Random Variable

Random Variable: Definition, Types and 10 Solved Examples

Definition of Random Variable

A random variable is a real-valued function defined on the sample space of a random experiment.

\( X:S\rightarrow \mathbb{R} \)

It assigns a numerical value to every possible outcome of a random experiment.

Example

Suppose two coins are tossed. The sample space is

\( S=\{HH,HT,TH,TT\}. \)

Let \(X\) denote the number of heads obtained. Then

\(X(HH)=2,\quad X(HT)=1,\quad X(TH)=1,\quad X(TT)=0.\)

Therefore,

\(X\in\{0,1,2\}.\)

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Types of Random Variables

1. Discrete Random Variable

A random variable is called discrete if it takes a finite or countably infinite number of values.

Its probability mass function (PMF) is

\(p_X(x)=P(X=x)\)

and it satisfies

\(p_X(x)\geq 0,\qquad \sum_x p_X(x)=1.\)

2. Continuous Random Variable

A random variable is called continuous if it can take any value in an interval of the real line.

It is described by a probability density function \(f_X(x)\) satisfying

\(f_X(x)\geq0\)

\(\displaystyle \int_{-\infty}^{\infty}f_X(x)\,dx=1.\)

For a continuous random variable,

\(\displaystyle P(a<X<b)=\int_a^b f_X(x)\,dx.\)

Also,

\(P(X=x)=0.\)

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10 University-Level Solved Examples

Question 1: Number of Heads in Three Coin Tosses

Three fair coins are tossed. Let \(X\) denote the number of heads obtained.

Find:

1. The probability distribution of \(X\)
2. \(E(X)\)
3. \(\operatorname{Var}(X)\)

Show Solution

The possible values of \(X\) are

\(X=0,1,2,3.\)

Since three fair coins are tossed, the total number of equally likely outcomes is

\(2^3=8.\)

The probability of obtaining \(x\) heads is

\[ P(X=x)=\frac{\binom{3}{x}}{8}. \]

\(x\)	0	1	2	3
\(P(X=x)\)	\(1/8\)	\(3/8\)	\(3/8\)	\(1/8\)

Now,

\[ E(X)=\sum xP(X=x). \]

\[ E(X) =0\left(\frac18\right) +1\left(\frac38\right) +2\left(\frac38\right) +3\left(\frac18\right). \]

\[ \boxed{E(X)=\frac32}. \]

Since \(X\sim Binomial(3,1/2)\),

\[ \operatorname{Var}(X)=np(1-p). \]

\[ =3\left(\frac12\right)\left(\frac12\right) =\boxed{\frac34}. \]

Type: Discrete Random Variable.

Question 2: Random Variable Defined on a Die

A fair die is rolled once. Define

\[ X= \begin{cases} 0,&\text{if the result is odd},\\ 2,&\text{if the result is even}. \end{cases} \]

Find the probability distribution, mean and variance of \(X\).

Show Solution

There are three odd numbers and three even numbers on a die.

\[ P(X=0)=\frac36=\frac12. \]

\[ P(X=2)=\frac36=\frac12. \]

Therefore,

\[ E(X)=0\left(\frac12\right) +2\left(\frac12\right)=1. \]

Next,

\[ E(X^2)=0^2\left(\frac12\right) +2^2\left(\frac12\right)=2. \]

Hence,

\[ \operatorname{Var}(X) =E(X^2)-[E(X)]^2. \]

\[ =2-1^2=1. \]

\[ \boxed{E(X)=1,\qquad \operatorname{Var}(X)=1}. \]

Type: Discrete Random Variable.

Question 3: Find the Unknown Constant in a PMF

A random variable \(X\) has probability mass function

\[ P(X=x)=kx,\qquad x=1,2,3,4. \]

Find:

1. \(k\)
2. \(P(X\geq3)\)
3. \(E(X)\)

Show Solution

Since the total probability must equal 1,

\[ \sum_{x=1}^{4}kx=1. \]

\[ k(1+2+3+4)=1. \]

\[ 10k=1. \]

\[ \boxed{k=\frac1{10}}. \]

Now,

\[ P(X\geq3) =P(X=3)+P(X=4). \]

\[ =\frac3{10}+\frac4{10} =\boxed{\frac7{10}}. \]

For the expectation,

\[ E(X)=\sum xP(X=x). \]

\[ E(X) =\frac1{10}(1^2+2^2+3^2+4^2). \]

\[ =\frac{30}{10}=3. \]

\[ \boxed{E(X)=3}. \]

Type: Discrete Random Variable.

