IID finite-variance weak law

Identical distribution and IID families: Let $(X_i{)}_{i\in I}$ be random elements with the same measurable target $(E,\mathcal{E})$. They are identically distributed if $\mathbb{P}(X_i\in B)=\mathbb{P}(X_j\in B)$ for all $i,j\in I$ and $B\in \mathcal{E}$, that is, their laws in def-law-or-distribution-of-a-random-element agree. They are independent and identically distributed (IID) if, in addition, the whole family is independent in def-independent-random-elements. Independence means mutual independence, not merely pairwise independence. No moment assumption is part of either definition. The empty family satisfies these universal conditions vacuously.

Chebyshev weak law for uncorrelated arrays: For each $n\ge 1$, let $X_{n,1},\dots ,X_{n,r_n}$ be square-integrable real random variables on one probability space, pairwise uncorrelated within the row, where $r_n\ge 0$ is finite. Set $S_n={\sum }_{k=1}^{r_n}{X}_{n,k}$ and let $b_n>0$ be deterministic. If $v_n:=b_n^{-2}{\sum }_{k=1}^{r_n}\mathrm{Var}(X_{n,k})⟶0,$ then $(S_n-\mathbb{E}{S}_n)\mathrm{/}{b}_n\to 0$ in $L^2$ and in probability. More precisely, its second moment is $v_n$, and its probability of absolute value at least $\epsilon >0$ is at most $v_n\mathrm{/}{\epsilon }^2$. No independence between rows is required.

Expectations factor over finite products of independent random variables: Let $n\ge 1$, let $X_0,\dots ,X_{n-1}$ be independent real random variables on a common probability space, and let $g_i:\mathbb{R}\to \mathbb{R}$ be Borel measurable for each $i<n$. 1. If every $g_i$ is nonnegative, then $\mathbb{E}[{\prod }_{i<n}{g}_i(X_i)]={\prod }_{i<n}\mathbb{E}[g_i(X_i)]$ in $[0,+\mathrm{\infty }]$. 2. If every $g_i(X_i)$ is integrable, then ${\prod }_{i<n}{g}_i(X_i)$ is integrable and the same factorization holds in $\mathbb{R}$.