The conditional expectation of exp(XY) given Y

Question

$ 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] $ ?

Answer 1

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 $g(Y)=\exp(\frac{1}{2}\sigma^2Y^2)$.

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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 $N(y\sigma^2,\sigma^2)$.