Square-Integrable Random Variables and the Mean-Square Inner Product

Let $(\Omega,\mathcal{F},P)$ be a probability space and $\mathbb{R}$ the set of real numbers.

Preliminaries. For random variables $X,Y$ on $(\Omega,\mathcal{F},P)$, the functions $X+Y$, $cX$ ($c\in\mathbb{R}$), $X^{2}$, and $XY$ are again random variables: differences and sums are handled by the rational-decomposition argument recorded in Stochastic Process, Independent Increments, and Inhomogeneous Poisson Process together with the generator criterion of Measurable Function and Real-Valued Measurable Function; for $X^{2}$, the level sets satisfy, for $a\ge0$, the identity that $X^{2}>a$ holds exactly when $X>\sqrt{a}$ or $X<-\sqrt{a}$ (with the nonnegative square root), while for $a<0$ the set is all of $\Omega$; and $XY=\frac{1}{4}\bigl((X+Y)^{2}-(X-Y)^{2}\bigr)$ reduces the product to sums and squares.

Square-integrability. A random variable $X$ is square-integrable if the expectation of the nonnegative random variable $X^{2}$ is finite, $\mathbb{E}[X^{2}]<\infty$. If $X$ and $Y$ are square-integrable, then: $cX$ is square-integrable; $X+Y$ is square-integrable, because pointwise $(X+Y)^{2}\le2X^{2}+2Y^{2}$ (as $2X^{2}+2Y^{2}-(X+Y)^{2}=(X-Y)^{2}\ge0$) and expectation is monotone and additive on nonnegative random variables by Linearity and Monotonicity of the Lebesgue Integral; and $XY$ is integrable, because pointwise $|XY|\le\frac{1}{2}(X^{2}+Y^{2})$ (as $X^{2}+Y^{2}-2|XY|=(|X|-|Y|)^{2}\ge0$). Moreover $X$ itself is integrable: $|X|\le\frac{1}{2}(1+X^{2})$ pointwise.

Mean-square inner product and norm. For square-integrable $X,Y$ define

using the nonnegative square root. By Linearity and Monotonicity of the Lebesgue Integral, $\langle\cdot,\cdot\rangle_{2}$ is symmetric and linear in each argument on the set of square-integrable random variables, and $\langle X,X\rangle_{2}=\mathbb{E}[X^{2}]=\lVert X\rVert_{2}^{2}\ge0$.

Mean-square distance and null equivalence. The mean-square distance between square-integrable $X$ and $Y$ is $\lVert X-Y\rVert_{2}$. It vanishes if and only if $P(X=Y)=1$: if $\mathbb{E}[(X-Y)^{2}]=0$, then for every $\varepsilon>0$, Markov's inequality applied to $(X-Y)^{2}$ gives $P(|X-Y|\ge\varepsilon)\le\varepsilon^{-2}\,\mathbb{E}[(X-Y)^{2}]=0$, and letting $\varepsilon=1/n$ along $n\in\mathbb{N}$ with countable additivity gives $P(X\ne Y)=0$; conversely, if $P(X=Y)=1$ then $(X-Y)^{2}=0$ off an event of probability $0$, so $\mathbb{E}[(X-Y)^{2}]=0$ by Linearity and Monotonicity of the Lebesgue Integral applied to the pointwise bound by simple functions vanishing off a null set (a nonnegative simple function supported on a null set has integral $0$). Random variables at mean-square distance $0$ are called versions of one another, or almost surely equal.