In this page, we will find the mean and variance of the beta distribution. The proof requires the following result that we state without proof.

A useful integral

For any constants \(a \gt 0\) and \(b \gt 0\),

\[ \int_0^1{x^{a - 1} (1 - x)^{b - 1}} dx \;\;=\;\; \frac{\Gamma(a) \Gamma(b)}{\Gamma(a + b)} \]

(This result can be used to prove that the beta distribution's pdf integrates to 1.)

Mean and variance of beta distribution

If a random variable, \(X\), has a beta distribution with pdf

\[ f(x) \;\;=\;\; \begin{cases} \dfrac {\Gamma(\alpha + \beta) }{\Gamma(\alpha)\Gamma(\beta)} x^{\alpha - 1} (1 - x)^{\beta - 1}& \text{if }0 \lt x \le 1 \\ 0 & \text{otherwise} \end{cases} \]

its mean and variance are

\[ E[X] \;=\; \frac{\alpha}{\alpha + \beta} \spaced{and} \Var(X) \;=\; \frac{\alpha\beta}{(\alpha + \beta)^2(\alpha + \beta + 1)} \]
\[ \begin{align} E[X] \;\;&=\;\; \int_0^1{x \times \frac {\Gamma(\alpha + \beta) }{\Gamma(\alpha)\Gamma(\beta)} x^{\alpha - 1} (1 - x)^{\beta - 1}} dx \\[0.2em] &=\;\; \frac {\Gamma(\alpha + \beta) }{\Gamma(\alpha)\Gamma(\beta)} \int_0^1{x^{\alpha} (1 - x)^{\beta - 1}} dx \\[0.3em] &=\;\; \frac {\Gamma(\alpha + \beta) }{\Gamma(\alpha)\Gamma(\beta)} \times \frac {\Gamma(\alpha + 1)\Gamma(\beta)}{\Gamma(\alpha + \beta + 1) } \\[0.3em] &=\;\; \frac {\Gamma(\alpha + 1)}{\Gamma(\alpha)} \times \frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha + \beta + 1)} \\[0.3em] &=\;\; \frac{\alpha}{\alpha + \beta} \end{align} \]

In a similar way,

\[ \begin{align} E\left[X^2\right] \;\;&=\;\; \int_0^1{x^2 \times \frac {\Gamma(\alpha + \beta) }{\Gamma(\alpha)\Gamma(\beta)} x^{\alpha - 1} (1 - x)^{\beta - 1}} dx \\[0.2em] &=\;\; \frac {\Gamma(\alpha + \beta) }{\Gamma(\alpha)\Gamma(\beta)} \int_0^1{x^{\alpha + 1} (1 - x)^{\beta - 1}} dx \\[0.3em] &=\;\; \frac {\Gamma(\alpha + \beta) }{\Gamma(\alpha)\Gamma(\beta)} \times \frac {\Gamma(\alpha + 2)\Gamma(\beta)}{\Gamma(\alpha + \beta + 2) } \\[0.3em] &=\;\; \frac {\Gamma(\alpha + 2)}{\Gamma(\alpha)} \times \frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha + \beta + 2)} \\[0.3em] &=\;\; \frac{\alpha(\alpha + 1)}{(\alpha + \beta)(\alpha + \beta + 1)} \end{align} \]

Simplifying \(\Var(X) = E[X^2] - \big(E[X]\big)^2\) gives the variance of \(X\).