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  "body": "\n## Statement\n\nLet $\\lambda \u003e 0$ and let $X$ have the [[Poisson distribution|0VEFSSKX18]] with parameter $\\lambda$. Then\n$$E[X] = \\sum_{k=0}^{\\infty} k \\, e^{-\\lambda} \\frac{\\lambda^k}{k!} = \\lambda.$$\n\n## Remarks\n\nProof: the term with $k = 0$ is zero. For $k \\ge 1$, $k \\lambda^k / k! = \\lambda \\cdot \\lambda^{k-1}/(k-1)!$. So the sum is $\\lambda e^{-\\lambda} \\sum_{j \\ge 0} \\lambda^j / j! = \\lambda e^{-\\lambda} e^{\\lambda} = \\lambda$.\nThe variance of $X$ is also $\\lambda$.\nSo the parameter $\\lambda$ is the mean number of events.\n\n## Explanations\n\n### L1 human\n\nThe parameter of a Poisson distribution is its average.\nIf a shop gets on average $4$ customers per hour, a Poisson model of the number of customers in one hour uses $\\lambda = 4$.\n",
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