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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, $\begin{eqnarray*} E(\xi\ |\ \xi I_{\{|\xi|\leq A\}}) &=& E(\xi I_{\{|\xi|\leq A\}}\ |\ \xi I_{\{|\xi|\leq A\}}) + E(\xi I_{\{|\xi| > A\}}\ |\ \xi I_{\{|\xi|\leq A\}})\\ &=& \xi I_{\{|\xi|\leq A\}}. \end{eqnarray*}$

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up vote 2 down vote accepted

Let $\xi_A=\xi\,\mathbf 1_{\{|\xi|\leqslant A\}}$. By definition, $\mathrm E(\xi\mid\xi_A)=u(\xi_A)$ for some measurable function $u$ such that, for every Borel set $B$, $\mathrm E(\xi:\xi_A\in B)=\mathrm E(u(\xi_A):\xi_A\in B)$.

  • For every $x\ne0$ such that $|x|\leqslant A$, $[\xi_A=x]=[\xi=x]$ hence $u(x)=x$.
  • For $x=0$, $[\xi_A=0]=[\xi=0]\cup[|\xi|\gt A]$ and the distribution of $\xi$ is symmetric hence the best predictor of $\xi$ on $[\xi_A=0]$ is $0$.
  • For every $x$ such that $|x|\gt A$, $[\xi_A=x]=\varnothing$ hence the value of $u(x)$ is irrelevant.

To sum up, a solution is $\mathrm E(\xi\mid\xi_A)=\xi_A$ almost surely.

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Thanks very much! – Eastsun Jun 11 '12 at 16:06

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