Negative Exponential Distribution

Negative Exponential Distribution

An exponential distribution is one of the most frequently used continious probability distributions in stastics. The negative exponential distribution is a special case of two parameter gamma distribution G(α,β), whenα = 1. The exponential distribution has most application in queuing theory and realibilitytheory.

Definition:

A random variable (rv) X having a probability density function (pdf) :

$$f(x) \ = \ f(x;\theta) \ = \ \theta \ {e^-}^{\theta x} \ \ \ \ \ ; \ x \ > \ 0$$

otherwise , 0

is said to have a negative exponential distribution with parameterθ (>0).

If a random variable X is distributed as a negative exponential distribution with parameterθ, then we write X ~ Expo (θ).

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

$$\int_0^{\infty} \ f(x) \ dx \ = \ \int_0^{\infty} \ \theta \ {e^-}^{\theta x} \ dx$$

$$= \ \left [ \ - \ {e^-}^{\theta x} \ \right ]_0^{\infty} \ = \ 1$$

The distribution function of the negative exponential distribution is given by

F(x) = p (X≤ x)

$$= \ \int_0^{x} \ f(u) \ du$$

$$= \ \theta \ \int_0^{x} \ {e^-}^{\theta u} \ du$$

$$= \ \left [ \ - \ {e^-}^{\theta u} \ \right ]_0^{x}$$

$$= \ 1 \ - \ {e^-}^{\theta x}$$

Hence,

F(x) = 0 ; x≤ 0

F(x) = 1 - e^-θx; x > 0

F(x) = 1 ; otherwise

The graph of the probability density function and the distribution function of negative exponential distribution are given in the following figure:

_{^{www.r-tutor.com}}

Note:

Let X ~ G (α,β). Ifα = 1 andβ = θ, then X ~ G (1,θ) = Expo (θ).

Moments of Negative Exponential Distributio

Moments about origin

The r^th moment about origin of negative exponential distribution is given by

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

$$= \ \int_0^{\infty} \ x^r \ f(x) \ dx$$

$$= \ \theta \ \int_0^{\infty} \ x^r \ {e^-}^{\theta x} \ dx$$

Put y =θx so that

$$dx \ = \ \frac{dy}{d\theta}$$

$$\therefore \ \mu_r^{'} \ = \ \theta \ \int_{0}^{\infty} \ \left ( \ \frac{y}{\theta} \ \right )^r \ {e^-}^{y} \ \frac{dx}{\theta}$$

$$= \ \frac{1}{\theta^r} \ \int_0^{\infty} \ {e^-}^{y} \ {y^(}^{r+1)-1} \ dy$$

$$= \ \frac{1}{\theta^r} \ \Gamma(r+1)$$

$$\mu_r^{'} \ = \ \frac{r!}{\theta^r} \ \ \ \ \ \ ; \ r \ = \ 1, \ 2, \ 3, \ 4$$

Therefore

$$\mu_1^{'} \ = \ \frac{1}{\theta} \ = \ Mean \ (Reciporcal \ of \ the \ parameter \ \theta)$$

$$\mu_2^{'} \ = \ \frac{2}{\theta^2}$$

$$\mu_3^{'} \ = \ \frac{3!}{\theta^3} \ = \ \frac{6}{\theta^3}$$

$$\mu_4^{'} \ = \ \frac{4!}{\theta^4} \ = \ \frac{24}{\theta^4}$$

Moments about Mean:

The first four Central moments of the exponential distribution are obtained as :

$$\mu_1 \ = \ 0$$

$$\mu_2 \ = \ \mu_2^{'} \ - \ (\mu_1^{'})^2$$

$$= \ \frac{2}{\theta^2} \ - \ \left ( \ \frac{1}{\theta} \ \right )^2 \ = \ \frac{1}{\theta^2} \ = \ Variance$$

$$\mu_3 \ = \ \mu_3^{'} \ - \ 3 \mu_2^{'} \ \mu_1^{'} \ + \ 2 (\mu_1^{'})^3$$

$$= \ \frac{6}{\theta^3} \ - \ 3 \ . \ \frac{2}{\theta^2} \ . \ \frac{1}{\theta} \ + \ 2 \ \left ( \ \frac{1}{\theta} \ \right )^3$$

$$= \ \frac{2}{\theta^3}$$

$$\mu_4 \ = \ \mu_4^{'} \ - \ 4 \mu_3{'} \ \mu_1^{'} \ + \ 6 \ \mu_2^{'} \ ( \mu_1^{'} )^2 \ - \ 3 \ (\mu_1^{'})^4$$

$$= \ \frac{24}{\theta^4} \ - \ 4 \ . \ \frac{6}{\theta^3} \ . \ \frac{1}{\theta} \ + \ 6 \ \frac{2}{\theta^2} \ . \ \left ( \frac{1}{\theta} \right )^2 \ - \ 3 \ . \left( \ \frac{1}{\theta} \ \right )^4$$

$$= \ \frac{9}{\theta^4}$$

Also, the beta and gamma coefficient of the exponential distribution are obtained as

$$\beta_1 \ = \ \frac{\mu_3^{2}}{\mu_2^{3}} = \ 4$$

Again,

$$\beta_2 \ = \ \frac{\mu_4}{\mu_2^{2}} \ = \ 9$$

$$ Therefore, \ \gamma_1 \ = \ \sqrt\beta_1 \ = \ 2$$

$$and, \gamma_2 \ = \ \beta_2 \ - 3 \ = \ 6$$

Moment Generating Function of Negative Exponential Distribution

Let X has a negative exponential distribution with probability density function (pdf)

f(x) =­θ e^-θx ; x > 0

The moment generating function of X is given by

$$M_X (t) \ = \ E({e^t}^{x})$$

$$= \ \int_0^{\infty} \ {e^t}^{x} \ f(x) \ dx$$

$$= \ \int_0^{\infty} \ {e^t}^{x} \ \theta \ {e^-}^{\theta x} \ dx$$

$$= \ \theta \ \int_0^{\infty} \ {e^-}^{(\theta - t) x} \ dx$$

$$= \ \frac{\theta}{\theta \ - t} \ \left [ \ - \ {e^-}^{(\theta-t)x} \ \right ]_0^{\infty}$$

$$= \ frac{\theta}{\theta \ - \ t}$$

$$= \ \frac{1}{\left ( 1 \ - \ \frac{t}{\theta} \ \right )}$$

$$\Rightarrow \ M_x(t) \ = \ {\left ( 1 \ - \ \frac{t}{\theta} \ \right )^-}^{1} \ \ \ \ ; \ t \ < \ \theta$$

To find moments:

Expanding M_x(t), we have

$$M_x(t) \ = \ 1 \ + \ \frac{t}{\theta} \ + \ \frac{t^2}{\theta^2} \ + \ \frac{t^3}{\theta^3} \ + \ . \ . \ . \ + \ \frac{t^r}{\theta^t} \ + \ . \ . \ .$$

$$\therefore \ \mu_r^{'} \ = \ Coefficient \ of \ \frac{t^r}{r!} \ in \ M_x(t).$$

$$= \ \frac{r!}{\theta^r} \ \ \ \ \ \ \ \ for \ r \ = \ 1, \ 2, \ 3, \ 4$$

Then, the moments of the exponential distribution can be easily obtained.

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.