Index: The Book of Statistical ProofsGeneral Theorems ▷ Probability theory ▷ Expected value ▷ Non-negative random variable

Theorem: Let $X$ be a non-negative random variable. Then, the expected value of $X$ is

$\label{eq:mean-cdf} \mathrm{E}(X) = \int_{0}^{\infty} (1 - F_X(x)) \, \mathrm{d}x$

where $F_X(x)$ is the cumulative distribution function of $X$.

Proof: Because the cumulative distribution function gives the probability of a random variable being smaller than a given value,

$\label{eq:cdf-Pr-leq} F_X(x) = \mathrm{Pr}(X \leq x) \; ,$

we have

$\label{eq:cdf-Pr-geq} 1 - F_X(x) = \mathrm{Pr}(X > x) \; ,$

such that

$\label{eq:mean-cdf-s1} \int_{0}^{\infty} (1 - F_X(x)) \, \mathrm{d}x = \int_{0}^{\infty} \mathrm{Pr}(X > x) \, \mathrm{d}x$

which, using the probability density function of $X$, can be rewritten as

$\label{eq:mean-cdf-s2} \begin{split} \int_{0}^{\infty} (1 - F_X(x)) \, \mathrm{d}x &= \int_{0}^{\infty} \int_{x}^{\infty} f_X(z) \, \mathrm{d}z \, \mathrm{d}x \\ &= \int_{0}^{\infty} \int_{0}^{z} f_X(z) \, \mathrm{d}x \, \mathrm{d}z \\ &= \int_{0}^{\infty} f_X(z) \int_{0}^{z} 1 \, \mathrm{d}x \, \mathrm{d}z \\ &= \int_{0}^{\infty} [x]_{0}^{z} \cdot f_X(z) \, \mathrm{d}z \\ &= \int_{0}^{\infty} z \cdot f_X(z) \, \mathrm{d}z \\ \end{split}$

and by applying the definition of the expected value, we see that

$\label{eq:mean-cdf-s3} \int_{0}^{\infty} (1 - F_X(x)) \, \mathrm{d}x = \int_{0}^{\infty} z \cdot f_X(z) \, \mathrm{d}z = \mathrm{E}(X)$

which proves the identity given above.

Sources:

Metadata: ID: P103 | shortcut: mean-nnrvar | author: JoramSoch | date: 2020-05-18, 23:54.