Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm trying to understand what it means for a discrete random variable to have a probability mass function (pmf) that is a function of another random variable. For example, one homework problem of mine starts with "suppose $X$ has Poisson distribution with parameter $Y$, where $Y$ has Poisson distribution with parameter $\mu$." Does this mean that to determine the probability that $X=0$, one would first "run" $Y$ to obtain a value, then plug that value into $X$'s pmf? If so, then to determine the probability that $X=1$, would we have to run $Y$ again?

share|cite|improve this question
up vote 3 down vote accepted

This means that the probability that $X=k$ knowing $Y$ is a given $p(k,Y)$, hence $\mathrm P(X=k)=\mathrm E(p(k,Y))$. In your case $p(k,y)=\mathrm e^{-y}y^k/k!$ hence $$ k!\,\mathrm P(X=k)=\mathrm E(\mathrm e^{-Y}Y^k)=\sum\limits_{n\ge0}\mathrm e^{-\mu}\frac{\mu^n}{n!}\mathrm e^{-n}n^k. $$ One sees that $$ \mathrm P(X=0)=\sum\limits_{n\ge0}\mathrm e^{-\mu}\frac{(\mu/\mathrm e)^n}{n!}=\mathrm e^{-\mu(1-1/\mathrm e)}. $$ Likewise, $$ \mathrm P(X=1)=\sum\limits_{n\ge1}\mathrm e^{-\mu}\frac{(\mu/\mathrm e)^n}{(n-1)!}=\mathrm e^{-\mu(1-1/\mathrm e)}\mu/\mathrm e. $$ Expectations are easier to compute than $\mathrm P(X=k)$ for a general $k$. For example, noting that the expectation of a Poisson random variable is its parameter, one gets directly $$ \mathrm E(X)=\mathrm E(Y)=\mu. $$ Likewise, for any positive integer $k$, $$ \mathrm E(X(X-1)\cdots(X-k+1))=\mathrm E(Y^k). $$

share|cite|improve this answer
So is the way to think of it, in the terms of my question, that we "run" $Y$ once, and that determines $X$'s pdf? (Though ahead of time, we don't know what will happen when we run $Y$, so your analysis is appropriate.) – Quinn Culver Oct 25 '11 at 22:37
One does not run $Y$ to get the PDF of $X$, whatever that means. If one wants to estimate the PDF of $X$ by repeating an experiment, one should generate i.i.d. random variables $Y_k$, then use each $Y_k$ to generate $X_k$ and, for example, estimate $P(X=15)$ by the proportion of indices $k$ such that $X_k=15$. – Did Oct 25 '11 at 22:42
Generally, $\Pr(X=k)$ would be the $k$th moment of a Poisson distribution with expectation $\mu/e$. This would be the $k$th-degree [Touchard polynomial]( evaluated at $/\mu/e$. – Michael Hardy Oct 25 '11 at 23:27
Correction: It would be $e^{-\mu}/(k!)$ times the said $k$th moment. (If it were just the $k$th moment, then we'd have probabilities greater than 1.) – Michael Hardy Oct 25 '11 at 23:42

This can also be stated in terms of conditional probabilities. You know that, if $Y$ is given/known, $X$ follows a Poisson distribution with that particular value as parameter. You can write this fact down as:

$$P(X=x \; |\; Y=y) = e^{-y} \; \frac{\; y^x}{x!}$$

Then, the joint probability is given by $P(X \; Y ) = P ( X \; | \; Y) \; P(Y) $, where $P(Y)$ is another Poisson with paramenter $\mu$; and from this you can compute the "marginal" $P(X)$;

$$P(X = x ) = \sum_{y=0}^{\infty} P(X=x \; |\; Y=y) \; P(Y=y) $$

share|cite|improve this answer

Definitely it ought to say that the conditional distribution of $X$ given $Y$ is a Poisson distribution with expected value $Y$. Without the word "conditional", one could take a statement about the distribution of $X$ to be about its marginal (i.e. "unconditional") distribution, and that would be wrong.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.