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  "body": "\n## Definition\n\nLet $T$ be a [[random variable|0PNRKMRC55]] with values in $[0, \\infty)$, and let $F$ be its [[cumulative distribution function|0H8H7ZTHW2]].\nThe *survival function* of $T$ is the function $S \\colon [0, \\infty) \\to [0, 1]$ given by\n$$S(t) = P(T \u003e t) = 1 - F(t).$$\n\n## Remarks\n\nThe function $S$ is nonincreasing and right-continuous, and $S(t) \\to 0$ as $t \\to \\infty$.\nIf $T$ has a density $f$, then $S(t) = \\int_t^\\infty f(u) \\, du$ and $f = -S'$ wherever $f$ is continuous.\nThe mean of $T$ is $E[T] = \\int_0^\\infty S(t) \\, dt$.\nSome authors use $P(T \\ge t)$ instead; the two agree at every $t$ with $P(T = t) = 0$.\n\nExample: if $T$ has the [[exponential distribution|0E99F51GVB]] with rate $\\lambda \u003e 0$, then $S(t) = e^{-\\lambda t}$.\nIndeed $P(T \u003e t) = \\int_t^\\infty \\lambda e^{-\\lambda u} \\, du = e^{-\\lambda t}$, and $\\int_0^\\infty e^{-\\lambda t} \\, dt = 1/\\lambda$ is the mean.\n\nNon-example: the function $G(t) = e^{-t} \\cos^2 t$ takes values in $[0, 1]$, has $G(0) = 1$ and tends to $0$, but it is not a survival function.\nIt satisfies $G(\\pi/2) = 0 \u003c e^{-\\pi} = G(\\pi)$, so it is not nonincreasing, while $P(T \u003e t)$ can only decrease as $t$ grows.\n",
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