Gamma Distribution | Gamma Function | Properties (2024)

4.2.4 Gamma Distribution

The gamma distribution is another widely used distribution. Its importance is largely due toits relation to exponential and normal distributions. Here, we will provide an introduction to thegamma distribution. In Chapters 6 and 11, we will discuss more properties of the gammarandom variables. Before introducing the gamma random variable, we need to introduce thegamma function.

Gamma function: The gamma function [10], shown by $ \Gamma(x)$, is an extension of the factorialfunction to real (and complex) numbers. Specifically, if $n \in \{1,2,3,...\} $, then$$ \Gamma(n) = (n-1)!$$More generally, for any positive real number $\alpha$, $\Gamma(\alpha)$ is defined as$$ \Gamma(\alpha) = \int_0^\infty x^{\alpha - 1} e^{-x} {\rm d}x, \hspace{20pt} \textrm{for }\alpha>0. $$
Figure 4.9 shows the gamma function for positive real values.

Figure 4.9: The Gamma function for some real values of $\alpha$.

Note that for $\alpha=1$, we can write$$\begin{align*}\Gamma(1) &= \int_0^\infty e^{-x} dx\\ &= 1.\end{align*}$$Using the change of variable $x = \lambda y$, we can show the following equation that is often useful when working withthe gamma distribution:$$ \Gamma(\alpha) = \lambda^{\alpha} \int_0^\infty y^{\alpha-1} e^{-\lambda y} dy \hspace{20pt} \textrm{for } \alpha,\lambda > 0.$$Also, using integration by parts it can be shown that$$ \Gamma(\alpha + 1) = \alpha\Gamma(\alpha), \hspace{20pt} \textrm{for } \alpha > 0.$$Note that if $\alpha = n$, where $n$ is a positive integer, the above equation reduces to$$ n! = n \cdot (n-1)!$$

Example
Answer the following questions:

1. Find $\Gamma(\frac{7}{2}).$
2. Find the value of the following integral:$$ I = \int_0^\infty x^{6} e^{-5x} dx.$$

Gamma Distribution:
We now define the gamma distribution by providing its PDF:

A continuous random variable $X$ is said to have a gamma distribution with parameters$\alpha > 0 \textrm{ and } \lambda > 0 $, shown as $X \sim Gamma(\alpha,\lambda)$, if its PDFis given by$$f_X(x) = \left\{\begin{array}{l l}\frac{\lambda^{\alpha} x^{\alpha-1} e^{-\lambda x}}{\Gamma(\alpha)} \hspace {5pt} x > 0\\0 \hspace{56pt} \textrm{otherwise}\end{array}\right.$$

If we let $\alpha = 1$, we obtain$$f_X(x) = \left\{\begin{array}{l l}\lambda e^{-\lambda x} \hspace{20pt} x > 0\\0 \hspace{41pt} \textrm{otherwise}\end{array}\right.$$Thus, we conclude $Gamma(1,\lambda) = Exponential(\lambda)$. More generally, if you sum $n$independent $Exponential(\lambda)$ random variables, then you will get a $Gamma(n,\lambda)$random variable. We will prove this later on using the moment generating function. The gammadistribution is also related to the normal distribution as will be discussed later. Figure 4.10shows the PDF of the gamma distribution for several values of $\alpha$.

Figure 4.10: PDF of the gamma distribution for some values of $\alpha$ and $\lambda$.

Example
Using the properties of the gamma function, show that the gamma PDF integrates to 1, i.e., showthat for $\alpha , \lambda > 0$, we have$$\int_0^\infty \frac{\lambda^{\alpha}x^{\alpha - 1} e^{-\lambda x}}{\Gamma(\alpha)} dx = 1.$$

In the Solved Problems section, we calculate the mean and variance for the gamma distribution. In particular,we find out that if $X \sim Gamma(\alpha,\lambda)$, then$$ EX = \frac{\alpha}{\lambda}, \hspace{20pt} Var(X) = \frac{\alpha}{\lambda^2}.$$