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Suppose $X$ and $I$ and independent, $X$ has a standard normal distribution and $I$ take values $1$ and $-1$ with equal probabilities.

Let $Y = IX$. How would I find the distribution of $Y$ and $X+Y$? and $E(Y|X)$, $E(Y^2|X)$?

Thanks so much!

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Multiplying I has 1/2 chance of mirroring X about 0. So, symmetric distributions about $x=0$ are still themselves. Y follows the same normal distribution. –  FrenzY DT. Aug 21 '12 at 11:16

1 Answer 1

If $U$ and $V$ are independent, and if $F$ is a positive (or bounded) measurable function on $\mathbb{R}^2$, then $E(F(U,V) | U) = G(U)$, where $G(u) = E(F(u,V))$.

It can help you to compute $E(Y|X)$ in your case. (for $E(Y^2|X)$, it's easy since $Y^2 = X^2$ which is $\sigma(X)$-measurable)

For the distribution of $Y$, just write the event $\{IX \in A\}$ as the disjoint union of the events $\{I = -1, X \in - A\}$ and $\{I = 1, X \in A\}$. Then use independence.

Same kind of reasoning for $X + Y = (1+I)X$.

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Maybe you should rename the variables in your first sentence, since the $X,Y$ in the question are not independent. –  Nate Eldredge Aug 21 '12 at 12:50
    
My first sentence is a general result, completely independant from the notation in the question. I can rename them if you think it can avoids confusion ... –  Ahriman Aug 21 '12 at 12:57

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