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 $E(Y\mid X)=0$.

  • $\begingroup$ Hi Did, thanks. So $E[X|Y] = \sqrt[3]{Y^2}$.But I don't understand your second hint for $E[Y|X]$ $\endgroup$ – lemontree Aug 29 '14 at 23:29
  • $\begingroup$ Assume you know that $X(\omega)=x$ (that is, you are given $x$ but not $\omega$). What is $\omega$? $\endgroup$ – Did Aug 30 '14 at 6:51
  • $\begingroup$ Then $\omega = +- \sqrt{x}$. What does actually X is Y-measurable mean? That $\forall x \in X(\omega): X^{-1}(\omega) \in Y^{-1}(\omega)$? $\endgroup$ – lemontree Aug 30 '14 at 12:00
  • $\begingroup$ Yes, $\omega=\pm\sqrt{x}$. What does this suggest regarding $E(Y\mid X)$? // "Actually", $X$ $\sigma(Y)$-measurable here means that $X$ is a measurable function of $Y$. $\endgroup$ – Did Aug 30 '14 at 12:02
  • $\begingroup$ Then for every $x\in \Omega$: $(Y|X=x)$ $\in$ {$-\sqrt{x}^3,\sqrt{x}^3$} thus $E[Y|X]=0$. // But why can't you say: $Y=\sqrt{X}^3$ hence Y is X-measurable (since $x^\frac{3}{2}$ is a measurable function) so $E[Y|X]=\sqrt{X}^3$ $\endgroup$ – lemontree Aug 30 '14 at 12:23

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