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$ X \sim N(0;\sigma^2) $ and Y is a random variable that is independent of X. How do I compute $ \mathbb{E}[e^{XY}|Y] $ ?

Gono
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SidiAli
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  • How do you get from $ E[e^{XY}|Y] $ to $ e^YE[e^X] $ ? You can't isolate Y from X like you did because it's not a sum. It's $ e^{XY} $ not $ e^Xe^Y $. – SidiAli Nov 04 '17 at 13:53
  • You are right. I was wrong. Then it just $ E[e^{XY}|Y=y]=E[e^{yX}]$ this is just moment generating function of $X$ evaluated at $y$. – Boby Nov 04 '17 at 13:57
  • I don't know what a moment generating function is. Is there a way to just compute that conditional expectation without doing that? – SidiAli Nov 04 '17 at 14:11

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Let $f(x,y)=e^{xy}$ and $g(y)=E[f(X,y)]$. By the Independence Lemma, $$ E[f(X,Y)|Y]=g(Y). $$ So let's compute $g(y)$: $$ g(y)=E[f(X,y)]=E[e^{yX}]=\exp(\frac{1}{2}\sigma^2y^2) $$ where we have noted $E[e^{yX}]$ is just the moment generating function of $X$ evaluated at $y$. The final answer is then $g(Y)=\exp(\frac{1}{2}\sigma^2Y^2)$.

To derive $E[e^{yX}]$ note that \begin{align*} E[e^{yX}]&=\int(2\pi\sigma^2)^{-1/2}\exp(yx)\exp(-\frac{x^2}{2\sigma^2})dx\\ &=\exp(\frac{1}{2}\sigma^2y^2)\int(2\pi\sigma^2)^{-1/2}\exp(-\frac{(x-y\sigma^2)^2}{2\sigma^2})dx\\ &=\exp(\frac{1}{2}\sigma^2y^2). \end{align*} The integral in the second line evaluates to $1$ because $x\mapsto(2\pi\sigma^2)^{-1/2}\exp(-\frac{(x-y\sigma^2)^2}{2\sigma^2})$ is the density function of $N(y\sigma^2,\sigma^2)$.

Kim Jong Un
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  • The rest is clear. But I'm having a problem understanding why $ E[f(X,Y)|Y]=g(Y) $ or $ E[e^{XY}|Y]= E[e^{XY}] $. X is independent of Y so it would make sense for me to say $ E[e^{X}|Y]= E[e^{X}] $ because $ e^X $ is independent of $Y$. But is $ e^{XY} $ independent of Y? – SidiAli Nov 04 '17 at 15:05
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    @SidiAli I don't think I implied anywhere that $e^{XY}$ is independent of $Y$. Where did you see it? – Kim Jong Un Nov 04 '17 at 15:14
  • You didn't say that. But I'm assuming that you can't say that $ E[Z|Y]=E[Z] $ unless Z is independent of Y. So why is $ E[e^{XY}|Y]=E[e^{XY}] $ if $ e^{XY} $ is not independent of $Y$ ? – SidiAli Nov 04 '17 at 15:53
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    @SidiAli I never claimed $E[e^{XY}|Y]=E[e^{XY}]$ either. $E[e^{XY}]$ is nonrandom whereas the answer for $E[e^{XY}|Y]$ I gave you is $g(Y)$ is random because it is a function of $Y$. I suggest you read the linked thread about the Independence Lemma (the statement at least) carefully. – Kim Jong Un Nov 04 '17 at 16:57