Gamma Distribution#

The gamma distribution is a continuous probability distribution that generalizes the exponential distribution, modeling the time until a specified number of independent events occur, and is parameterized by a shape parameter \(\alpha\) and a scale parameter \(\beta\).

A gamma random variable is denoted as follows:

\[ X \sim \text{Gamma}(\alpha, \beta) \]

The Gamma Function#

The gamma distribution is based on the gamma function:

\[ \Gamma(\alpha) = \int_0^\infty t^{\alpha - 1} e^{-t} dt, \ \ \alpha > 0 \]

The gamma function can be evaluated as followed:

\[\begin{split} \begin{align} \Gamma(1) &= \int_0^\infty t^{1-1} e^{-t} dt \\ &= \int_0^\infty e^{-t} dt \\ &= \left[ -e^{-t} \right]_0^\infty \\ &= 0 - (-1) \\ &= 1 \\ \end{align} \end{split}\]

The gamma function has a special property:

\[\begin{split} \begin{align} \Gamma(\alpha + 1) &= \int_0^\infty t^{\alpha + 1-1} e^{-t} dt \\ &= \int_0^\infty t^\alpha e^{-t} dt \\ &= -t^\alpha e^{-t} \Big|_0^\infty + \alpha \int_0^\infty t^{\alpha-1} e^{-t} dt \\ &= \alpha \int_0^\infty t^\alpha e^{-t} dt \\ &= \alpha \Gamma({\alpha}) \\ \end{align} \end{split}\]

Note:

  • In line (3), we perform integration by parts where \(u=t^\alpha\) and \(dv=e^{-t} dt\)

A special case in the gamma function is:

\[ \Gamma\left(\frac{1}{2}\right) = \sqrt{\pi} \]

Probability Distribution Function (PDF)#

The pdf for the gamma distribution is given as followed:

\[\begin{split} f_X(x) = \begin{cases} \frac{1}{\Gamma(\alpha) \beta^\alpha} x^{\alpha-1} e^{-\frac{x}{\beta}} , & \text{if } x>0 \\ 0, & \text{otherwise} \end{cases} \end{split}\]

Expectation#

The expected value of the gamma distribution can be derived using by performing the following:

\[\begin{split} \begin{align} EX &= \int_0^\infty x f_X(x) dx \\ &= \int_0^\infty x \frac{1}{\Gamma(\alpha) \beta^\alpha} x^{\alpha-1} e^{-\frac{x}{\beta}} dx \\ &= \frac{1}{\Gamma(\alpha) \beta^\alpha} \int_0^\infty x^\alpha e^{-\frac{x}{\beta}} dx \\ &= \frac{1}{\Gamma(\alpha) \beta^\alpha} \int_0^\infty \beta (u \beta)^\alpha e^{-u} du \\ &= \frac{\beta}{\Gamma(\alpha)} \int_0^\infty x^\alpha e^{-u} du \\ &= \frac{\beta}{\Gamma(\alpha)} \int_0^\infty x^{(\alpha + 1) - 1} e^{-u} du \\ &= \frac{\beta \Gamma(\alpha + 1)}{\Gamma(\alpha)} \\ &= \frac{\alpha \beta \Gamma(\alpha)}{\Gamma(\alpha)} \\ &= \alpha \beta \\ \end{align} \end{split}\]

Note:

  • In line (4), we substitute in \(u = \frac{x}{\beta}\) or \(x=\beta u\)

  • In line (7), we recognize that the integral is the definition of the gamma function with the argument \(\alpha + 1\)

Variance#

A similar argument reveals that the variance of the gamma distribution is as followed:

\[ Var(X) = \alpha \beta^2 \]