# Distribution Function Of Random Variables

Probability / Saturday, September 22nd, 2018
(Last Updated On: November 16, 2018)

# Distribution Function Definition

Let X be an random variable, then the function such that F:R→R defined by F(X) = P(X ≤ x) is called distribution function of random variable X.

Note 1: F(-∞) = 0 and F(∞) = 1

Note 2: If F(X) is a distribution function of a random variable X if a < b, then P(a < x ≤ b) = F(b) – F(a)

Note 3: P(a ≤ x ≤ b) = P(x = a) + F(b) – F(a)

## Distribution function of discrete random variables

Definition

Let X be a random variable whose space is {x1, x2, x3, … , xn}. Let P( X = xi) = pi then the distribution function F:R→R of the discrete random variable X is given by
$\text{F}\left( \text{X} \right)\text{ }=\text{ P}(\text{X}\le \text{x})\text{ }=\sum\limits_{{{x}_{i}}\le x}{{{p}_{i}}}$
and F(X) is called the Probability Mass Function (PMF) of the discrete random variable X.

Note: Let P( X = xi) = pi then the definition of probability, it follows that Σ pi = 1.

## Distribution function of continuous random variables

Definition

Let X be a continuous random variable with probability density function f(x), we define F:R→R is given by
$F(x)=\int\limits_{-\infty }^{x}{f(x)dx}$
is called the distribution function of the continuous random variable X.

Note: From the definition of F(x) we have the following property,
$(i)F(-\infty )=0$
$(ii)F(\infty )=1$
$(iii)P(a\le x\le b)=F(b)-F(a)=\int\limits_{a}^{b}{f(x)dx}$
$=\int\limits_{-\infty }^{b}{f(x)dx-}\int\limits_{-\infty }^{a}{f(x)dx}$

 Example 01

Verify that p(x) defined by

Find the probability that the variate having this density will fall in the interval [1.5, 2.5]. Also, evaluate the cumulative distribution function F(2.5).

Solution:

Evidently, the given p(x) ≥ 0. Further,
$P(-\infty <x<\infty )=\int\limits_{-\infty }^{\infty }{p(x)dx}$
$\Rightarrow P(-\infty <x<\infty )=\int\limits_{-\infty }^{0}{p(x)dx}+\int\limits_{0}^{\infty }{p(x)dx}$
$\Rightarrow P(-\infty <x<\infty )=0+\int\limits_{0}^{\infty }{{{e}^{-x}}dx}$
$\therefore P(-\infty <x<\infty )=1$
The two conditions verified above show that p(x) is a probability density function.

Next, we find that
$P(1.5<x<2.5)=\int\limits_{1.5}^{2.5}{p(x)dx}$
$\Rightarrow P(1.5<x<2.5)=\int\limits_{1.5}^{2.5}{{{e}^{-x}}dx}$
$\therefore P(-\infty <x<\infty )=-\left( {{e}^{-2.5}}-{{e}^{-1.5}} \right)=0.141$
Lastly we find that
$P(2.5)=\int\limits_{-\infty }^{2.5}{p(x)dx=\int\limits_{0}^{2.5}{{{e}^{-x}}dx}}$
$\Rightarrow P(2.5)=-\left( {{e}^{-2.5}}-1 \right)=0.9179$

 Example 02

For the distribution given by the cumulative function

Find density function. Also, evaluate
$(i)P(0.5<X<0.75)$
$(ii)P(X\le 0.5)$
$(iii)P(X>0.75)$
Solution:

We know that, If F(x) is the cumulative distribution function (cdf) the probability density function f(x) can be obtained by
$f(x)=\frac{d}{dx}F(x)=F'(x)$
Given, cumulative distributive function

Also,
$P(0.5<X<0.75)=F(0.75)-F(0.5)$
$\Rightarrow P(0.5<X<0.75)={{\left( 0.75 \right)}^{2}}-{{\left( 0.5 \right)}^{2}},\left[ \because F(t)=\int{f(x)dx} \right]$
$\therefore P(0.5<X<0.75)=0.3125$
$Now,P(X\le 0.5)=F(0.5)={{\left( 0.5 \right)}^{2}}=0.25$
$Again,P(X>0.75)=1-P(X\le 0.75)=1-F(0.75)$
$\therefore P(X>0.75)=1-{{\left( 0.75 \right)}^{2}}=0.4325$

 Example 03

The pdf of a random variable X is given by

Find (i) cumulative distribution function and (ii) P(X ≥ 1.5)

Solution:

By using formula
$F(t)=P(X\le t)=\int\limits_{-\infty }^{t}{f(x)dx}$

For -∞ < t < 0, we get

$F(t)=\int\limits_{-\infty }^{0}{f(x)dx=}\int\limits_{-\infty }^{0}{0dx=}0$
For 0 ≤ t ≤ 1, we get
$F(t)=\int\limits_{-\infty }^{0}{f(x)dx+\int\limits_{0}^{t}{f(x)dx}}$
$\Rightarrow F(t)=0+\int\limits_{0}^{t}{xdx}=\frac{{{t}^{2}}}{2}$
For 1 < t ≤ 2, we get
$F(t)=\int\limits_{-\infty }^{0}{f(x)dx+\int\limits_{0}^{1}{f(x)dx}}+\int\limits_{1}^{t}{f(x)dx}$
$\Rightarrow F(t)=\int\limits_{-\infty }^{0}{0dx+\int\limits_{0}^{1}{xdx}}+\int\limits_{1}^{t}{\left( 2-x \right)dx}$
$\therefore F(t)=1-\frac{1}{2}{{\left( t-2 \right)}^{2}}$
For t > 2, we get
$F(t)=\int\limits_{-\infty }^{0}{f(x)dx+\int\limits_{0}^{1}{f(x)dx}}+\int\limits_{1}^{2}{f(x)dx}+\int\limits_{2}^{t}{f(x)dx}$
$\Rightarrow F(t)=\int\limits_{-\infty }^{0}{0dx+\int\limits_{0}^{1}{xdx}}+\int\limits_{1}^{2}{\left( 2-x \right)dx}+\int\limits_{2}^{t}{0dx}$
$\therefore F(t)=\frac{1}{2}+\left( 2-\frac{3}{2} \right)=1$
Thus the required cdf is

$Now,P(X\ge 1.5)=\int\limits_{1.5}^{\infty }{f(x)dx}$
$\therefore P(X\ge 1.5)=\int\limits_{1.5}^{2}{\left( 2-x \right)dx}+\int\limits_{2}^{\infty }{0dx=\frac{1}{8}}$