Assume we have a probability space with $\Omega = [-\frac{1}{2}, \frac{1}{2}], \; \mathcal{F} = \mathcal{B}([-\frac{1}{2}, \frac{1}{2}])$, P being the Lesbegue measure and two random variables
given by
$$
X(\omega) = \omega^2 \; and \; Y(\omega) = \omega^3
$$
How do you calculate $E[Y|X] \; and \; E[X|Y]$ in this specific case. I started by calculating the densites of both random varibales which should be $f_X = \frac{1}{2\sqrt{x}}$ and
$f_Y = \frac{x^{-\frac{2}{3}}}{3}$ unless I didn't make a mistake there. But now I'm stuck trying to calculate the conditional densites of $f_{X|Y}$ and $f_{Y|X}$.
Is an explicit calculation the right way to do it in this case? Or is there another way to obtain the answer by making use of the properties of the conditional expectation? Thanks in advance for any help with this problem.
Best Answer
Hints: $\omega \to \omega^{3}$ is a homemomorphism so $\sigma (Y)$ is nothing but the Borel sigma algebra of $[-1,1]$. Hence $X$ is measurable w.r.t. $\sigma (Y)$ which shows that $E(X|Y)=X$.
$E(Y|X)=0$ by symmetry.