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  "body": "\n## Statement\n\nLet $X_1, \\dots, X_n$ be [[independent|0XCE070V8B]] random variables with $X_i \\sim \\operatorname{Poisson}(\\lambda_i)$ and $\\lambda_i \u003e 0$. Then\n$$X_1 + \\dots + X_n \\sim \\operatorname{Poisson}(\\lambda_1 + \\dots + \\lambda_n).$$\n\n## Remarks\n\nProof with [[generating functions|0REMT01N36]]: $G_{X_i}(z) = e^{\\lambda_i (z - 1)}$, so the sum has $G(z) = e^{(\\lambda_1 + \\dots + \\lambda_n)(z - 1)}$. This is the generating function of $\\operatorname{Poisson}(\\lambda_1 + \\dots + \\lambda_n)$.\nA direct proof for $n = 2$ uses the binomial theorem: $\\sum_{k=0}^{m} e^{-\\lambda_1} \\frac{\\lambda_1^k}{k!} e^{-\\lambda_2} \\frac{\\lambda_2^{m-k}}{(m-k)!} = e^{-(\\lambda_1 + \\lambda_2)} \\frac{(\\lambda_1 + \\lambda_2)^m}{m!}$.\nConversely, given $X_1 + X_2 = m$, the conditional law of $X_1$ is $\\operatorname{Bin}(m, \\lambda_1 / (\\lambda_1 + \\lambda_2))$.\n\n## Explanations\n\n### L1 human\n\nIf calls come to two help desks as independent Poisson counts, the total number of calls is again a Poisson count. Its mean is the sum of the two means.\n",
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