Truncation weak law for independent arrays

Zero truncation at a positive level: For a real random variable $X$ and a deterministic level $A>0$, its zero truncation is $X^{(A)}=X{\mathbf{1}}_{\{\mathrm{\mid }X\mathrm{\mid }\le A\}}\mathrm{.}$ The threshold event is measurable because $X$ is measurable and $[-A,A]$ is Borel; its indicator and the product are measurable by thm-arithmetic-and-lattice-operations-preserve-measurability. Thus $X^{(A)}$ is a real random variable as in def-random-element-and-real-random-variable. It equals $X$ at both cutoff endpoints and is zero outside the interval. Since $\mathrm{\mid }{X}^{(A)}\mathrm{\mid }\le A$, for every $0<p<\mathrm{\infty }$ its absolute $p$th moment is at most $A^p\mathbb{P}(\mathrm{\Omega })=A^p$. This is not clipping to the endpoints.

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.

Measurable coordinatewise functions preserve independence: Let $(X_i{)}_{i\in I}$ be an independent family of random elements $X_i:(\mathrm{\Omega },\mathcal{F},\mathbb{P})\to (S_i,{\mathrm{\Sigma }}_i)$. For each $i$, let $g_i:(S_i,{\mathrm{\Sigma }}_i)\to (T_i,{\mathcal{T}}_i)$ be measurable. Then the family $(g_i\circ X_i{)}_{i\in I}$ is independent.

Convergence in probability: For real random variables $(X_n)$ and $X$ on one probability space, write $X_n\to X$ in probability when, for every $\epsilon >0$, $\mathbb{P}(\mathrm{\mid }{X}_n-X\mathrm{\mid }>\epsilon )⟶0.$ This is precisely def-convergence-in-measure for the probability measure.

Finite and countable subadditivity of measures: Let $\mu$ be a measure and let $(E_k{)}_{k\in \mathbb{N}}$ be measurable. Then $\mu \left({\bigcup }_{k\in \mathbb{N}}{E}_k\right)\le {\sum }_{k=0}^{\mathrm{\infty }}\mu (E_k)\mathrm{.}$ For every $m\in \mathbb{N}$ one also has $\mu \left({\bigcup }_{k<m}{E}_k\right)\le {\sum }_{k<m}\mu (E_k),$ including $m=0$, where both sides are $0$.

For $n\ge 1$, if $X_0,\dots ,X_{n-1}$ are independent real random variables and $g_i:\mathbb{R}\to \mathbb{R}$ are Borel measurable with each $g_i(X_i)$ integrable, then ${\prod }_{i<n}{g}_i(X_i)$ is integrable and $\mathbb{E}[{\prod }_{i<n}{g}_i(X_i)]={\prod }_{i<n}\mathbb{E}[g_i(X_i)]$ in $\mathbb{R}$ (Expectations factor over finite products of independent random variables).