Let $X_1, \ldots, X_n$ be square-integrable random variables in $L^2(\Omega, \mathcal{F}, \mathbb{P})$ with the usual inner product (and induced norm):
$$ \langle X, Y \rangle = \int XY \; d\mathbb{P}$$ I am tring to understand what it means for them to be independent and identically distributed from a "functional / inner product" perspective. For simplicity, assume $\mathbb{E} X_i = \langle X_i, \mathbf{1}_\Omega \rangle = 0$ and $X_i \neq 0$ almost everywhere for each $i$.
If they are identically distributed, this means that the pushforward measure is the same for each $X_i$. In other words, $\mathbb{P} \circ X^{-1}_i = \mathbb{P} \circ X^{-1}_j$, which I think in turn implies that $X_i = X_j$ almost everywhere. Let's call the common function $X$.
However, if they are independent this implies that they have covariance $0$, i.e.
$$ \langle X_i, X_j \rangle = 0 $$
However, since each $X_i = X$ and $X_i \neq 0$ almost everywhere, this would seem to imply that
$$\langle X_i, X_j \rangle = \langle X, X \rangle = \lVert X \rVert^2 > 0 $$ which seems to be a contradiction. Can someone clarify where my logic is going wrong?
Best Answer
It's certainly not true that $\ \mathbb{P}\circ X_i^{-1}=\ \mathbb{P}\circ X_j^{-1}\ $ implies that $\ X_i=X_j\ $ almost everywhere. To take a simple counterexample, let $\ \Omega=\mathbb{R}^2\ $, $\ \mathcal{F}\ $ be the Borel subsets of $\ \Omega\ $, $$ \mathbb{P}(A)=\frac{1}{2\pi}\iint_A e^{-\frac{x_1^2+x_2^2}{2}}dx_1dx_ 2 $$ for $\ A\in\mathcal{F}\ ,$ and $$ X_1\big(x_1,x_2\big)=x_1\ ,\\ X_2\big(x_1,x_2\big)=x_2\ \ $$ for $\ \big(x_1,x_2\big)\in\Omega\ $. Then $\ X_1,X_2\ $ are independent, identically distributed random variables in $\ L^2(\Omega,\mathcal{F},\mathbb{P})\ $ (their common distribution is the standard normal), but $$ \left\{\big(x_1,x_2\big)\in\Omega\,\big|\,X_1\big(x_1,x_2\big)=X_2\big(x_1,x_2\big)\right\}=\left\{\big(x_1,x_2\big)\in\Omega\,\big|\,x_1=x_2\right\}\ , $$ which has probability $\ 0\ $.