2
$\begingroup$

Another exercise with conditional expectation that I have problems with.

Let $\Omega=[-1,1]$, $\mathcal{F}=\mathcal{B}(\Omega)$, $\mathbb{P}=\frac{1}{2}\lambda$. Let X be a $\mathcal{F}$-measurable random variable, $\mathcal{G}=\sigma(\{A \in \mathcal{F}, A=-A\})$, with $-A=\{\omega \in \Omega: -\omega \in A\}$. Calculate $E[X|\mathcal{G}]$.

So first I got the hint that $X=\frac{X(\omega)+X(-\omega)}{2}+\frac{X(\omega)-X(-\omega)}{2}$.

Let $X_1=\frac{X(\omega)+X(-\omega)}{2}$, $X_2=\frac{X(\omega)-X(-\omega)}{2}$

So I consider $E[X\mid\mathcal{G}]=E[X_1\mid\mathcal{G}]+E[X_2\mid\mathcal{G}]$

$\mathcal{G}$ is the $\sigma$-algebra of all sets $\subset \Omega$ which are symmetric regarding 0.

So $\forall A\in \mathcal{G}$: $X(A)=X(-A) \Rightarrow E[X_2\mid\mathcal{G}]=0$ and $E[X_1\mid\mathcal{G}]=X$

Then I need help.

Thanks and good evening to all, Zitrone.

$\endgroup$
2
$\begingroup$

Once you get $\mathbb E[X_2\mid\mathcal G]=0$ and $\mathbb E[X_1\mid\mathcal G]=X_1$ you are done by linearity of conditional expectation.

To check this, pick $A\in\mathcal G$ and check that $\int_A X_2\mathrm d\mathbb P=0$ and for the second equality check that $X_1$ is $\mathcal G$-measurable.

$\endgroup$
  • $\begingroup$ Is there a mistake? $E[X_1\mid\mathcal{G}]=0$ but $E[X_1\mid\mathcal{G}]=X_1$, too? $\endgroup$ – lemontree Sep 3 '14 at 0:14
  • $\begingroup$ There was a confusion between $X_1$ and $X_2$. Fixed now. $\endgroup$ – Davide Giraudo Sep 3 '14 at 8:53
  • $\begingroup$ I know: $\forall A \in \mathcal{G}: X_1(A)=X \in \mathcal{B}(\mathbb{R})$and $X$ is $\mathcal{F}$-measurable, i.e. $\forall A\in \mathcal{B}(\Omega): X^{-1}(A) \in \mathcal{B}(\mathbb{R})$, but for the proof that $X_1$ is $\mathcal{G}$-measurable, i have to prove that $X_1^{-1}(A) \in \mathcal{G} \forall A \in \mathcal{B}(\mathbb{R})$. But I don't know how to prove that. $\endgroup$ – lemontree Sep 3 '14 at 14:27
  • $\begingroup$ You can write $X$ as a pointwise limit of simple functions, say $X_n$. Then $X_1(\omega)=\lim_n X_n(\omega)+X_n(-\omega)$ for each $\omega$. $\endgroup$ – Davide Giraudo Sep 3 '14 at 15:57
  • $\begingroup$ @DavideGiraudo I can't understand your hint to show that $X_1$ is $\mathcal G$-measurable. First: simple functions are not necessarily $\mathcal G$-measurable. Second: suppose for now they are. If the sum of $\mathcal G$-measurable r.v.'s is $\mathcal G$-measurable, then we are done, since we take the pointwise limit. The problem is that I don't think it's true, since if it was, then we could apply the very same argument to $X_2$ (because the function $x \mapsto -x$ is clearly $\mathcal G$-measurable). $\\$ Did I misunderstand your hint? $\endgroup$ – aerdna91 Jan 13 '15 at 15:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.