# What is the PDF of a product of a continuous random variable and a discrete random variable?

Let $N$ be a discrete random variable which takes values in [0, ..., M], M > 0, with known PDF $P(N=n)$. Let also the continuous random variable $Z = \sum_{i=1}^{N}X_{i}$ as the sum of i.i.d. $X_{i}$ continuous variables with known and same PDF $f_{X_{i}}(x_{i}) = f_{X}(x), i=1\ldots,N$. The joint PDF of $N$ and $Z$ is $f_{Z,N}(z,n) = f_{Z|N}(z|n) P(N=n)$. What is the $f_{Y}(y) = f_{NZ}(nz)$ (i.e., Y = NZ)? Thanks!

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## 1 Answer

First of all, a continuous and a discrete random variable don't have a joint PDF, i.e. their joint distribution is not absolutely continuous with respect to $2$-dimensional Lebesgue measure.

The cdf of $Y = NZ$ is $F_Y(y) = P(NZ \le y) = \sum_{n=0}^m P(N=n) P(nZ \le y | N=n)$. Note that if $N=0$ has nonzero probability, so will $Y=0$, so $Y$ doesn't have a pdf either (it has a mixed distribution with a discrete part and a continuous part).

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Thanks a lot! The $P(nZ \leq y | N=n)$ is $P(Z \leq \frac{y}{n}|N=n)$ with $n >0$, thus, $P(nZ \leq y | N=n) = F_{Z}(\frac{y}{n})$? Is this correct? – Christos Feb 27 '13 at 22:00
Yes, for $n \ne 0$ it is. On the other hand, $P(0Z \le y | N = 0) = 1$ for $y \ge 0$, $0$ for $y < 0$. – Robert Israel Feb 27 '13 at 23:09
Thanks a lot .... – Christos Feb 28 '13 at 14:31
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