Question 4: Binomial Random Variable

A component has probability \(0.2\) of being defective. Five independently selected components are inspected.

Let \(X\) be the number of defective components. Find \(P(X=2)\), \(E(X)\), and \(\operatorname{Var}(X)\).

Show Solution

Here,

\[ X\sim Binomial(5,0.2). \]

Using

\[ P(X=x)=\binom{n}{x}p^x(1-p)^{n-x}, \]

we obtain

\[ P(X=2) =\binom52(0.2)^2(0.8)^3. \]

\[ =10(0.04)(0.512) =\boxed{0.2048}. \]

The mean is

\[ E(X)=np=5(0.2)=\boxed{1}. \]

The variance is

\[ \operatorname{Var}(X)=np(1-p). \]

\[ =5(0.2)(0.8)=\boxed{0.8}. \]

Type: Discrete Random Variable.

Question 5: Geometric Random Variable

The probability that a machine produces an acceptable item is \(0.8\). Items are inspected independently until the first acceptable item appears.

Let \(X\) denote the number of inspections required. Find \(P(X=4)\), \(E(X)\), and \(\operatorname{Var}(X)\).

Show Solution

Success probability is

\[ p=0.8. \]

Therefore, failure probability is

\[ 1-p=0.2. \]

For the first success to occur on the fourth trial, the first three trials must fail.

\[ P(X=4)=(0.2)^3(0.8). \]

\[ \boxed{P(X=4)=0.0064}. \]

For a geometric random variable,

\[ E(X)=\frac1p. \]

\[ E(X)=\frac1{0.8}=\boxed{1.25}. \]

Also,

\[ \operatorname{Var}(X) =\frac{1-p}{p^2}. \]

\[ =\frac{0.2}{0.8^2} =\boxed{0.3125}. \]

Type: Discrete Random Variable.

Question 6: Find the Constant in a Probability Density Function

A continuous random variable \(X\) has density

\[ f_X(x)= \begin{cases} kx,&0<x<2,\\ 0,&\text{otherwise}. \end{cases} \]

Find \(k\) and \(P(0.5<X<1.5)\).

Show Solution

Since the total area under the density must be 1,

\[ \int_0^2 kx\,dx=1. \]

\[ k\left[\frac{x^2}{2}\right]_0^2=1. \]

\[ 2k=1. \]

\[ \boxed{k=\frac12}. \]

Therefore,

\[ f_X(x)=\frac{x}{2},\qquad0<x<2. \]

Now,

\[ P(0.5<X<1.5) = \int_{0.5}^{1.5}\frac{x}{2}\,dx. \]

\[ = \left[\frac{x^2}{4}\right]_{0.5}^{1.5}. \]

\[ =\frac{2.25-0.25}{4} =\boxed{\frac12}. \]

Type: Continuous Random Variable.

Question 7: Mean and Variance from a PDF

Let

\[ f_X(x)= \begin{cases} 2x,&0<x<1,\\ 0,&\text{otherwise}. \end{cases} \]

Find \(E(X)\) and \(\operatorname{Var}(X)\).

Show Solution

The expectation is

\[ E(X)=\int_0^1 x(2x)\,dx. \]

\[ =2\int_0^1x^2\,dx =2\left[\frac{x^3}{3}\right]_0^1. \]

\[ \boxed{E(X)=\frac23}. \]

Next,

\[ E(X^2) =\int_0^1x^2(2x)\,dx. \]

\[ =2\int_0^1x^3\,dx =\frac12. \]

Therefore,

\[ \operatorname{Var}(X) = E(X^2)-[E(X)]^2. \]

\[ =\frac12-\left(\frac23\right)^2. \]

\[ =\frac12-\frac49 =\boxed{\frac1{18}}. \]

Type: Continuous Random Variable.

