4.10 Conditional Probability Distribution

If P(X=x_i, Y=y_j) is the probability function of discrete random variable X and Y, then the conditional probability function of X given Y=y_j is defined as

$P(X=x_i/Y=y_j)=\frac{P(X=x_1,Y=y_j)}{P(Y=y_i)}$

And the conditional probability function of Y, given X=x_i is denoted by

$P(Y=y_i/X=x_j)=\frac{P(X=x_1,Y=y_j)}{P(Y=x_i)}$

The conditional probability function of X given Y, and the conditional probability function of Y given X are also denoted by f(x/y) and f(y/x), respectively.

The conditional probability distributions are univariate probability distribution.

4.11 Independent Random Variables

If P(X=x, Y=y)=f(x, y) is the joint pdf of discrete random variables X and Y and marginal pdfs f₁(x) and f₂(y) such that P(X=x, Y=y)=f₁(x) f₂(y).

Then X and Y are said to be independent random variables.

Example 4.18: The two-dimensional random variables (X, Y) has the joint density function

$f(x,y)=\frac{x+2y}{27},x=0,1,2;y=0,1,2;$

Find the conditional distribution of Y for X=x. Also find the conditional distribution of X given Y=1?

Solution: The joint probability distribution of X and Y in the tabular form is:

XY	1	2	3
0	0	$\frac{1}{27}$	$\frac{2}{27}$
1	$\frac{2}{27}$	$\frac{3}{27}$	$\frac{4}{27}$
2	$\frac{4}{27}$	$\frac{5}{27}$	$\frac{6}{27}$

We have P(X=x)=f(x).

The conditional probability distribution of Y for X=x is $f(y/x)=\frac{f(x,y)}{f(x)}$ where f(x, y) is the joint probability distribution of X and Y. Hence we get

$\begin{array}{ll}f(y=0)(x=0)\hfill & =\frac{f(X=x,y)}{f(x)}=\frac{f(0,0)}{f(x=0)}=0\hfill \\ f(y=1)(x=0)\hfill & =\frac{f(0,1)}{f(x=0)}=\frac{1/27}{3/27}=\frac{1}{3}\left(\mathrm{Since}f(x=0)=\frac{3}{27}\right)\hfill \\ f(y=2)(x=0)\hfill & =\frac{f(0,2)}{f(x=0)}=\frac{2/27}{3/27}=\frac{2}{3}\hfill \end{array}$

We have

$f(x=1)=\frac{2}{27}+\frac{3}{27}+\frac{4}{27}=\frac{9}{27}$

Hence

$\begin{array}{ll}f(y=0)(x=1)\hfill & =\frac{f(0,1)}{f(x=1)}=\frac{2/27}{9/27}=\frac{2}{9}\hfill \\ f(y=1)(x=1)\hfill & =\frac{f(1,1)}{f(x=1)}=\frac{3/27}{9/27}=\frac{3}{9}\hfill \\ f(y=2)(x=1)\hfill & =\frac{f(2,1)}{f(x=1)}=\frac{4/27}{9/27}=\frac{4}{9}\hfill \end{array}$

Since

$f(x=2)=\frac{4}{27}+\frac{5}{27}+\frac{6}{27}=\frac{15}{27}$

We get

$\begin{array}{ll}f(y=0)(x=2)\hfill & =\frac{f(2,0)}{f(x=2)}=\frac{4/27}{15/27}=\frac{4}{15}\hfill \\ f(y=1)(x=2)\hfill & =\frac{f(2,1)}{f(x=2)}=\frac{5/27}{15/27}=\frac{5}{15}\hfill \\ f(y=2)(x=2)\hfill & =\frac{f(2,2)}{f(x=2)}=\frac{6/27}{15/27}=\frac{6}{15}\hfill \end{array}$

Therefore the conditional distribution of Y for X=x is

XY	1	2	3
0	0	$\frac{1}{3}$	$\frac{2}{3}$
1	$\frac{2}{9}$	$\frac{3}{9}$	$\frac{4}{9}$
2	$\frac{4}{15}$	$\frac{5}{15}$	$\frac{6}{15}$

From the table, the conditional distribution of X given Y=1 is

xy	$\frac{f(x,y)}{f(y=1)}$
0	$\frac{1}{9}$
1	$\frac{3}{9}$
2	$\frac{5}{29}$

4.12 Joint Probability Function of Continuous Random Variables

Definition 4.14

If x and y are two continuous random variables, then the function f(x, y) given by

$p(a_1\le x\le b_1,a_2\le y\le b_2)={\int }_{a_1}^{b_1}{\int }_{a_2}^{b_2}f(x,y)\mathrm{d}y\mathrm{d}x$

is called the joint probability function of X and Y, if and only if it satisfies the following conditions:

1. $f(x,y)\ge 0$, for $-\infty <x<\infty$, $-\infty <y<\infty$

2. ${\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }f(x,y)\mathrm{d}y\mathrm{d}x=1$

The joint probability function of X and Y is also called the joint pdf of the continuous random variables X and Y.

