Kolmogorov maximal inequality

Disjoint groups of an independent sigma-algebra family remain independent: Let $({\mathcal{F}}_i{)}_{i\in I}$ be an independent family of sigma-algebras on a probability space, and let $J_0,\dots ,J_{m-1}\subseteq I$ be pairwise disjoint index sets. For each $r<m$, define ${\mathcal{G}}_r:=\sigma \left({\bigcup }_{i\in J_r}{\mathcal{F}}_i\right)\mathrm{.}$ Then the sigma-algebras ${\mathcal{G}}_0,\dots ,{\mathcal{G}}_{m-1}$ are independent.

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).

Variance and covariance identities for random variables: Let $X,Y$ be square-integrable real random variables on one probability space. Then $\mathrm{Var}(X)=\mathbb{E}[X^2]-\mathbb{E}[X{]}^2,$ $\mathrm{Cov}(X,Y)=\mathbb{E}[XY]-\mathbb{E}[X]\mathbb{E}[Y]\mathrm{.}$ Moreover, covariance is symmetric and bilinear on finite linear combinations. On finite full-power-set probability spaces these formulas reduce to the published finite identities.

Arithmetic and lattice operations preserve measurability whenever they are defined: Let $(X,\mathcal{A})$ be a measurable space and let $f,g:X\to \bar{\mathbb{R}}$ be measurable. Then: 1. $cf$ is measurable for every real scalar $c$; 2. $\max (f,g)$, $\min (f,g)$, $\mathrm{\mid }f\mathrm{\mid }$, $f^{+}$, and $f^{-}$ are measurable; 3. if $f+g$ is pointwise defined, then $f+g$ is measurable; 4. with the total function-product convention declared in that theorem, the pointwise product $fg$ is measurable.