Uniform or Rectangular Distribution

Introduction

If a random variable x assumes n discrete values x₁, x₂, x₃, ... , x_neach with an equal probability 1\n, then the distribution of X is called an uniform distribution. For example, let us consider an random experiment of throwing a balanced die and let X be the number of points obtained on face of the die. Then X assume the different values 1, 2, 3, 4, 5 and 6 and the different values of the random variable X are equally likely i.e.$$ the \ random \ variable \ X \ assumes \ each \ of \ its \ values \ 1, \ 2, \ 3, \ 4, \ 5 \ and \ 6 \ with \ an \ equal \ probability \ \frac{1}{6}.$$ Thus, the distribution of X is

which is called a uniform distribution. The graph of the probability function p(x) shows the uniform or equal height for all the values of the random variable X as shown in the uniform or equal height for all the values of the random variable X as shown in the figure given below

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Figure : Graph of uniform distribution

Since the graph of p(x) fits into a rectangle, the uniform distribution is also called a rectangular distribution.

Definition

A random variable X is said to follow a uniform or rectangular distribution over the range [1, n], if its probability mass function is given by

$$ P ( X = x ) \ = \ p(x) \ = \ \frac{1}{n} \ \ \ \ \ \ ; \ x \ = 1, \ 2, \ 3, \ ..., \ n$$

other wise 0.

Here, n is called the parameter of the discrete uniform distribution. We shall use the notation X ~ U (x ; n) or R (x ; n) to denote the random variable X follows uniform or rectangular distribution with parameter n.

Moments of Uniform or Rectangular Distribution

Suppose a random variable X has a uniform distribution defined over the range [1, n] i.e. X ~ U (x ; n). Then the mean of the distribution is given by

$$\mu_{1}^{'} \ = \ E(X) \ = \ \sum_{x=0}^{n} \ x \ p(x)$$

$$= \ \sum_{x=1}^{n} \ x \ \frac{1}{n}$$

$$= \ \frac{1}{n} \ \sum_{x=1}^{n} \ x$$

$$= \ \frac{1}{n} [1 + 2 + ... + n]$$

$$= \ \frac{n (n+1)}{2n}$$

$$= \ \frac{n + 1}{2}$$

Similarly,

$$\mu_{2}^{'} \ = \ E(X^2) \ = \ \sum_{x=1}^{n} \ x^2 \ p(x)$$

$$= \ \sum_{x=1}^{n} \ x^2 \ \frac{1}{n}$$

$$= \ \frac{1}{n} \ \sum_{x=1}^{n} \ x^2$$

$$= \ \frac{1}{n} [1^2 + 2^2 + ... + n^2]$$

$$= \ \frac{ n (n+1) (2n + 1)}{6n}$$

$$= \ \frac{(n+1) (2n + 1)}{6}$$

$$\therefore \ The \ variance \ of \ the \ distribution \ is$$

V(x) = E(X²) - [E(X)]²

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

$$= \ \frac{(n+1) (n-1)}{12}$$

$$= \ \frac{n^2 - 1}{12}$$

Moment Generating Function of Uniform Distribution

Let X ~ U(x; n). Then the moment generating function of the uniform or rectangular distribution is given by

$$ M_x(t) \ = \ E (e^{tx}) \ = \ \sum_{x=1}^{n} \ e^{tx} p(x)$$

$$ =\ \sum_{x=1}^{n} \ e^{tx} \frac{1}{n}$$

$$= \ \frac{1}{n}\ \sum_{x=1}^{n} \ e^{tx}$$

$$= \ \frac{1}{n} [e^t + e^{2t} + ... + e^{nt}]$$

$$\therefore \ M_x(t) \ = \ \frac{e^t (1-e^{nt})}{n (1-e^t)}$$

Also, the moments about origin of distributioncan be obtained by

$$\mu_{r}^{'} \ = \ \frac{d^r M_x(t)}{dt^r} |_{t = 0}$$

In partical

$$\mu_{1}^{'} \ = \ \frac{d M_x(t)}{dt} |_{t = 0}$$

$$= \ \frac{d}{dt} \left ( \frac{e^t + e^{2t} + ... + e^{nt}}{n} \right )|_{t=0}$$

$$= \ \frac{1}{n} \ [e^t + e^{2t} + ... + e^{nt}] \ |_{t=0}$$

$$= \ \frac{1}{n} \ {1 + 2 + ... + n}$$

$$= \ \frac{n+2}{2}$$

Similarly,

$$\mu_{2}^{'} \ = \ \frac{1}{n} \ [1^2 + 2^2 + ... + n^2]$$

$$= \ \frac{(n+1) (2n+1)}{6}$$

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

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

$$= \ \frac{n^2 - 1}{12}$$

Recurrence Relation of Uniform Distribution

Let X ~ U (x ; n ) then the probability mass function of X is given by

$$p(x) \ = \ \frac{1}{n} \ \ \ \ ;x \ = \ 1, \ 2, \ ..., \ n$$

$$and \ p(x+1) \ = \ \frac{1}{n}$$

$$\rightarrow \ p(x+1) \ = \ p(x) \ \ \ \ ; \ for \ all \ x \ = \ 1, \ 2, \ ..., \ n$$