4.13 Joint Probability Distribution Function of Continuous Random Variables

Definition 4.15

If X and Y are continuous random variables, the function given by

$\begin{array}{ll}f(x,y)\hfill & =P(X\le x,Y\le y)\hfill \\ \hfill & ={\int }_{-\infty }^y{\int }_{-\infty }^{x}f(s,t)\mathrm{d}s\mathrm{d}t\mathrm{for}-\infty <x<\infty ,-\infty <y<\infty \hfill \end{array}$

where f(s, t) is the value of the joint pdf of X and Y at (s, t) is called the joint probability distribution function or joint distribution function of X and Y.

We have $f(x,y)=\frac{{\partial }^2}{\partial x\partial y}F(x,y)$

Hence the joint distribution function is obtained by integrating the joint probability function, i.e.,

$f(x,y)={\int }_{-\infty }^y{\int }_{-\infty }^{x}f(t_1,t_2)\mathrm{d}{t}_1\mathrm{d}{t}_2$

4.14 Marginal Distribution Function

Let X and Y be continuous random variables. If f(x, y) is the joint distribution function of the random variables X and Y, then the marginal distribution function of X denoted by F(x) is given by

$F(x)={\int }_{-\infty }^x\left({\int }_{-\infty }^{\infty }f(x,y)\mathrm{d}y\right)\mathrm{d}x$

And the marginal distribution function of Y denoted by F(Y) is given by

$F(Y)={\int }_{-\infty }^y\left({\int }_{-\infty }^{\infty }f(x,y)\mathrm{d}x\right)\mathrm{d}y$

where f(x, y) is the joint probability function of X and Y.

4.14.1 Marginal Density Functions

Let X and Y be continuous random variables and f(x, y) be the value of their joint probability density at (x, y). Then the function

$f_1(x)={\int }_{-\infty }^{\infty }f(x,y)\mathrm{d}x\mathrm{for}-\infty <t<\infty$

is called the marginal density function of X. Correspondingly, the function given by

$f_2(x)={\int }_{-\infty }^{\infty }f(x,y)\mathrm{d}y\mathrm{for}-\infty <y<\infty$

is called the marginal density of Y.

Example 4.19: The joint probability of continuous random variables X and Y is given by

$\begin{array}{lll}f(x,y)\hfill & =8xy,\hfill & 0\le x\le 1,0\le y\le x\hfill \\ \hfill & =0,\hfill & \mathrm{otherwise}\hfill \end{array}$

Find the marginal pdfs f(x) and f(y)?

Solution: We have

$\begin{array}{ll}f(x)\hfill & ={\int }_{-\infty }^{\infty }f(x,y)\mathrm{d}y(\mathrm{By}\mathrm{definition})\hfill \\ \hfill & ={\int }_0^{x}8xy\mathrm{d}y\hfill \\ \hfill & =8x{\left|\frac{y^2}{2}\right|}_x^0\hfill \\ \hfill & =4x^3,0\le x\le 1\hfill \end{array}$

and

$\begin{array}{ll}f(y)\hfill & ={\int }_{-\infty }^{\infty }f(x,y)\mathrm{d}x\hfill \\ \hfill & ={\int }_0^{1}8xy\mathrm{d}x\hfill \\ \hfill & =8y{\left|\frac{x^2}{2}\right|}_0^1\hfill \\ \hfill & =4y,0\le y\le x\hfill \end{array}$

4.15 Conditional Probability Density Functions

Definition 4.16

If X and Y are continuous random variables and f(x, y) is the value of the joint density function of X and Y at (x, y) then the conditional density of Y given X=x is defined as

$\begin{array}{ll}f(y/x)\hfill & =\frac{\partial }{\partial y}F\left(y/x\right)\hfill \\ \hfill & =\frac{f(x,y)}{f(x)},f(x)\ne 0\mathrm{for}-\infty <x<\infty \hfill \end{array}$

And the conditional density of X given Y=y is defined as

$f\left(x/y\right)=\frac{f(x,y)}{f(y)},f(y)\ne 0$

Example 4.20: The joint density function of continuous random variables X and Y is given by f(x, y)=2, 0<x<y<1. Find the marginal and conditional pdfs?

