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.


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.

  • $\begingroup$ 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

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