Variance of Beta Distribution/Proof 2
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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
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 $1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\alpha^3 + \alpha^2 \beta + \alpha^2 + \alpha \beta - \alpha^3 - \alpha^2 \beta - \alpha^2} {\paren {\alpha + \beta}^2 \paren {\alpha + \beta + 1} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\alpha \beta} {\paren {\alpha + \beta}^2 \paren {\alpha + \beta + 1} }\) |
$\blacksquare$