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Suppose $X$ follows a Poisson distribution with mean $\lambda$ and $Y$ follows an exponential distribution with mean $\displaystyle\frac{1}{\mu}$. How can I find distribution of $Z=XY$?

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I will downvote this question. Please accept answers. You have been told several time in the past to do so. And, looking at the answers you have got, atleast a few of them fully address your questions. – user21436 Apr 25 '12 at 7:17
I'd just try to calculate it using $\mathbb P_Z(A) = \mathbb P(Z\in A) = \sum_n \mathbb P(XY \in A \mid X= n)\mathbb P(X = n)$ – martini Apr 25 '12 at 7:18
@Sasha Done. Thank You for letting me know. – user21436 Apr 26 '12 at 0:29
up vote 0 down vote accepted

Assuming $X$ and $Y$ are independent, the CDF for $z \ge 0$ is $$ F(z) = P(X Y \le z) = e^{-\lambda} + \sum_{x=1}^\infty e^{-\lambda} \frac{\lambda^x}{x!} (1 - e^{-\mu z/x}) = 1 - \sum_{x=1}^\infty \frac{\lambda^x}{x!} e^{-\lambda-\mu z/x}$$

The sum is very unlikely to be expressible in closed form.

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Consider $$ F_Z(z) = \mathbb{P}(Z \leqslant z) = \mathbb{P}(X Y \leqslant z) \stackrel{\color\maroon{\text{conditional tower}}}{=} \mathbb{E}\left( \mathbb{P}(X Y \leqslant z |X) \right) = \mathrm{e}^{-\lambda} \mathbf{1}_{z \geqslant 0} + \sum_{x=1}^\infty \frac{\mathrm{e}^{-\lambda}}{x!} F_Y\left( \frac{z}{x} \right) $$

Notice that $F_Z(z)$ has a jump discontinuity at the origin: $$ F_Z(0+) - F_Z(0-) = \mathrm{e}^{-\lambda} $$ meaning that $Z$ is not an absolutely continuous random variable. You can see this rather well of the cumulative function histogram of the simulated product: enter image description here

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