Solution: The marginal density function of X given y is

$f(x)={\int }_{-\infty }^{\infty }f(x,y)\mathrm{d}y={\int }_x^{1}2\mathrm{d}y={\left[2y\right]}_x^1=2(1-x)$

The marginal density function of Y given x is

$f(y)={\int }_{-\infty }^{\infty }f(x,y)\mathrm{d}x={\int }_0^{1}2\mathrm{d}x={\left[2x\right]}_0^1=2$

Conditional pdf of X given y is

$f\left(y/x\right)=\frac{f(x,y)}{f(x)}=\frac{2}{2(1-x)}=\frac{1}{1-x},0<x<1$

Conditional pdf of Y given x is

$f(x/y)=\frac{f(x,y)}{f(y)}=\frac{2}{2}=1$

Remarks

The conditional distribution of a random variable entering into a system is its distribution calculated under the condition that the other random variable has assumed a definite value.

The random variables X and Y are said to be mutually independent if the conditional probability distribution of one does not depend on the value of the other, i.e.,

$f(x/y)f(x)\mathrm{or}f(y/x)=f(y)$

The above mentioned properties of one-dimensional (univariate) and two-dimensional (bivariate) random variables can also be established for multidimensional random variables.

Exercise 4.3

1. Find the constant k so that the function

$\begin{array}{lll}f(x)\hfill & =\frac{1}{k},\hfill & a\le x\le b\hfill \\ \hfill & =0,\hfill & \mathrm{otherwise}\hfill \end{array}$

is a density function. Also find cumulative distributive function (c.d.f) of the random variable X?

Ans: $k=b–a,f(x)=\frac{x-a}{b-a}$

2. Distribution function of a random variable X is

$\begin{array}{lll}f(x)\hfill & =x,\hfill & 0<x<1\hfill \\ \hfill & =2-x,\hfill & 1\le x\le 2\hfill \\ \hfill & =0,\hfill & x\le 2\hfill \end{array}$

Compute the cumulative distribution function of X.

Ans: $\begin{array}{lll}f(x)\hfill & =\frac{x-a}{b-a}\hfill & 0<x<1\hfill \\ \hfill & =2x-\frac{x^2}{2}-1,\hfill & 1\le x\le 2\hfill \\ \hfill & =1,\hfill & x\ge 2\hfill \end{array}$

3. A continuous random variable X that can assume any value between x=2 and x=5 has a density function given by f(x)=k(1+x). Find k, P(X<4)?

Ans: $k=\frac{2}{27};P(X<4)=\frac{16}{27}$

4. Suppose that the duration in minutes of a long-distance telephone conversation follows an exponential density function.

$f(x)=\frac{1}{5}{e}^{–x/5},\mathrm{for}x>0$

Find the probability that duration of a conversation

a. Will exceed 5 minutes?

b. Will be less than 3 minutes?

Ans: (a) e⁻¹ (b) $1-e^{-3/5}$

5. If X is a continuous random variable with distribution

$\begin{array}{lll}f(x)\hfill & =kx,\hfill & 0\le x\le 5\hfill \\ \hfill & -0,\hfill & \mathrm{otherwise}\hfill \end{array}$

Find

a. k

b. P(1<x<3)

c. P(2≤x≤4)

d. P(x≤3)

Ans: (a) $k=\frac{2}{25}$; (b) $\frac{8}{25}$; (c) $\frac{12}{25}$; (d) $\frac{9}{25}$

6. Let X be a continuous random variate with pdf

$\begin{array}{lll}f(x)\hfill & =ax,\hfill & 0\le x\le 1\hfill \\ \hfill & =a,\hfill & 1\le x\le 2\hfill \\ \hfill & =–ax+3a,\hfill & 2\le x\le 3\hfill \\ \hfill & =0,\hfill & \mathrm{otherwise}\hfill \end{array}$

a. Determine the constant a.

b. Compute P(X<1.5)

Ans: (a) $a=\frac{1}{2}$; (b) $\frac{1}{2}$

7. The distribution function of a random variable X is given by

$\begin{array}{lll}F(x)\hfill & =1-(1+x)e^{-x},\hfill & x\ge 0\hfill \\ \hfill & =0,\hfill & \mathrm{for}x<0\hfill \end{array}$

Find the corresponding density function of X?

