Probability theory and stochastic processes Proposition

Sum of independent Poisson random variables

Builds on Poisson distribution Generating function of a sum of independent random variables Independence of random variables

Statement

Let $X_1, \dots, X_n$ be independent random variables with $X_i \sim \operatorname{Poisson}(\lambda_i)$ and $\lambda_i > 0$ . Then $X_1 + \dots + X_n \sim \operatorname{Poisson}(\lambda_1 + \dots + \lambda_n).$

Remarks

Proof with generating functions: $G_{X_i}(z) = e^{\lambda_i (z - 1)}$ , so the sum has $G(z) = e^{(\lambda_1 + \dots + \lambda_n)(z - 1)}$ . This is the generating function of $\operatorname{Poisson}(\lambda_1 + \dots + \lambda_n)$ . A direct proof for $n = 2$ uses the binomial theorem: $\sum_{k=0}^{m} e^{-\lambda_1} \frac{\lambda_1^k}{k!} e^{-\lambda_2} \frac{\lambda_2^{m-k}}{(m-k)!} = e^{-(\lambda_1 + \lambda_2)} \frac{(\lambda_1 + \lambda_2)^m}{m!}$ . Conversely, given $X_1 + X_2 = m$ , the conditional law of $X_1$ is $\operatorname{Bin}(m, \lambda_1 / (\lambda_1 + \lambda_2))$ .

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Generating function of a sum of independent random variables statement
Sum of independent Poisson random variables statement

Tag	`0YH1M56SBN`
URI	`m2://0YH1M56SBN`
Kind	statement, proposition
Writing	W0 stub
MSC	60E05
Status	This copy of the site has no build data. Run `m2 status 0YH1M56SBN`.
Uses	Poisson distribution, Generating function of a sum of independent random variables, Independence of random variables

Sum of independent Poisson random variables

Statement

Remarks

Reading path

Further reading

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