{
  "aliases": [
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  "body": "\n## Statement\n\nLet $X$ be a random variable with values in $[0, \\infty)$. Then\n$$E[X] = \\int_0^\\infty P(X \u003e t) \\, dt,$$\nwith both sides in $[0, \\infty]$.\nIf $X$ takes values in $\\{0, 1, 2, \\dots\\}$, then\n$$E[X] = \\sum_{n=1}^{\\infty} P(X \\ge n).$$\n\n## Remarks\n\nThe proof writes $X = \\int_0^\\infty 1_{\\{X \u003e t\\}} \\, dt$ and exchanges expectation and integral by Tonelli's theorem.\nIn terms of the [[cumulative distribution function|0MXMS3AE13]] $F$, the first formula reads $E[X] = \\int_0^\\infty (1 - F(t)) \\, dt$.\nFor $X \\sim \\operatorname{Exp}(\\lambda)$, the formula gives $E[X] = \\int_0^\\infty e^{-\\lambda t} \\, dt = 1/\\lambda$.\n\n## Explanations\n\n### L1 human\n\nYou can compute an average by adding the chances that $X$ passes each level.\nFor a count $X$, each $n \\le X$ adds $1$ to $X$, and it does so with probability $P(X \\ge n)$.\n",
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