Question 8: Uniform Random Variable

A waiting time \(X\), measured in minutes, is uniformly distributed over the interval \([2,10]\).

Find \(P(4<X<7)\), \(E(X)\), and \(\operatorname{Var}(X)\).

Show Solution

Since

\[ X\sim U(2,10), \]

the density is

\[ f_X(x)=\frac{1}{10-2}=\frac18. \]

Therefore,

\[ P(4<X<7) = \int_4^7\frac18\,dx. \]

\[ =\frac{7-4}{8} =\boxed{\frac38}. \]

The mean of a uniform random variable is

\[ E(X)=\frac{a+b}{2}. \]

\[ =\frac{2+10}{2} =\boxed{6}. \]

Its variance is

\[ \operatorname{Var}(X) =\frac{(b-a)^2}{12}. \]

\[ =\frac{(10-2)^2}{12} =\boxed{\frac{16}{3}}. \]

Type: Continuous Random Variable.

Question 9: Exponential Random Variable

The lifetime \(X\) of an electronic device, measured in years, has probability density function

\[ f_X(x)=2e^{-2x},\qquad x\geq0. \]

Find \(P(X>2)\), \(E(X)\), and \(\operatorname{Var}(X)\).

Show Solution

We have

\[ P(X>2) = \int_2^\infty2e^{-2x}\,dx. \]

Since

\[ \int2e^{-2x}\,dx=-e^{-2x}, \]

therefore

\[ P(X>2) = \left[-e^{-2x}\right]_2^\infty. \]

\[ =e^{-4}. \]

\[ \boxed{P(X>2)\approx0.0183}. \]

Here the exponential rate is

\[ \lambda=2. \]

Therefore,

\[ E(X)=\frac1{\lambda} =\boxed{\frac12}. \]

Also,

\[ \operatorname{Var}(X) =\frac1{\lambda^2} =\boxed{\frac14}. \]

Type: Continuous Random Variable.

Question 10: Transformation of a Random Variable

Let \(X\) have probability density

\[ f_X(x)= \begin{cases} 2x,&0<x<1,\\ 0,&\text{otherwise}. \end{cases} \]

Define

\[ Y=X^2. \]

Find the probability density function of \(Y\).

Show Solution

Since

\[ 0<X<1, \]

we have

\[ 0<Y<1. \]

Since

\[ Y=X^2, \]

we obtain

\[ X=\sqrt{Y}. \]

The transformation formula is

\[ f_Y(y) = f_X(x) \left|\frac{dx}{dy}\right|. \]

Since

\[ x=\sqrt y, \]

we get

\[ \frac{dx}{dy} = \frac1{2\sqrt y}. \]

Therefore,

\[ f_Y(y) = 2\sqrt y \left(\frac1{2\sqrt y}\right). \]

\[ \boxed{f_Y(y)=1,\qquad0<y<1}. \]

Hence,

\[ \boxed{Y\sim U(0,1)}. \]

Type: Continuous Random Variable.

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Important Formulas

Discrete Random Variable

\[ \boxed{E(X)=\sum_x xP(X=x)} \]

\[ \boxed{E(X^2)=\sum_xx^2P(X=x)} \]

\[ \boxed{\operatorname{Var}(X)=E(X^2)-[E(X)]^2} \]

Continuous Random Variable

\[ \boxed{ E(X)=\int_{-\infty}^{\infty}xf_X(x)\,dx } \]

\[ \boxed{ E(X^2)=\int_{-\infty}^{\infty}x^2f_X(x)\,dx } \]

\[ \boxed{ \operatorname{Var}(X)=E(X^2)-[E(X)]^2 } \]

Cumulative Distribution Function

For both discrete and continuous random variables,

\[ \boxed{F_X(x)=P(X\leq x)} \]

For a continuous random variable,

\[ F_X(x) = \int_{-\infty}^{x}f_X(t)\,dt. \]