Tail sum formula for expectation

Statement

Let $X$ be a random variable with values in $[0, \infty)$. Then $$E[X] = \int_0^\infty P(X > t) \, dt,$$ with both sides in $[0, \infty]$. If $X$ takes values in $\{0, 1, 2, \dots\}$, then $$E[X] = \sum_{n=1}^{\infty} P(X \ge n).$$

Remarks

The proof writes $X = \int_0^\infty 1_{\{X > t\}} \, dt$ and exchanges expectation and integral by Tonelli's theorem. In terms of the cumulative distribution function $F$, the first formula reads $E[X] = \int_0^\infty (1 - F(t)) \, dt$. For $X \sim \operatorname{Exp}(\lambda)$, the formula gives $E[X] = \int_0^\infty e^{-\lambda t} \, dt = 1/\lambda$.

Explanations

L1 human

You can compute an average by adding the chances that $X$ passes each level. For a count $X$, each $n \le X$ adds $1$ to $X$, and it does so with probability $P(X \ge n)$.