Prove that if $X$ is an integrable random variable, it has a symmetric distribution if and only if: $$E(X|X^2)=0$$ Can anyone check my solution?
Firstly $$EX=0$$
Then we have:
$$E(E(X|X^2))=EX=0$$
Secondly:
$$E(X|X^2)=0 $$
So:
$$E(X)=E(E(X|X^2)=E(0)=0$$
I will be grateful for help!Thanks
The problem of your approach is that you confuse the assertions "the distribution of $X$ is symmetric" and "$X$ is centered".
Notice that if $B$ is a symmetric Borel set (i.e. $B=-B$), then $\{X\in B\}$ is measurable for the $\sigma$-algebra generated by $X^2$. Conversely, an element of $\sigma(X^2)$ is of the form $\{X\in B\}$ for some symmetric $B$.
Assume that $\mathbb E[X\mid X^2]=0$. Then for each symmetric $B$, we have $\mathbb E[X\mathbb 1_{\{X\in B\}}]=0$.
Conversely, if $X$ is symmetric, then we can compute $\mathbb E[X\mathbb 1_{\{X\in B\}}]$ where $B$ is symmetric and link this with $\mathbb E[\mathbb E[X \mid X^2]\mathbb 1_{\{X\in B\}}]$.