Variance of Beta Distribution

Theorem

Let $X \sim \map \Beta {\alpha, \beta}$ for some $\alpha, \beta > 0$, where $\Beta$ is the Beta distribution.

Then:

$\var X = \dfrac {\alpha \beta} {\paren {\alpha + \beta}^2 \paren {\alpha + \beta + 1} }$

Proof 1

From the definition of the Beta distribution, $X$ has probability density function:

$\map {f_X} x = \dfrac {x^{\alpha - 1} \paren {1 - x}^{\beta - 1} } {\map \Beta {\alpha, \beta} }$

From Variance as Expectation of Square minus Square of Expectation:

$\ds \var X = \int_0^1 x^2 \map {f_X} X \rd x - \paren {\expect X}^2$

So:

\(\ds \var X\)

\(=\)

\(\ds \frac 1 {\map \Beta {\alpha, \beta} } \int_0^1 x^{\alpha + 1} \paren {1 - x}^{\beta - 1} \rd x - \frac {\alpha^2} {\paren {\alpha + \beta}^2}\)

Expectation of Beta Distribution

\(\ds \)

\(=\)

\(\ds \frac {\map \Beta {\alpha + 2, \beta} } {\map \Beta {\alpha, \beta} } - \frac {\alpha^2} {\paren {\alpha + \beta}^2}\)

Definition 1 of Beta Function

\(\ds \)

\(=\)

\(\ds \frac {\map \Gamma {\alpha + 2} \, \map \Gamma \beta} {\map \Gamma {\alpha + \beta + 2} } \cdot \frac {\map \Gamma {\alpha + \beta} } {\map \Gamma \alpha \, \map \Gamma \beta} - \frac {\alpha^2} {\paren {\alpha + \beta}^2}\)

Definition 3 of Beta Function

\(\ds \)

\(=\)

\(\ds \frac {\alpha \paren {\alpha + 1} } {\paren {\alpha + \beta} \paren {\alpha + \beta + 1} } \cdot \frac {\map \Gamma \alpha \, \map \Gamma \beta \, \map \Gamma {\alpha + \beta} } {\map \Gamma \alpha \, \map \Gamma \beta \, \map \Gamma {\alpha + \beta} } - \frac {\alpha^2} {\paren {\alpha + \beta}^2}\)

Gamma Difference Equation

\(\ds \)

\(=\)

\(\ds \frac {\paren {\alpha^2 + \alpha} \paren {\alpha + \beta} } {\paren {\alpha + \beta}^2 \paren {\alpha + \beta + 1} } - \frac {\alpha^2 \paren {\alpha + \beta + 1} } {\paren {\alpha + \beta}^2 \paren {\alpha + \beta + 1} }\)

\(\ds \)

\(=\)

\(\ds \frac {\alpha ^3 + \alpha^2 \beta + \alpha^2 + \alpha \beta - \alpha^3 - \alpha^2 \beta - \alpha^2} {\paren {\alpha + \beta}^2 \paren {\alpha + \beta + 1} }\)

\(\ds \)

\(=\)

\(\ds \frac {\alpha \beta} {\paren {\alpha + \beta}^2 \paren {\alpha + \beta + 1} }\)

$\blacksquare$

Proof 2

From the definition of Variance as Expectation of Square minus Square of Expectation:

$\var X = \expect {X^2} - \paren {\expect X}^2$

From Expectation of Beta Distribution:

$\expect X = \dfrac \alpha {\alpha + \beta}$

From Raw Moment of Beta Distribution:

$\ds \expect {X^n} = \prod_{r \mathop = 0}^{n - 1} \frac {\alpha + r} {\alpha + \beta + r}$

So:

\(\ds \expect {X^2}\)

\(=\)

\(\ds \prod_{r \mathop = 0}^1 \frac {\alpha + r} {\alpha + \beta + r}\)

Raw Moment of Beta Distribution

\(\ds \)

\(=\)

\(\ds \paren {\frac {\alpha} {\alpha + \beta} } \paren {\frac {\alpha + 1} {\alpha + \beta + 1} }\)

Therefore:

\(\ds \expect {X^2} - \paren {\expect X}^2\)

\(=\)

\(\ds \paren {\frac {\alpha} {\alpha + \beta} } \paren {\frac {\alpha + 1} {\alpha + \beta + 1} } - \paren {\dfrac \alpha {\alpha + \beta} }^2\)

\(\ds \)

\(=\)

\(\ds \paren {\frac {\alpha} {\alpha + \beta} } \paren {\frac {\alpha + 1} {\alpha + \beta + 1} } \paren {\frac {\alpha + \beta} {\alpha + \beta} } - \paren {\dfrac \alpha {\alpha + \beta} }^2 \paren {\frac {\alpha + \beta + 1} {\alpha + \beta + 1} }\)

multiplying by $1$

\(\ds \)

\(=\)

\(\ds \frac {\alpha^3 + \alpha^2 \beta + \alpha^2 + \alpha \beta - \alpha^3 - \alpha^2 \beta - \alpha^2} {\paren {\alpha + \beta}^2 \paren {\alpha + \beta + 1} }\)

\(\ds \)

\(=\)

\(\ds \frac {\alpha \beta} {\paren {\alpha + \beta}^2 \paren {\alpha + \beta + 1} }\)

$\blacksquare$