Ans: $\begin{array}{lll}f(x)\hfill & =x{e}^{–x},\hfill & x\ge 0\hfill \\ \hfill & =0,\hfill & x<0\hfill \end{array}$

8. Two random variables X and Y have the probability function

$\begin{array}{lll}f(x,y)\hfill & =A{e}^{-(2x+y)},\hfill & x,y\ge 0\hfill \\ \hfill & =0,\hfill & \mathrm{otherwise}\hfill \end{array}$

a. Evaluate A

b. Find the marginal pdfs.

Ans: (a) A=2; (b) $\begin{array}{lll}f(x)\hfill & =2e^{-2x},\hfill & x\ge 0,\hfill \\ \hfill & =0,\hfill & \mathrm{otherwise}\hfill \end{array}$, $\begin{array}{lll}f(y)\hfill & =e^{-y},\hfill & y\ge 0\hfill \\ \hfill & =0,\hfill & y<0\hfill \end{array}$

9. If $\begin{array}{lll}f(x,y)\hfill & =2–x–y,\hfill & 0\le x\le 1,0\le y\le 1\hfill \\ \hfill & =0,\hfill & \mathrm{elsewhere}\hfill \end{array}$
Find

a. Marginal probability function?

b. Conditional probability function?

Ans: (a) $f(x)=\frac{3}{2}-x;f(y)=\frac{3}{2}-y$

   (b) $f(y/x)=\frac{2-x-y}{\frac{3}{2}-x},f(x/y)=\frac{2-x-y}{\frac{3}{2}-y}$

10. Let X and Y have joint density function

$\begin{array}{lll}f(x,y)\hfill & =e^{-(x+y)},\hfill & x>0,y>\infty \hfill \\ \hfill & =0,\hfill & \mathrm{otherwise}\hfill \end{array}$

Find P(0<x<1/y=2)?

Ans: $\frac{e-1}{e}$

11. The joint density function of a bivariate distribution is given as

$\begin{array}{lll}f(x,y)\hfill & =x+y,\hfill & 0\le x\le 1,0\le y\le 1\hfill \\ \hfill & =0,\hfill & \mathrm{elsewhere}\hfill \end{array}$

Determine the marginal distributions of x and y.

Ans: $x+\frac{1}{2},0\le x\le 1,y+\frac{1}{2},0\le y\le 1$

12. Let X and Y be the joint distributed with pdf

$\begin{array}{lll}f(x,y)\hfill & =\frac{1}{4}(1+xy),\hfill & \left|x\right|<1,\left|y\right|<1\hfill \\ \hfill & =0,\hfill & \mathrm{elsewhere}\hfill \end{array}$

Show that X and Y are independent?

13. The joint probability function of two random variables X and Y is

$\begin{array}{lll}f(x,y)\hfill & =c(1+xy),\hfill & 0\le x\le 6,0\le y\le 5\hfill \\ \hfill & =0,\hfill & \mathrm{elsewhere}\hfill \end{array}$

Find

a. c

b. F(0.1, 0.5)

c. f(x, 3)

Ans: (a) $\frac{1}{255}$; (b) 0.0006102;
   (c) $\begin{array}{lll}f(x,3)\hfill & =\frac{1}{255}(1+3x),\hfill & 0\le x\le 6\hfill \\ \hfill & =0,\hfill & \mathrm{otherwise}\hfill \end{array}$

14. Let X and Y be two continuous random variables with joint pdf

$\begin{array}{lll}f(x,y)\hfill & =c(x-y),\hfill & 0\le x\le 2,-x\le y\le x\hfill \\ \hfill & =0,\hfill & \mathrm{elsewhere}\hfill \end{array}$

a. Evaluate c

b. find f(x)

c. find f(y/x)

Ans: (a) $c=\frac{1}{8}$; (b) $\frac{x^2}{4},0<x<2$; (c) $\frac{x-y}{2x^2},0<x<2,-x<y<x$

15. Joint distribution of X and Y is given by

$f(x,y)=4xy{e}^{-(x^2+y^2)},x\ge 0,y\ge 0$

Show that X and Y are independent. Find the conditional density of X given Y=y?

Ans: $2x{e}^{-x^2}$

16. The joint distribution of two random variables X and Y is given by the following table:

XY	2	3	4
1	0.06	0.15	0.09
2	0.14	0.35	0.21

Determine the individual distributions of X and Y. Also verify that X and Y are stochastically independent.

