**Examples of uniform distribution is given below :****Example 1 :** If X is uniformly distributed with mean 1 and variance

$\frac{4}{3}$, find P(X < 0)

**Solution :** Let X $\sim$ U[a, b] so that

$p(x)$ = $\frac{1}{b - a}$, a < x < b

Given : Mean = $\frac{1}{2}$(b + a) = 1

b + a = 2

Var (X) = $\frac{1}{12}$ (b - a)$^{2}$ = $\frac{4}{3}$

b - a = $\pm$ 4

On solving we get a = -1 and b = 3 => (a < b)

p(x) = $\frac{1}{4}$; -1 < x < 3

P(X < 0) = $\int_{-1}^{0}$ p(x) dx

= $\frac{1}{4}$ |x|$_{-1}^{0}$

= $\frac{1}{4}$

**Example 2 :** If X is a random variable with a continuous distribution function F, then prove that F(X) has a uniform distribution on [0, 1].

Prove that P[ a $\leq$ F(x) $\leq$ b] = b - a, 0 $\leq$ (a, b) $\leq$ 1

**Solution :** Since F is a distribution function, it is non-decreasing. Let Y = F(X)

Then the distribution function G of Y is given by:

G$_{Y}$(y) = P(Y $\leq$ y)

= P [F(X) $\leq$ y]

= P [ X $\leq$ F$^{-1}$(y)]

Inverse of function exists since F is non decreasing and given to be continuous.

G$_{Y}$(y) = F[F$^{-1}$(y)]

since F is the distribution function of X.

Thus G$_{Y}$(y) = y

Therefore the pdf of Y = F(X) is given by ;

g$_{Y}$(y) = $\frac{d}{dy}$[G$_{Y}$(y)] = 1

Since F is a distribution function Y = F(X) takes the value in the range [0, 1].

Hence g$_{Y}$(y) = 1, 0 $\leq$ y $\leq$ 1

Y is a uniform variate on [0, 1].

Since Y = F(X) $\sim$ U[0, 1]

P[ a $\leq$ F(X) $\leq$ b] = P(a $\leq$ Y $\leq$ b]

= $\int_{a}^{b}$ g(y) dy

= $\int_{a}^{b}$ 1. dy

= |y|$_{a}^{b}$

= b - a