Expectation of Beta Distribution/Proof 1

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Theorem

Let $X \sim \BetaDist \alpha \beta$ for some $\alpha, \beta > 0$, where $\operatorname{Beta}$ denotes the beta distribution.

Then:

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


Proof

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 the definition of the expected value of a continuous random variable:

$\displaystyle \expect X = \int_0^1 x \, \map {f_X} x \rd x$

So:

\(\ds \expect X\) \(=\) \(\ds \frac 1 {\map \Beta {\alpha, \beta} } \int_0^1 x^\alpha \paren {1 - x}^{\beta - 1} \rd x\)
\(\ds \) \(=\) \(\ds \frac {\map \Beta {\alpha + 1, \beta} } {\map \Beta {\alpha, \beta} }\) Definition 1 of Beta Function
\(\ds \) \(=\) \(\ds \frac {\map \Gamma {\alpha + 1} \, \map \Gamma \beta} {\map \Gamma {\alpha + \beta + 1} } \cdot \frac {\map \Gamma {\alpha + \beta} } {\map \Gamma \alpha \, \map \Gamma \beta}\) Definition 3 of Beta Function
\(\ds \) \(=\) \(\ds \frac \alpha {\alpha + \beta} \cdot \frac {\map \Gamma \alpha \, \map \Gamma \beta \, \map \Gamma {\alpha + \beta} } {\map \Gamma \alpha \, \map \Gamma \beta \, \map \Gamma {\alpha + \beta} }\) Gamma Difference Equation
\(\ds \) \(=\) \(\ds \frac \alpha {\alpha + \beta}\)

$\blacksquare$