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  "body": "\n## Proof\n\nWe compute the probability mass function of the sum directly.\n\n1. Let $S = X_1 + \\dots + X_n$ and $\\Lambda = \\lambda_1 + \\dots + \\lambda_n$. Then $\\Lambda \u003e 0$.\n2. By the definition of the [[Poisson distribution|0VEFSSKX18]], $P(X_i = k) = e^{-\\lambda_i} \\lambda_i^k / k!$ for each integer $k \\ge 0$. These values add up to $1$, so $X_i \\in \\set{0, 1, 2, \\dots}$ almost surely.\n3. Let $N$ be the event that every $X_i$ is a nonnegative integer. By step 2, its complement is a finite union of null events, so $P(N) = 1$.\n4. On $N$, the sum $S$ is a nonnegative integer. So $S \\in \\set{0, 1, 2, \\dots}$ almost surely.\n5. Fix an integer $m \\ge 0$. Let $K_m$ be the set of $n$-tuples $(k_1, \\dots, k_n)$ of nonnegative integers with $k_1 + \\dots + k_n = m$. It is a finite set.\n6. Up to a null event, $\\set{S = m}$ is the disjoint union over $(k_1, \\dots, k_n) \\in K_m$ of the events $\\set{X_1 = k_1, \\dots, X_n = k_n}$. So $P(S = m) = \\sum_{K_m} P(X_1 = k_1, \\dots, X_n = k_n)$.\n7. By [[independence|0XCE070V8B]] with $B_i = \\set{k_i}$, each term is $\\prod_{i=1}^{n} P(X_i = k_i) = e^{-\\Lambda} \\prod_{i=1}^{n} \\lambda_i^{k_i} / k_i!$.\n8. Multiply and divide by $m!$. Then\n$$P(S = m) = \\frac{e^{-\\Lambda}}{m!} \\sum_{(k_1, \\dots, k_n) \\in K_m} \\frac{m!}{k_1! \\cdots k_n!} \\lambda_1^{k_1} \\cdots \\lambda_n^{k_n}.$$\n9. The fraction $m! / (k_1! \\cdots k_n!)$ is the [[multinomial coefficient|00K0RF9BFY]]. By the [[multinomial theorem|0XFC1VQYEK]], the sum equals $(\\lambda_1 + \\dots + \\lambda_n)^m = \\Lambda^m$.\n10. So $P(S = m) = e^{-\\Lambda} \\Lambda^m / m!$ for each integer $m \\ge 0$. With step 4, the [[probability mass function|0WF4EJP91W]] of $S$ is that of $\\operatorname{Poisson}(\\Lambda)$.\n11. Hence $X_1 + \\dots + X_n \\sim \\operatorname{Poisson}(\\lambda_1 + \\dots + \\lambda_n)$.\n\n## Idea\n\nThe sum equals $m$ exactly when the counts $k_1, \\dots, k_n$ add up to $m$. Independence turns each such case into a product of Poisson weights, and the multinomial theorem collects all the cases into $\\Lambda^m$.\n",
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