3.1 The Discrete Case
Definition 3.1.1. Let \(X\) and \(Y\) be independent discrete random variables and \(Z = X+Y\). Then \[P(Z=z) = \sum _{x} P(X=x)\,P(Y=z-x),\] the sum running over all \(x\) for which both terms are positive. The function \(p_Z\) is called the convolution of \(p_X\) and \(p_Y\).
Proof. The event \(\{Z=z\}\) is the disjoint union over \(x\) of the events \(\{X=x\}\cap \{Y=z-x\}\), so \[P(Z=z) = \sum _x P\left (X=x,\ Y=z-x\right ) = \sum _x P(X=x)P(Y=z-x),\] the last step by independence. □
Example 3.1.2. If \(X\sim \) POI\((\mu )\) and \(Y\sim \) POI\((\lambda )\) are independent, derive the distribution of \(Z = X+Y\).
Solution. Since \(X\) and \(Y\) are non-negative, \(x\) runs from \(0\) to \(z\): \[P(Z=z) = \sum _{x=0}^{z}\frac {e^{-\mu }\mu ^{x}}{x!}\cdot \frac {e^{-\lambda }\lambda ^{z-x}}{(z-x)!} = \frac {e^{-(\mu +\lambda )}}{z!}\sum _{x=0}^{z}\frac {z!}{x!\,(z-x)!}\,\mu ^{x}\lambda ^{z-x}.\] The sum is a binomial expansion: \[\sum _{x=0}^{z}\binom {z}{x}\mu ^{x}\lambda ^{z-x} = (\mu +\lambda )^{z},\] so \[P(Z=z) = \frac {e^{-(\mu +\lambda )}(\mu +\lambda )^{z}}{z!},\qquad z=0,1,2,\dots \] which is the probability function of POI\((\mu +\lambda )\). Hence \(Z\sim \) POI\((\mu +\lambda )\).
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