Gamma Distribution

Before discussing the gamma distribution, we first define the gamma distribution and state some properties of the gamma function, for the sake of completeness.

The gamma function (or factorial function ) with parameter n denoted byΓn is defined as

$$\Gamma(n) \ = \ \int_{0}^{\infty} \ {e^-}^{x} \ {x^n}^{-1} \ d(x) \ \ \ \ \ \ ; \ n \ > \ 0.$$

This integral is an improper integral which converges for n > 0.

The gamma function has the following properties:

$$1.\ \ \Gamma(1) \ = \ \int_{0}^{\infty} \ {e^-}^{x} \ dx \ = \ 1$$

$$2. \ \ \Gamma(n) \ = \ (n - 1) \ \Gamma(n - 1)$$

proof:

$$\Gamma(n) \ = \ {e^-}^{x} \ {x^n}^{-1} \ d(x)$$

Integrating by parts,

$$= \ n-1 \ \int_{0}^{\infty} \ {e^-}^{x} \ dx \ - \ \int_{0}^{\infty} \ \left [ \ (n-1) \ {x^n}^{-2} \ \int_{0}^{\infty} \ {e^-}^{x} \ dx \ \right ] \ dx$$

$$= \ {e^-}{^x} \ {x^n}^{-1} \mid |_{0}^{\infty} \ + \ (n-1) \ \int_{0}^{\infty} \ {e^-}^{x} \ \ {x^(}^{n-1)-1} \ dx$$

$$= \ (n-1) \ \Gamma(n-1)$$

$$3. \ \ \Gamma(n) \ = \ (n-1) \ ! \ for \ positive \ integral \ n.$$

proof:

We have,

$$\Gamma(n) \ = \ (n-1) \ \Gamma(n-1)$$

$$ = \ (n-1) \ (n-2) \ \Gamma(n-2)$$

$$= \ (n-1) \ (n-2) \ ... \ 2. \ 1. \ \Gamma1$$

$$= \ (n-1) \ (n-2) \ ... \ 2. \ 1.$$

$$ = \ (n-1)!$$

$$Since \Gamma(1) \ = \ 1, \ then \ 0! \ = 1.$$

Gamma distribution is one of the continious probability distribution of non-negative random variables.

Defination:

A random variable (r.v) X having a probability density function (pdf)

$$f(x) \ = \ f(x; \alpha) \ = \ \frac{ {e^-}^{x} \ {x^\alpha}^{-1}}{\Gamma\alpha} \ \ \ \ \ ;0 \ < \ x \ < \infty, \ \alpha \ < \ 0.$$

otherwise 0.

is said to have a gamma distribution with one parameter α.

If a random variable X follows a gamma distribution with parameterα, we write X ~ gamma (α ) or simply X ~ G(α. ) orγ(α ).

It can easily be proved that, the function f(x) is a probability density function (pdf), since,

$$\int_{0}^{\infty} \ f(x) \ dx \ = \ \frac{1}{\Gamma\alpha} \ \int_{0}^{\infty} \ {e^-}^{x} \ {x^\alpha}^{-1} \ dx$$

$$= \ \frac{1}{\Gamma\alpha} \ . \ \Gamma\alpha$$

$$= \ 1$$

The gamma distribution is positively skewed. The graph of gamma distribution depends upon the values of the parameterα. The gamma curves for different values ofα i.e.α = 1, α = 2 andα = 5 are shown in the figure below.

_{^{TRUE) plot(function(x) {dlnorm(x, meanlog=1)}, 0, 10, col="blue", add=TRUE) legend(x = "topright", legend = c("mu = 0", "mu = 0.5", "mu = 1"), lty = c(1, 1, 1), col = c("black", "red", "blue"))}}

Figure: Gamma curve

The distribution function of one parameter gamma distribution is given by

$$F(x) \ = \ \int_{0}^{x} \ f(u) \ du$$

$$= \ \frac{1}{\Gamma\alpha} \ \int_{0}^{x} \ {e^-}^{u} \ {u^\alpha}^{-1} \ du \ \ \ \ ; \ u \ > \ 0.$$

$$ = \ 0 \ \ \ \ \ \ \ \ \ \ ; \ otherwise$$

This F(x) is called incomplete gamma function.