Ans:

X=x	1	2
f(x)	0.3	0.7

Y=y	2	3	4
f(y)	0.2	0.5	0.3

17. Determine the value of k for which the function given by

$f(x,y)=kxy,\mathrm{for}x=1,2,3,y=1,2,3$

can serve as a joint probability distribution.

Ans: $k=\frac{1}{36}$

18. Given the joint pdf

$\begin{array}{lll}f(x,y)\hfill & =\frac{3}{5}x(y+x),\hfill & 0\le x\le 1,0\le y\le 2\hfill \\ \hfill & =0,\hfill & \mathrm{elsewhere}\hfill \end{array}$

of two random variables X and Y. Find P{(X, Y)∈A} where A is the region

$\left\{(x,y):0<X<\frac{1}{2},1<Y<2\right\}$

Ans: $\frac{11}{80}$

19. Find the joint probability density of the two random variables X and Y whose joint distribution is given by

$\begin{array}{lll}f(x,y)\hfill & =(1-e^{-x})(1-e-y),\hfill & x>0,y>0\hfill \\ \hfill & =0,\hfill & \mathrm{otherwise}\hfill \end{array}$

Also use the joint probability density to determine P(1<X<3, 1<Y<2).

Ans: $\begin{array}{lll}f(x,y)\hfill & =e^{-(x+y)},\hfill & x>0,y>0\hfill \\ \hfill & =0,\hfill & \mathrm{elsewhere}\hfill \end{array}$

4.16 Mathematical Expectation and Moments

In this section we introduce special constants, which serve to give a quantitative description of random variables. The constants of particular importance are mathematical expectation, variance, and moments of various orders. We begin our study by defining mathematical expectation, which is the central value of a variable distribution.

Mathematical Expectation: We shall first give a definition of mathematical expectation for discrete random variable.

Definition 4.17

Let X be discrete random variable. Let x₁, x₂, …, x_n denote possible values of X and let p₁, p₂, …, p_n denote the corresponding probabilities. Then the mathematical expectation denoted by E(X) is defined as

$E(X)=\sum _{i=1}^n{x}_i{p}_{i}i=1,2,3,\dots ,n$

If X is a continuous random variable having f(x) is its pdf, then the mathematical expectation of X is defined as

$E(X)={\int }_{-\infty }^{\infty }xf(x)\mathrm{d}x$

Mathematical expectation is also called the expectation or mean of the probability distribution.

Expectation is also denoted by μ (or by $\overline{x}$) and is defined by $\sum xP(X=x)$ (or by $\sum (X)=\sum xf(x)$ where f(x)=P(X=x)).

Example: Consider an experiment in which a die is thrown we have S={1, 2, 3, 4, 5, 6}. Let X be the number of points obtained in a single throw. Then the corresponding probabilities are $\frac{1}{6},$ $\frac{1}{6},$ $\frac{1}{6},$ $\frac{1}{6},$ $\frac{1}{6}$, and $\frac{1}{6}$.

The expected value of X is

$\begin{array}{l}E(X)=1\cdot \frac{1}{6}+2\cdot \frac{1}{6}+3\cdot \frac{1}{6}+4\cdot \frac{1}{6}+5\cdot \frac{1}{6}+6\cdot \frac{1}{6}\hfill \\ \frac{21}{6}=\frac{7}{2}\hfill \end{array}$

4.16.1 Properties of Mathematical Expectation

Theorem 1

If X is a random variable and k is a real number, then

1. E(k)=k

2. E(K x)=k E(x)

3. E(x+k)=E(x)+k

Proof

1. If X is a discrete random variable, then

$E(k)=\sum _{i=1}^{n}k{p}_i=k\sum _{i=1}^n{p}_i=k\cdot 1=k$

If X is a continuous random variable, then

$E(k)={\int }_{-\infty }^{\infty }kf(x)\mathrm{d}x=k{\int }_{-\infty }^{\infty }f(x)\mathrm{d}x=k·1=k$

Thus E(k)=k

2. Let X be a discrete random variable. Then

$\begin{array}{ll}E(kx)\hfill & =\sum _{i=1}^{n}k{x}_i{p}_i\hfill \\ \hfill & =k\sum _{i=1}^n{x}_i{p}_i\hfill \\ \hfill & =kE(x)\hfill \end{array}$

If X is a continuous random variable, then

$\begin{array}{ll}E(kx)\hfill & ={\int }_{-\infty }^{\infty }kxf(x)\mathrm{d}x\hfill \\ k\hfill & ={\int }_{-\infty }^{\infty }xf(x)\mathrm{d}x\hfill \\ \hfill & =kE(x)\hfill \end{array}$

