]=\mu$ when $\mathbb{E}\mathrm{\mid }{Y}_1\mathrm{\mid }<\mathrm{\infty }$. If $\mathbb{E}[Y_1^2]<\mathrm{\infty }$, applying [F2] to $q$ also gives $\mathrm{Var}(Y_i)={\sigma }^2$ by [F4].

By [F6], ${\bar{Y}}_n=n^{-1}{\sum }_{i=1}^n{Y}_i$. Linearity [F3] and step 1.1 therefore give $\mathbb{E}[{\bar{Y}}_n]=n^{-1}{\sum }_{i=1}^n\mathbb{E}[Y_i]=\mu$.

Assume $\mathbb{E}[Y_1^2]<\mathrm{\infty }$. Since $\mathrm{\mid }y\mathrm{\mid }\le 1+y^2$ for every real $y$, each $Y_i$ is integrable as well as square-integrable. The first $n$ terms are pairwise independent by [F1], so [F5] and step 1.1 give $\mathrm{Var}({\sum }_{i=1}^n{Y}_i)=n{\sigma }^2$. Linearity [F3] and the variance definition [F4] give $\mathrm{Var}(n^{-1}{\sum }_{i=1}^n{Y}_i)=n^{-2}\mathrm{Var}({\sum }_{i=1}^n{Y}_i)$; hence $\mathrm{Var}({\bar{Y}}_n)=n^{-2}(n{\sigma }^2)={\sigma }^2\mathrm{/}n$.

By [F2], ${\bar{Y}}_n=S_n\mathrm{/}n$. The hypotheses of [F3] hold by [F1] and the given integrability, so $\mathbb{E}\mathrm{\mid }{\bar{Y}}_n-\mu \mathrm{\mid }\to 0$ and ${\bar{Y}}_n\to \mu$ in probability.

Under the same hypotheses, [F4] gives $S_n\mathrm{/}n\to \mu$ almost surely. Rewriting with [F2] yields ${\bar{Y}}_n\to \mu$ almost surely.

Since $\mathrm{\mid }y\mathrm{\mid }\le 1+y^2$ for every real $y$, the given second moment implies $\mathbb{E}\mathrm{\mid }{Y}_1\mathrm{\mid }<\mathrm{\infty }$. Applying [F6] to the original IID sequence gives ${\bar{Y}}_n\to \mu$ almost surely by [F1].

The identity map on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$ is measurable, so its pointwise product with itself is Borel measurable by [F3]; call this square map $q$. Applying [F4] with $q$ preserves independence. For any Borel set $B$, $\mathbb{P}(Y_i^2\in B)=\mathbb{P}(Y_i\in q^{-1}(B))$, so the common-law clause of [F2] gives a common law for the $Y_i^2$. By [F5], their common finite expectation is $\mathbb{E}[Y_1^2]$. Thus [F6] applies to this IID sequence and gives $n^{-1}{\sum }_{i=1}^n{Y}_i^2\to \mathbb{E}[Y_1^2]$ almost surely.

Expanding the centered squares in [F1] gives $s_n^2=\frac{n}{n-1}(n^{-1}{\sum }_{i=1}^n{Y}_i^2-{\bar{Y}}_n^2)$. Steps 1.1 and 1.2 imply that the bracket tends almost surely to $\mathbb{E}[Y_1^2]-{\mu }^2$; also $n\mathrm{/}(n-1)\to 1$. By [F7] and [F8], $\mathbb{E}[Y_1^2]-{\mu }^2=\mathbb{E}[(Y_1-\mu {)}^2]={\sigma }^2$. Hence $s_n^2\to {\sigma }^2$ almost surely.

If ${\sigma }^2>0$, step 2.1 gives $s_n^2\mathrm{/}{\sigma }^2\to 1$ almost surely. Both $s_n$ and $\sigma$ are nonnegative, so continuity of the square root gives $s_n\mathrm{/}\sigma \to 1$. By [F1], ${\widehat{se}}_n\mathrm{/}(\sigma \mathrm{/}\sqrt{n})=s_n\mathrm{/}\sigma$, proving the relative standard-error consistency.

The hypotheses of [F3] hold by [A1], [F1], and the given positive finite variance. Since [F2] gives ${\sum }_{i=1}^n{Y}_i=n{\bar{Y}}_n$, [F3] yields $Z_n:=\sqrt{n}({\bar{Y}}_n-\mu )\mathrm{/}\sigma \Rightarrow N(0,1)$.

The map $x\mapsto \sigma x$ is continuous because $\sigma >0$. Applying [F4] to step 1.1 gives $\sigma Z_n\Rightarrow \sigma Z$ for $Z\sim N(0,1)$. By [A1] and [F5], the law of $\sigma Z$ is $N(0,{\sigma }^2)$, and $\sigma Z_n=\sqrt{n}({\bar{Y}}_n-\mu )$. This is the stated convergence.