Need to confirm my thinking on a probability problem

[newbie]

Hi this is a homework problem, so I am not looking for an answer. Just want to confirm my thinking is right, because this seems too simple. We are given
\xi is a random variable following the standard normal distribution. Find E\left( {\xi |{\xi ^2}} \right).
It seem given {{\xi ^2}}>0 we only have two possibilities for \xi of opposite sign with equal probability. So E\left( {\xi |{\xi ^2}} \right)=0

Am I missing something???

[Macbi]

Hi this is a homework problem, so I am not looking for an answer. Just want to confirm my thinking is right, because this seems too simple. We are given
\xi is a random variable following the standard normal distribution. Find E\left( {\xi |{\xi ^2}} \right).
It seem given {{\xi ^2}}>0 we only have two possibilities for \xi of opposite sign with equal probability. So E\left( {\xi |{\xi ^2}} \right)=0

Am I missing something???

I don't think your missing anything. But that is a really weird question, maybe you should double check that you've read it correctly?

(Or maybe your professor doesn't like conditioning on events with probability zero, so if you answer the question they'll laugh at you because you should have written "undefined" or something dumb like that.)

[z4lis]

How was E(X|Y) defined for you?

[newbie]

Indirectly, assume S is a sigma algebra induced by the random variable {\xi ^2} then \int_A {\xi P\left( {d\omega } \right) = \int_A {E\left( {\xi |{\xi ^2}} \right)P\left( {d\omega } \right)} \;\forall A \in S}

[z4lis]

Right, so I believe it suffices to assume that A is of the form \xi^2 \in B for a measurable subset B of the real line. Now you want to use the symmetry of the normal distribution.

[newbie]

Yep, that is the way I was viewing it. Thanks for the confirmation.