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Important properties of Poisson random variables Addition of two Poisson random variables (Example 4.5) Let X and Y be independent Poisson random variables with parameters λ and µ respectively. Then fX (s) = eλ(s−1) and fY (s) = eµ(s−1) , from Example 4.1. Then Theorem 4.6 tells us that fX+Y (s) = eλ(s−1) eµ(s−1) = e(λ+µ)(s−1) . Using Theorem 4.2 above, we can now conclude that X + Y is a Poisson random variable with mean λ + µ. Thinning of a Poisson random variable (Example 6.1) Let N be a Poisson random variable with mean λ, and let {Xn ; n ∈ N} be a sequence of independent (of each other and of N ) Bernoulli random variables with parameter p, i.e. P (Xn = 1) = p and P (Xn = 0) = 1 − p. Now consider S= N X Xn . n=1 We can think of this as taking a Poisson distributed number of objects, each of which has some property with probability p, and selecting only those which have this property. The p.g.f. of N is fN (s) = eλ(s−1) by Example 4.1 and the p.g.f. of each Xn is 1 − p + ps. So using Lemma 6.1, the p.g.f. of S is eλ(1−p+ps−1) = eλp(s−1) which we recognise as the p.g.f. of a Poisson random variable with mean λp. We recall (Theorem 4.2) that the p.g.f. determines the distribution, so we have shown that the thinned Poisson random variable S also has a Poisson distribution. 1