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Lets' $p$ and $q$ are probability mass functions of two discrete random variables. I need examples of functions $F(p,q)$ that $r = F(p,q)$ and $r$ is a probability mass function for some random variable. Thank you.

Edit: I mean that $r$ should be probability mass function for arbitrary distributions $p$ and $q$.

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The question does not make sense. And sounds similar to one asked a few days ago. – Xi'an Feb 21 '12 at 8:58
by the way people at math stack exchange upvote my question link – ashim Feb 21 '12 at 14:47
Do you by any chance mean that $r$ is a probability mass function on the distributions $p$ and $q$ themselves? – jbowman Feb 21 '12 at 15:15
also at – Henry Feb 21 '12 at 17:38

The question is why? The most simple example is to take $F(p,q)=p$. This can be generalised to $F(p,q)=\lambda p+(1-\lambda)q$, where $\lambda\in[0,1]$. The resulting $r$ is then called mixture distribution.

Another not interesting example would be to totally ignore $p$ and $q$ let $F$ give any probability mass function you like. Without any additional restrictions this question can become an exercise in futility.

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This question is rather vague so I'm not sure this is what you're looking for, but I have one example that many people may not be familiar with:

Let $p$ be a density corresponding to a distribution symmetric around 0 and let $G$ be the CDF of a distribution symmetric around 0. Then

$$ h(x) = 2 p(x) G \left( w(x) \right) $$

is a density for any odd function $w(\cdot)$.

To see why this is true let $X$ have CDF $G$ and let $Y$ have density $p$. Then it follows from symmetry and the fact that $w(\cdot)$ is an odd function that $\frac{1}{2} = P(X - w(Y) \leq 0)$. By the law of total expectation it follows that

$$ 1 = 2 \cdot E_{Y}[P(X-w(Y) \leq 0)] = 2\cdot E_{Y}[P(X \leq w(Y))] = 2\cdot\int_{-\infty}^{\infty} p(y)G \left(w(y)\right) dy $$

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Let $P \sim$ Poisson($\lambda_1$) & $Q \sim$ Poisson($\lambda_2$). P and Q are independent.

$G(P,Q) = P+Q=R \sim $ Poisson($ \lambda_1+ \lambda_2$).

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I am not sure this deals with the "probability mass function" part of the question. You might need to translate $R=P+Q$ into something like $r(n) = \sum_i p(i) q(n-i)$ – Henry Feb 21 '12 at 8:20
I didn't get your explanation. – vinux Feb 21 '12 at 9:44

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