Finding the pdf of $Z=XY$, where the joint pdf of $(X,Y)$ is known

I've been trying to figure this out for a while. Suppose $X$ and $Y$ are random variables with joint pdf (probability density function) $$f(x,y) = \frac{x+y}{15}, \quad\text{for }x = 0,1,2,\;\text{ and }\; y = 1,2.$$ Find the pdf for $Z = XY$.

So I started off by saying:

$$f_Z (z) = \mathbf{P}(XY \leq z) = \mathbf{P}\left(X \leq \frac{z}{Y}\right) = \int_0^2\int_2^1 \frac{x+y}{15} dy \, dx.$$

I'm a bit confused how to do this with two known limits. I'm not sure where they fit in.

• Welcome to MSE! This was a bit hard to parse/read. The site uses Mathjax, essentially "tex between dollar signs." I did some editing, but it is in an edit review queue. Maybe have a look later if I got what you wrote in plain english right. I kept the somewhat random use of "$X$" vs "$x$" in your equations as you had typed them, but note that usually "$X$" is a RV; "$x$" a realization. I warmly recommend having a brief look at this for any further submissions: meta.math.stackexchange.com/questions/5020/…. – gnometorule Feb 13 '13 at 6:43

Since $X$ and $Y$ are discrete random variables, the quickest road might be to enumerate the set $S$ of values of $Z=XY$ and to compute the probability that $Z=n$ for each $n$ in $S$.

Thus, $S = \{0,1,2,4\}$ and, for example, $[Z=0]$ is the disjoint union of $[X=0,Y=1]$ and $[X=0,Y=2]$ hence $\mathbb P(Z=0)=$ $____$.

• I'm just confused about what the limits of the integrals would be considering x=0,1,2, and y=1,2. how would I set up the integral for dydx? – Caty Feb 13 '13 at 8:07
• Your problem seems to concern discrete random variables (which take only a finite number of values, not values in a whole continuous set). Therefore integrals will not come into play at all. – Greg Martin Feb 13 '13 at 8:16
• What @GregMartin said. – Did Feb 13 '13 at 8:31

You only use integrals when the probability distribution is defined over a range of values rather than at a finite set of values. In other words, the distribution you've described is a probability mass function, but you've used a method suited towards a probability density function.

In order for integrals to be meaningful, your distribution would have to look something like this: $$f(x,y) = \begin{cases} \frac{x+y}{5} & 0\leq x \leq 2 \text{ and } 1 \leq y \leq 2\\ 0 & \text{ everywhere else} \end{cases}$$ If you'd like to solve the problem as stated, you should the method that Did has explained.

Supposing that your density function is for continuous random variables. e.g. $X = [0,2], Y = [0,3]$. Note that the range was changed so that the density function integrates to $1$.

Your confusion is that you forgot that you are no longer working in the $Z$ space, but rather in the $X$ space:

$$P(XY \le z) = P(X \le z/Y) = \int_0^{z/Y}f_X(x)\,dx,$$

where

$$f_X(x) = \int_0^3f_{X,Y}(x,y)\,dy.$$