Remark 1:

A continuous random variable X is said to have a gamma distribution with two parametersα andβ if its probability density function (pdf) is

$$f(x) \ = \ f(x; \alpha, \beta) \ = \ \frac{\beta^\alpha}{\Gamma\alpha} \ {e^-}^{\beta \alpha} \ {x^\alpha}^{-1} \ \ \ \ ; \ 0 \ < \ x \ < \ \infty, \ \alpha \ > \ 0, \ \beta \ > \ 0$$

$$= \ 0 \ ; \ otherwise$$

Another form of two parameter gamma distribution is given by the probability density function

$$f(x) \ = \ f \left ( x; \ \alpha, \ \beta \ = \ \frac{1}{\theta} \right ) \ = \\frac{{e^-}^{\frac{x}{\theta}} {x^\alpha}^{-1}}{\theta^\alpha \ \Gamma\alpha}$$

$$Here, \ X \ ~ \ G(\alpha, \ \beta). \ If \ \beta \ = \ 1 \ or \ \theta \ = \ 1, \ then \ X \ ~ \ G(\alpha, \ 1) \ = \ G(\alpha).$$

Remark 2:

Ifα = 1 in G(α, β) then the gamma distribution reduces to an exponential distribution with parameterβ.

Moments of Gamma Distribution

Let X ~ G(α) i.e. X follosw a gamma distribution with the parameterα. The r^th moment about origin of gamma distribution is given by

$$\mu_{r}^{'} \ = \ E(X^r) $$

$$ = \ \int_{0}^{\infty} \ f(x) \ dx$$

$$= \ \int_{0}^{\infty} \ x^r \ \frac{{e^-}^{x} \ {x^\alpha}^{-1}}{\Gamma\alpha} \ dx$$

$$= \ \frac{1}{\Gamma\alpha} \ \int_{0}^{\infty} \ {e^-}^{x} \ {x^(}^{\alpha + r)-1} \ dx$$

$$= \ \frac{\Gamma(\alpha + r)}{\Gamma\alpha} \ \ \ \ ; \ r \ = 1, \ 2, \ 3, \ 4$$

Therefore,

$$\mu_{1}^{'} \ = \ \alpha \ = \ mean$$

$$\mu_{2}^{'} \ = \ \alpha(\alpha+1)$$

$$\mu_{3}^{'} \ = \ \alpha(\alpha+1) \ (\alpha+2)$$

$$\mu_{4}^{'} \ = \ \alpha(\alpha+1) \ (\alpha+2) \ (\alpha+3)$$

Similarly, the first four moments about mean of the gamma distribution are obtained as follows:

$$\mu_1 \ = \ 0$$

$$\mu_{2} \ = \ \mu_{2}^{'} - (\mu_{1}^{'})^2$$

$$=\ \alpha(\alpha+1)\ -\ (\alpha)^2 \ = \ \alpha \ = \ Variance$$

$$\mu_{3} \ = \ \mu_{3}^{'} \ - \ 3\mu_{2}^{'} \mu_{1}^{'} \ + \ 2(\mu_{1}^{'})^3$$

$$= \\alpha(\alpha+1) \ (\alpha+2) \ - \ 3\ \alpha(\alpha+1) \ \alpha \ + \ 2 \ \alpha^3 \ = \ 2 \alpha$$

$$\mu_{4} \ = \ \mu_{4}^{'} \ - \ 4\mu_{3}^{'} \ \mu_{1}^{'} \ + \ 6 \ \mu_{2}^{'} \ (\mu_{1}^{'})^2 \ - \ 3 \ (\mu_{1}^{'})^4$$

$$= \ \alpha(\alpha+1) \ (\alpha+2) \ (\alpha+3) \ - 4\ \alpha(\alpha+1) \ (\alpha+2) \ \alpha \ + \ 6 \\alpha(\alpha+1) \ (\alpha)^2 \ - \ 3 (\alpha)^4$$

$$= \ 3 \ (\alpha)^2 \ + \ 6 \ \alpha$$

It is important to note that for a one parameter gamma distribution G(α), the mean and variance are equal toα, like in the Poisson distribution. That is

Mean = Variance =α.

Bibliography

Sukubhattu N.P. (2013). Probability & Inference - II. Asmita Books Publishers & Distributors (P) Ltd., Kathmandu.

Larson H.J. Introduction to Probability Theory and Statistical Inference. WileyInternational, New York.