Expectation of Gamma Distribution/Proof 1

Theorem

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

The expectation of $X$ is given by:

$\expect X = \dfrac \alpha \beta$

Proof

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

$\map {f_X} x = \dfrac {\beta^\alpha x^{\alpha - 1} e^{-\beta x} } {\map \Gamma \alpha}$

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

$\ds \expect X = \int_0^\infty x \map {f_X} x \rd x$

So:

\(\ds \expect X\)

\(=\)

\(\ds \frac {\beta^\alpha} {\map \Gamma \alpha} \int_0^\infty x^\alpha e^{-\beta x} \rd x\)

\(\ds \)

\(=\)

\(\ds \frac {\beta^\alpha} {\map \Gamma \alpha} \int_0^\infty \left({\frac t \beta}\right)^\alpha e^{-t} \frac {\rd t} \beta\)

substituting $t = \beta x$

\(\ds \)

\(=\)

\(\ds \frac {\beta^\alpha} {\beta^{\alpha + 1} \map \Gamma \alpha} \int_0^\infty t^\alpha e^{-t} \rd t\)

\(\ds \)

\(=\)

\(\ds \frac {\map \Gamma {\alpha + 1} } {\beta \map \Gamma \alpha}\)

Definition of Gamma Function

\(\ds \)

\(=\)

\(\ds \frac {\alpha \map \Gamma \alpha} {\beta \map \Gamma \alpha}\)

Gamma Difference Equation

\(\ds \)

\(=\)

\(\ds \frac \alpha \beta\)

$\blacksquare$