# What does it mean for a pmf to be a function of a random variable?

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?

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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 $k$, $$\mathrm E(X(X-1)\cdots(X-k+1))=\mathrm E(Y^k).$$

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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" $P(X)$;

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

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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.