I'm studying advanced probability theorem by myself and have encountered a exercise:
Let $A>0$ be a constant, $\xi$ be a $r.v.$ such that $E|\xi|<\infty$ and
$P(\xi\leq x) = P(-\xi\leq x),\quad x\in\mathbb{R}$
Compute the conditional expectation $E(\xi\ |\ \xi I_{\{|\xi|\leq A\}})$.
I have no idea of this problem.
Any help would be appreciated.
I written a solution (more mathematical in my opinion) follow did's. Tell me if it has anything wrong.
For any $B\in \mathscr{B}_{\mathbb{R}}$, we have
- If $0\in B$, then $\{\xi I_{\{|\xi|\leq A\}}\in B\} = \{\xi\in B,|\xi|\leq A\}\cup\{|\xi|>A\}$. Hence
$\begin{eqnarray*} & &\int_{\{\xi I_{\{|\xi|\leq A\}}\in B\}}\xi I_{\{|\xi|> A\}} dP \\ &=& \int_{ \{\xi\in B,|\xi|\leq A\}}\xi I_{\{|\xi|> A\}} dP +\int_{\{|\xi|>A\}}\xi I_{\{|\xi|> A\}} dP\\ &=& 0. \end{eqnarray*}$
The last integral is zero due to the symmetric of $\xi$.
- If $0\notin B$, it is easy to see that
$ \int_{\{\xi I_{\{|\xi|\leq A\}}\in B\}}\xi I_{\{|\xi|> A\}} dP = 0. $
Consequently, we have $E(\xi I_{\{|\xi|> A\}}\ |\ \xi I_{\{|\xi|\leq A\}}) = 0, a.s.$. Therefor, $\begin{eqnarray*} E(\xi\ |\ \xi I_{\{|\xi|\leq A\}}) &=& E(\xi I_{\{|\xi|\leq A\}}\ |\ \xi I_{\{|\xi|\leq A\}}) + E(\xi I_{\{|\xi| > A\}}\ |\ \xi I_{\{|\xi|\leq A\}})\\ &=& \xi I_{\{|\xi|\leq A\}}. \end{eqnarray*}$