I know that $E(X^{2}|X)=X^2$ is true since $X^2$ is $\sigma(X)$-measurable. But I think that maybe $E(X|X^2)=X$ is not true since it could happen that $X$ be not $\sigma(X^2)$-measurable. But how to obtain a counterexample and, is my reasoning correct? Because I know that if $Y$ is $\mathcal{G}$-measurable where $\mathcal{G}$ is a sub-$\sigma$-algebra of $\mathcal{F}$ then $E(Y|\mathcal{G})=Y$ but what about the opposite? Any sugestions?
Asked
Active
Viewed 50 times
2
-
1Suppose $X = \pm 1$... – wzzx Apr 28 '22 at 20:31
-
2Consider $X$ taking values $\pm 1$ with probabilities $a, b$ (say $a = b$). Then $X^2 = 1,$ and $\sigma(X^2) = {\varnothing, \Omega},$ so $E(X \mid X^2) = E(X) = 0,$ which does not equal $X$ at all. – William M. Apr 28 '22 at 20:32
-
Thanks to everyone. I see, it was something easy. – Iesus Dave Sanz Apr 28 '22 at 20:33
-
1For another set of counterexamples: if the distribution of $X$ is symmetric about $0$, $\mathbb E[X \mid X^2] = 0$. – Robert Israel Apr 28 '22 at 22:12