Non-trivial examples of $E[X|X^2] = X$ and $E[Y|Y^2] = 0$.

Question

I want to find a few examples of non-trivial random variables $X, Y$ such that $E[X|X^2] = X$ and $E[Y|Y^2] = 0$.

From what i gather for $E[X|X^2] = X$ i need to find a function such that if $X=u$ then $f(-u)=0$. Then for $E[Y|Y^2] = 0$ if $Y=u$ then $f(u)=f(-u)$.

Answer 1

Assume that $X$ is any positive random variable. Then $X=\sqrt{X^2}$ is $\sigma(X^2)$-measurable so the conditional expectation of $X$ given $X^2$ is $X$.

Let $Y$ be any $L^2$ symmetric random variable. Then, for any bounded function $g$, $Yg(Y^2)$ is symmetric so has mean $0$, which is the mean value of $0\cdot g(Y^2)$, and $0$ is $\sigma(Y^2)$-measurable, so the conditional expectation of $Y$ given $Y^2$ is zero.

Answer 2

if you look at

A problem in Jun Shao mathematical statistics:exercises and solutions P33

example of conditions that $E(X|X^2)=0$

$E(X|X^2=t)=\sqrt(t) \bigg(\frac{f(\sqrt(t))-f(-\sqrt(t))}{f(\sqrt(t))+f(-\sqrt(t))} \bigg)$

$E(X|X^2=t)=0$ hence $f(\sqrt(t))-f(-\sqrt(t))=0$

so it is enough

$f(t)=f(-t)$ so you need to choose a symmetric distribution around Zero.

example of conditions that $E(X|X^2)=X$

$E(X|X^2=t)=\sqrt(t) $ if

$\bigg(\frac{f(\sqrt(t))-f(-\sqrt(t))}{f(\sqrt(t))+f(-\sqrt(t))} \bigg)=1$

so $2f(-\sqrt(t))=0$ so $X$ need to be positive.