Index: The Book of Statistical Proofs ▷ General 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.

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