# Compute $E(e^{XY}\mid\sigma(X))$, where $Y$ is exponentially distributed and independent of $X$

Suppose that $X,Y$ are independent random variables on $(\Omega, F, P)$. Assume there is a number $a <1$ such that $P(X \le a) = 1$. Also assume that $Y$ is exponentially distributed with mean one. Calculate the expected value of $[e^{XY} \ | \ \sigma(X)]$.

I'm really not sure where to even begin, here. We've been going over conditional expectations in class, but haven't really talked about any ways to actually calculate them. I see that X is a bounded random variable, and that since Y is exponentially distributed, $P(Y \in B) = \chi_B e^{-x}$ for any $B$ in the sigma-field of Y. I just don't know how to put any of these facts together to actually find the expectation.

If $Pr(X<\lambda)=1$ then \begin{align} E(e^{XY}|\sigma(X))&=\int_0^{\infty}e^{Xy}\lambda e^{-\lambda y} dy\\ &=\int_0^{\infty}e^{(X-\lambda)y}\lambda dy\\ &=\frac{\lambda}{\lambda-X} \end{align}
If $Pr(X<\lambda)<1$ then the conditional expectation does not exist, since $\sigma(X)$ includes realizations for $X$ for which the integral does not converge.
• @gesa This is not a bad question at all. Intuitively, when you condition on $X$ it is as if you know the value of $X$, hence it is not random anymore. Therefore the only "source of randomness" in $e^{XY}|\sigma(X)$ is through $Y$. Here we have independence therefore when we condition on $X$ the distribution of $Y$ remains unchanged. – Math-fun Feb 18 '15 at 10:03
• Four small $x$ should read capital $X$. – Did Feb 19 '15 at 21:06