17 Gamma Function

The Gamma Function is a function used frequently in statistical theory. It has properties that permit simplified calculations and is used in defining many probability distributions, particularly within the Exponential family of distributions.

17.1 Definition

\[\Gamma(x) = \int\limits_0^{\infty}t^{x-1}e^{-t}dt\]

Note that the definition defines a function of \(x\) that is integrated over \(t\). Thus, for each value of \(x\), a curve is defined, and the Gamma function calculates the area under the curve defined by \(x\).

17.2 Theorems for the Gamma Function

17.2.1 Lemma

\[\left[ -t^{x-1} e^{-t} \right]_{t = 0}^{t = \infty} = 0\]

Proof:

\[\begin{aligned} \left[ -t^{x-1} e^{-t} \right]_{t = 0}^{t = \infty} &= \lim_{t\to\infty}\big(-t^{x - 1} e^{-t}\big) - 0^{x - 1} e^{-0} \\ &= \lim_{t\to\infty}\big(-t^{x - 1} e^{-t}\big) - 0 \\ &= \lim_{t\to\infty}\big(-t^{x - 1} e^{-t}\big) \\ &= - \lim_{t\to\infty}\frac{t^{x - 1}}{e^t} \\ &= - \lim_{t\to\infty} \left\{exp\left[ (x - 1) \ln t - t\right] \right\} \\ &= - \lim_{t\to\infty} \left\{ exp \left[(x - 1) \cdot t \cdot \Big( \frac{\ln t}{t} - 1\Big) \right]\right\} \\ ^{[1]} &= \lim_{t\to\infty} \left\{ exp \left[(x - 1) \cdot t \cdot \Big( 0 - 1\Big) \right]\right\} \\ &= \lim_{t\to\infty} \left\{ exp \left[- (x - 1) \cdot t \right]\right\} \\ &= \lim_{t\to\infty} \frac{1}{e^{(x - 1) \cdot t}} \\ &= 0 \end{aligned}\]

  1. L’H^{o}pital’s Rule: \(\lim_{x\to u}\frac{f(x)}{g(x)} = \lim_{x\to u} \frac{f'(x)}{g'(x)}\). This implies \(\lim_{x\to \infty} \frac{\ln x}{x} = \lim_{x\to \infty} \frac{\frac{1}{x}}{1} = \lim_{x \to \infty} \frac{1}{x} = 0\)

17.2.2 Theorem: The Reduction Relation

\(\Gamma(x) = (x-1) \cdot \Gamma(x)\)

Proof:

The proof relies on integration by parts. Let:

\[\begin{aligned} u &= t^{x-1} \\ du &= (x - 1) \cdot t^{(x-2)} \\\\ v &= -e^{-t} \\ dv &= e^{-t} dt \end{aligned}\]

Integration by parts yields:

\[\begin{aligned} \Gamma(x) &= \int\limits_0^{\infty}t^{x-1}e^{-t}dt \\ &= u \cdot v - \int\limits_0^\infty v \cdot du \\ &= \Big[t^{x-1} \cdot - e^{-t}\Big]_{t=0}^{t=\infty} - \int\limits_0^\infty-e^{-t} \cdot (x-1)\cdot t^{(x-2)} dt \\ &= -\Big[t^{x-1} \cdot e^{-t}\Big]_{t=0}^{t=\infty} - (- (x - 1)) \int\limits_0^\infty e^{-t} \cdot t^{(x-2)} dt \\ ^{[1]} &= -0 + (x - 1) \int\limits_0^\infty e^{-t} \cdot t^{(x-2)} dt \\ &= (x - 1) \int\limits_0^\infty e^{-t} \cdot t^{((x-1)-1)} dt \\ &= (x - 1) \int\limits_0^\infty t^{((x-1)-1)} \cdot e^{-t} dt \\ &= (x - 1) \cdot \Gamma(x - 1) \end{aligned}\]

