Index: The Book of Statistical ProofsProbability Distributions ▷ Univariate continuous distributions ▷ Beta distribution ▷ Mean

Theorem: Let $X$ be a random variable following a beta distribution:

$\label{eq:beta} X \sim \mathrm{Bet}(\alpha, \beta) \; .$

Then, the mean or expected value of $X$ is

$\label{eq:beta-mean} \mathrm{E}(X) = \frac{\alpha}{\alpha + \beta} \; .$

Proof: The expected value is the probability-weighted average over all possible values:

$\label{eq:mean} \mathrm{E}(X) = \int_{\mathcal{X}} x \cdot f_X(x) \, \mathrm{d}x \; .$ $\label{eq:beta-pdf} f_X(x) = \frac{1}{\mathrm{B}(\alpha, \beta)} \, x^{\alpha-1} \, (1-x)^{\beta-1}, \quad 0 \leq x \leq 1$

where the beta function is given by a ratio gamma functions:

$\label{eq:beta-fct} \mathrm{B}(\alpha, \beta) = \frac{\Gamma(\alpha) \cdot \Gamma(\beta)}{\Gamma(\alpha+\beta)} \; .$

Combining \eqref{eq:mean}, \eqref{eq:beta-pdf} and \eqref{eq:beta-fct}, we have:

$\label{eq:beta-mean-s1} \begin{split} \mathrm{E}(X) &= \int_{0}^{1} x \cdot \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha) \cdot \Gamma(\beta)} \, x^{\alpha-1} \, (1-x)^{\beta-1} \, \mathrm{d}x \\ &= \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)} \cdot \frac{\Gamma(\alpha+1)}{\Gamma(\alpha+1+\beta)} \int_{0}^{1} \frac{\Gamma(\alpha+1+\beta)}{\Gamma(\alpha+1) \cdot \Gamma(\beta)} \, x^{(\alpha+1)-1} \, (1-x)^{\beta-1} \, \mathrm{d}x \; . \end{split}$

Employing the relation $\Gamma(x+1) = \Gamma(x) \cdot x$, we have

$\label{eq:beta-mean-s2} \begin{split} \mathrm{E}(X) &= \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)} \cdot \frac{\alpha \cdot \Gamma(\alpha)}{(\alpha+\beta) \cdot \Gamma(\alpha+\beta)} \int_{0}^{1} \frac{\Gamma(\alpha+1+\beta)}{\Gamma(\alpha+1) \cdot \Gamma(\beta)} \, x^{(\alpha+1)-1} \, (1-x)^{\beta-1} \, \mathrm{d}x \\ &= \frac{\alpha}{\alpha+\beta} \int_{0}^{1} \frac{\Gamma(\alpha+1+\beta)}{\Gamma(\alpha+1) \cdot \Gamma(\beta)} \, x^{(\alpha+1)-1} \, (1-x)^{\beta-1} \, \mathrm{d}x \end{split}$

and again using the density of the beta distribution, we get

$\label{eq:beta-mean-s3} \begin{split} \mathrm{E}(X) &= \frac{\alpha}{\alpha+\beta} \int_{0}^{1} \mathrm{Bet}(x; \alpha+1, \beta) \, \mathrm{d}x \\ &= \frac{\alpha}{\alpha+\beta} \; . \end{split}$
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Metadata: ID: P228 | shortcut: beta-mean | author: JoramSoch | date: 2021-04-29, 09:12.