Expectation of Gamma Distribution

From ProofWiki
Jump to navigation Jump to search

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 1

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:

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

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

$\blacksquare$


Proof 2

By Moment Generating Function of Gamma Distribution, the moment generating function of $X$ is given by:

$\map {M_X} t = \paren {1 - \dfrac t \beta}^{-\alpha}$

for $t < \beta$.

From Moment in terms of Moment Generating Function:

$\expect X = \map { {M_X}'} 0$

From Moment Generating Function of Gamma Distribution: First Moment:

$\map { {M_X}'} t = \dfrac {\beta^\alpha \alpha} {\paren {\beta - t}^{\alpha + 1} }$

Hence setting $t = 0$:

\(\ds \map { {M_X}'} 0\) \(=\) \(\ds \frac {\beta^\alpha \alpha} {\paren {\beta - 0}^{\alpha + 1} }\)
\(\ds \) \(=\) \(\ds \frac {\beta^\alpha \alpha} {\beta^{\alpha + 1} }\)
\(\ds \) \(=\) \(\ds \frac \alpha \beta\)

$\blacksquare$