  1. 17.2.1

17.2.3 Corollary

\[\Gamma(x) = \frac{1}{x} \cdot \Gamma(x + 1)\]

Proof:

Theorem 17.2.2 establishes

\[ \Gamma(x) = (x - 1) \cdot \Gamma(x - 1)\]

Let \(y = x + 1\). Then

\[\begin{aligned} \Gamma(y) &= (y - 1) \cdot \Gamma(y - 1) \\ \Rightarrow \Gamma(x + 1) &= (x + 1 - 1) \cdot \Gamma(x + 1 - 1) \\ &= x \cdot \Gamma(x) \\ \Rightarrow \frac{1}{x} \cdot \Gamma(x + 1) &= \Gamma(x) \\ \Rightarrow \Gamma(x) &= \frac{1}{x} \cdot \Gamma(x + 1) \end{aligned}\]

This allows the recurrence relation to move toward \(\Gamma(0)\) for any value of \(x\). Note, however, that \(\Gamma(0)\) is undefined. Thus, solutions for the Gamma Function may be determined for positive integers, since \(\Gamma(1)\) can be solved. On the other hand, \(\Gamma(-1)\) can not be solved, and the recurrence relation results in a zero denominator. Hence, the Gamma Function is defined for all \(x \in \mathbb{R}\) so long as \(x \not\in \mathbb{Z^-}\)

17.2.4 Theorem:

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

Proof:

\[\begin{aligned} \Gamma\Big(\frac{1}{2}\Big) &= \int\limits_0^\infty t^{-\frac{1}{2}} e ^{-t} dt \\ &= \frac{2}{2} \int\limits_0^\infty t^{-\frac{1}{2}} e ^{-t} dt \\ &= 2 \int\limits_0^\infty \frac{1}{2} \cdot t ^{-\frac{1}{2}} e^{-(\sqrt{t})^2} dt \\ &= 2 \int\limits_0^\infty \frac{1}{2} \cdot t ^{-\frac{1}{2}} e^{-(\sqrt{t})^2} dt \\ ^{[1]} &= 2 \int\limits_0^\infty \frac{1}{2} \cdot x ^{-1} e^{-x^2} \cdot 2\cdot x \ dx \\ &= 2 \int\limits_0^\infty \frac{2x}{2x} e^{-x^2} dx \\ &= 2 \int\limits_0^\infty e^{-x^2} dx \\ ^{[2]} &= 2 \cdot \frac{\sqrt{\pi}}{2} \\ &= \sqrt{\pi} \end{aligned}\]

  1. \(x = \sqrt{t}\); \(t = x^2\); and \(dt = 2x dx\)
  2. Theorem 18.1.2

17.2.5 Theorem:

Let \(c\) be a constant such that \(c > 0\). Then

\[ \int_0^\infty t^x e^{-ct} \cdot dt = \frac{\Gamma(x+1)}{c^{x+1}} \]

Proof:

\[\begin{aligned} \int\limits_0^\infty t^x e^{-ct} dt \\ ^{[1]} &= \int\limits_0^\infty \big(\frac{y}{c}\big) ^ x e ^{-c \frac{y}{c}} \frac{dy}{c} \\ &= \frac{1}{c^x} \cdot \frac{1}{c} \int\limits_0^\infty y ^ x e ^{-y} dy \\ &= \frac{1}{c^{x + 1}} \int\limits_0^\infty y ^ {(x+1)-1} e ^{-y} dy \\ ^{[2]} &= \frac{1}{c^{x + 1}} \cdot \Gamma(x + 1) \\ &= \frac{\Gamma(x + 1)}{c^{x + 1}} \end{aligned}\]

  1. Let \(y = ct\). Then \(y = \frac{y}{c}\) and \(dt = \frac{dy}{c}\).
  2. \(\int\limits_0^\infty y^{(x+1)-1} e^{-y} dy\) satisfies the form of a Gamma function. 17.1

17.3 References