What is the expected value of an Exponential r.v. raised to a Poisson r.v.?

Question

Let $X \sim \operatorname{Exp}(\lambda)$ and $Y \sim \operatorname{Poisson}(\mu)$, where $X$ and $Y$ are independent and $\mu < \lambda$. Compute $\mathop{\mathbb{E}}[X^Y]$ using conditional expectation. We are given that the expectation of the moment generation function ($\mathop{\mathbb{E}}[e^{tX}]$) of $X$ is $\frac{\lambda}{\lambda - t}$, though I'm not sure where to use this.

So far, what I've been able to do is $\mathop{\mathbb{E}}[X^Y] = \mathop{\mathbb{E}}[X^Y|Y=y]$ for all $y$. But I'm not sure how to proceed as there are both continuous and discrete distributions here.

Answer 1

A correct starting point is instead: $$\mathop{\mathbb{E}}[X^Y] = \sum_{y=0}^\infty \mathop{\mathbb{E}}[X^Y|Y=y] \mathop{\mathbb{P}}[Y=y] = \sum_{y=0}^\infty \mathop{\mathbb{E}}[X^y] \mathop{\mathbb{P}}[Y=y] $$

Answer 2

$\mathbb{E}[X^Y] = \sum_{y=0}^\infty\mathbb{E}[X^Y|Y=y]\mathbb{P}(Y=y)$

$ = \sum_{y=0}^\infty\mathbb{E}[X^y] {\mu^y e^{-\mu} \over y!}$

$ = e^{-\mu} \left( 1 + \mathbb{E}[X]\mu + \mathbb{E}[X^2] {\mu^2 \over 2!} + \mathbb{E}[X^3] {\mu^3 \over 3!} + \ldots\right)$

You are given that $\mathbb{E}[e^{tX}] = {\lambda \over \lambda - t}$

Write $\mathbb{E}[e^{\mu X}] = \mathbb{E}[1 + \mu X + {(\mu X)^2 \over 2!} + {(\mu X)^3 \over 3!} + \ldots]$

$ = 1 + \mathbb{E}[X]\mu + \mathbb{E}[X^2] {\mu^2 \over 2!} + \mathbb{E}[X^3] {\mu^3 \over 3!} + \ldots = {\lambda \over \lambda - \mu}$

Plugging this back into the first equation you get:

$\mathbb{E}[X^Y] = e^{-\mu} {\lambda \over \lambda - \mu}$