Statistical Models: Gamma Function

In mathematics, the $\Gamma$ function is an extension of the factorial function to the real and complex domains.

If $n$ is a positive integer, then: $\Gamma(n) = (n - 1)!$

In the complex domain, $\Gamma(z) = \int_0^{\infty} t^{z - 1} \exp(-t) dt$

Properties

$\Gamma(x + 1) = x\Gamma(x)$

Proof:

$$\begin{align} \Gamma(x + 1) &= \int_0^{\infty} t^x \exp(-t) dt \\ &= \int_0^{\infty}- t^x d \exp(-t) \\ &= -t \exp(-t) |_0^{\infty} - \int_0^{\infty} \exp(-t) d(-t^x) \text{ (integration by parts)}\\ &= 0 - 0 + \int_0^{\infty} \exp(-t) x t^{x - 1} dt \\ &= x \int_0^{\infty} \exp(-t) t^{x - 1} dt \\ &= x \Gamma(x) \end{align}$$

When $x$ is a positive integer, $\Gamma(n + 1) = n!$, which aligns perfectly with the factorial. Therefore, the $\Gamma$ function can be regarded as a generalization of the factorial to non-integer domains.

This has an advantage: when encountering integrals of the form $\int_0^{\infty} t^n \exp(-t)dt$ in the future, they can be equivalently expressed as computing the factorial $(n + 1)!$, which is much simpler than direct integration.

Alternative Forms of the $\Gamma$ Function

$\Gamma(z + 1) = \int_0^{\infty} x^{z} \exp(-x) dx$. Using the substitution $x = t^2$, we get:

$$\begin{align} \Gamma(z + 1) & = \int_0^{\infty} t^{ 2z } \exp(- t^2) d(t^2) \\ & = 2 \int_0^{\infty} t^{2z + 1} \exp(-t^2) dt \end{align}$$

If we set $z = -\frac{1}{2}$, we can compute a special value:

$$\Gamma(\frac{1}{2}) = 2 \int_0^{\infty} \exp(-t^2) dt = \sqrt{\pi} \text{ (Gaussian integral property)}$$

That is, the values of $\Gamma(x)$ for positive integers can all be computed by converting them to factorials, and now the value for the non-integer $x = \frac{1}{2}$ has also been computed.

Continuing with the property $\Gamma(x + 1) = x\Gamma(x)$, we can also compute the values in the following cases:

1. $\Gamma(\frac{3}{2}) = \frac{1}{2} \Gamma(\frac{1}{2}) = \frac{\sqrt{\pi}}{2}$
2. $\Gamma(\frac{5}{2}) = \frac{3}{2} \Gamma(\frac{3}{2}) = \frac{3}{4} \sqrt{\pi}$
3. $\Gamma(\frac{7}{2}) = \frac{5}{2} \Gamma(\frac{5}{2}) = \frac{15}{8}\sqrt{\pi}$
4. …
5. $\Gamma(n + \frac{1}{2}) = \frac{(2n - 1)!!}{2^n} \sqrt{\pi}$

Now, the values of both $\Gamma(n)$ and $\Gamma(n + \frac{1}{2})$ can be computed directly.