Consider the discrete random variables $X$ and $Y$ whose PMFs are:

$\mathbb{P}_X(X=0) = 2/3$ $~~~~~~~~~~~~~~~~~\mathbb{P}_Y(Y=0) = 2/3$

$\mathbb{P}_X(X=2) = 1/6$ $~~~~~~~~~~~~~~~~~\mathbb{P}_Y(Y=2) = 1/6$

$\mathbb{P}_X(X=-2) = 1/6$ $~~~~~~~~~~~~~~\mathbb{P}_Y(Y=-2) = 1/6$

and the variable $Z = X\cdot Y$. Intuitively, the PMF of $Z$ is:

$\mathbb{P}_Z(Z=0) = 8/9$

$\mathbb{P}_Z(Z=4) = 1/18$

$\mathbb{P}_Z(Z=-4)= 1/18$

However, I can not find in the literature a procedure to determine the PMF of the product of two (independent) discrete random variables. Is there a general method for this? If so, please provide references.

  • 2
    $\begingroup$ You cannot determine the pmf of the product without knowing the joint pmf of $X$ and $Y$, that is, you need to know the probability that $X$ equals $i$ and simultaneously $Y$ equals $j$, for all choices of $i$ and $j$. $\endgroup$ Commented Oct 10, 2016 at 1:10
  • 2
    $\begingroup$ Givent that $X$ and $Y$ are independent it is possible to determine those joint probabilities right? $\endgroup$ Commented Oct 10, 2016 at 8:24
  • $\begingroup$ You're right: A quick Google search turns up accounts of distributions of products only for continuous random variables. (+1) $\endgroup$
    – whuber
    Commented Oct 10, 2016 at 16:28

1 Answer 1


If $X,Y$ are independent discrete random variables with supports $\cal X, \cal Y$ (i.e. places where they are non-zero, then $$P(XY=c) = \sum_{x \in \cal X, ~~y \in \cal Y \\\text{ s.t. } xy=c} P(X=x,Y=y) = \sum_{x \in \cal X,~~ y \in \cal Y\\ \text{ s.t. } xy=c} P(X=x)P(Y=y)$$.

The first equality is true for any discrete random variables. The second is true for independent ones.

  • $\begingroup$ Also, note that the support of $XY$ is $\{ xy: x \in \cal X, y \in \cal Y\}$, so these are the $c$'s where $P(XY=c)$ is non-zero. $\endgroup$
    – Batman
    Commented Oct 10, 2016 at 17:13
  • $\begingroup$ do you have any textbook or literature sources for the formula that you give? I'm having some trouble reconciling a few things, and it would be nice to see a more complete discussion. $\endgroup$
    – tel
    Commented Oct 18, 2016 at 18:05
  • $\begingroup$ Any introductory probability textbook which deals with discrete random variables should cover this (or present enough material to derive it). $\endgroup$
    – Batman
    Commented Oct 18, 2016 at 18:52
  • $\begingroup$ You'd think so, but I've looked through half a dozen textbooks since yesterday and I haven't been able to find it. They only cover the continuous case. Can you think of a specific reference that would be likely to have this material? $\endgroup$
    – tel
    Commented Oct 18, 2016 at 18:58
  • $\begingroup$ Bertsekas and Tsiklis' Probability or maybe courses.engr.illinois.edu/ece313/notes_webpage.html. All I've used is that $P(V \in A) = \sum_{v \in A} P(V=v)$ here with $V=XY$ and $A=\{(x,y): x \in \cal X, y \in \cal Y, xy=c\}$ and applied that result to get the first equality. The second equality is just the definition of independence. $\endgroup$
    – Batman
    Commented Oct 18, 2016 at 19:10

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