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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 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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