Survival function

Remarks

The function $S$ is nonincreasing and right-continuous, and $S(t) \to 0$ as $t \to \infty$. If $T$ has a density $f$, then $S(t) = \int_t^\infty f(u) \, du$ and $f = -S'$ wherever $f$ is continuous. The mean of $T$ is $E[T] = \int_0^\infty S(t) \, dt$. Some authors use $P(T \ge t)$ instead; the two agree at every $t$ with $P(T = t) = 0$.

Example: if $T$ has the exponential distribution with rate $\lambda > 0$, then $S(t) = e^{-\lambda t}$. Indeed $P(T > t) = \int_t^\infty \lambda e^{-\lambda u} \, du = e^{-\lambda t}$, and $\int_0^\infty e^{-\lambda t} \, dt = 1/\lambda$ is the mean.

Non-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. It satisfies $G(\pi/2) = 0 < e^{-\pi} = G(\pi)$, so it is not nonincreasing, while $P(T > t)$ can only decrease as $t$ grows.