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.

  • $\begingroup$ I know that this is probably a very silly question, but how did you know to do this to calculate the expectation? I guess I don't understand why we can set up that first equation that way - why does the expected value equal the integral of the random variables multiplied by the distribution of one of them? Is there a specific formula or idea that you're using here? (I'm sorry again for the dumb question) $\endgroup$ – poppy3345 Feb 18 '15 at 9:38
  • $\begingroup$ @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. $\endgroup$ – Math-fun Feb 18 '15 at 10:03
  • $\begingroup$ Thank you so much! I see now where that came from. $\endgroup$ – poppy3345 Feb 18 '15 at 19:12
  • $\begingroup$ Four small $x$ should read capital $X$. $\endgroup$ – Did Feb 19 '15 at 21:06
  • $\begingroup$ @Did Many thanks for the comment. I just corrected it. $\endgroup$ – Math-fun Feb 19 '15 at 21:07

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