I have problems with the following exercise:
Let $\Omega=[-\frac{1}{3},\frac{1}{3}]$, $\mathcal{F}=\mathcal{B}(\Omega)$ the Borel-$\sigma$-algebra on $\Omega$ and P the Lebesgue-measure.
Let $X(\omega)=\omega^2$, $Y(\omega)=\omega^3$ $\forall \omega \in \Omega$. Calculate $E[X|Y]=E[X|\sigma(Y)]$ and $E[Y|X]$.
My thoughts:
Let $Z:=E[E[X|Y]$
$E[Z]=E[X] \iff$ $\int_{A}Zd\mathbb{P} = \int_AXd\mathbb{P}$ $\forall A \in \mathcal{F}$
$\Rightarrow$ $Z=X$ $a.s.$ $\Rightarrow$ $E[X|Y]=Y^\frac{2}{3}$?
E[Y|X] analog
Note that $X=\sqrt[3]{Y^2}$ hence $X$ is $\sigma(Y)$-measurable and $E(X\mid Y)=$ $____$.
In the other direction, assume you are given $X(\omega)$, what can you say about $\omega$? Use the answer to show that $E(Y\mid X)=0$.