Conditional expectation of symmetric Sigma algebra

Question

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.

Answer 1

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.