Remark

In particular we have

$E(–x)=–E(x)$

3. If X is a discrete random variable then

$\begin{array}{ll}E(X+k)\hfill & =\sum _{i=1}^n(k+x_i)p_i\hfill \\ \hfill & =\sum _{i=1}^n{x}_i{p}_i+\sum _{i=1}^{n}k{p}_i\hfill \\ \hfill & =\sum _{i=1}^n{x}_i{p}_i+k\sum _{i=1}^n{p}_i\hfill \\ \hfill & =E(X)+k\cdot 1=E(X)+k\hfill \end{array}$

If X is a continuous random variable then we have

$\begin{array}{ll}E(x+k)\hfill & ={\int }_{-\infty }^{\infty }(x+k)f(x)\mathrm{d}x\hfill \\ \hfill & ={\int }_{-\infty }^{\infty }xf(x)\mathrm{d}x+{\int }_{-\infty }^{\infty }kf(x)\mathrm{d}x\hfill \\ \hfill & ={\int }_{-\infty }^{\infty }xf(x)\mathrm{d}x+k{\int }_{-\infty }^{\infty }f(x)\mathrm{d}x\hfill \\ \hfill & =E(X)+k\cdot 1=E(X)+k\hfill \end{array}$

Hence proved.

Remark

The mathematical expectation of X need not be finite. For example, consider

The probability of the discontinuous random variable k defined by

$p(k)=\frac{e^{-1}}{k!},k=0,1,2,\dots$

The expectation E(k!) is

$\begin{array}{ll}E(k!)\hfill & =\sum _{k=0}^{\infty }k!\frac{e^{-1}}{k!}=\sum _{k=0}^{\infty }{e}^{-1}\hfill \\ \hfill & =e^{-1}\sum _{k=0}^{\infty }1\mathrm{which}\mathrm{is}\mathrm{not}\mathrm{finite}.\hfill \end{array}$

Theorem 2

If X and Y are random variables, then

E(X+Y)=E(X)+E(Y)

Proof

Let X and Y be two discrete random variables.

Let x₁, x₂, …, x_n be the values assumed by the random variable X. Let p₁, p₂, …, p_n denote the corresponding probabilities.

Let y₁, y₂, …, y_n be the values assumed by the random variable Y and ${p\prime }_1,{p\prime }_2,\dots ,{p\prime }_n$ denote the corresponding probabilities. Also let P(X=x_i, Y=y_j)=p_ij

By definition, we have

$\begin{array}{ll}E(X+Y)\hfill & =\sum _{i=1}^m\sum _{j=1}^n(x_i+y)p_{ij}\hfill \\ \hfill & =\sum _{i=1}^m\sum _{j=1}^n{x}_i{p}_{ij}+\sum _{i=1}^m\sum _{j=1}^n{y}_j{p}_{ij}\hfill \\ \hfill & =\sum _{i=1}^m\left[\sum _{j=1}^n{p}_{ij}\right]+\sum _{j=1}^n{p}_{ij}\left[\sum _{i=1}^m{p}_{ij}\right]\hfill \\ \hfill & =\sum _{i=1}^m{x}_i{p}_i+\sum _{j=1}^n{y}_j{p}_j\hfill \\ \hfill & =E(X)+E(Y)\hfill \end{array}$ (4.1)

(4.1)

Let X and Y be continuous random variables and let f(x, y) be the pdf of X and Y. Then

$\begin{array}{ll}E(X+Y)\hfill & ={\int }_{-\infty }^{\infty }(x+y)f(x,y)\mathrm{d}x\mathrm{d}y\hfill \\ \hfill & ={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }xf(x,y)\mathrm{d}x\mathrm{d}y+{\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }yf(x,y)\mathrm{d}x\mathrm{d}y\hfill \\ \hfill & ={\int }_{-\infty }^{\infty }x\left[{\int }_{-\infty }^{\infty }(x,y)\mathrm{d}y\right]\mathrm{d}x+{\int }_{-\infty }^{\infty }y\left[{\int }_{-\infty }^{\infty }f(x,y)\mathrm{d}x\right]\mathrm{d}y\hfill \\ \hfill & ={\int }_{-\infty }^{\infty }x{f}_1(x)\mathrm{d}x+{\int }_{-\infty }^{\infty }y{f}_2(y)\mathrm{d}y\hfill \\ \hfill & (f_1(x),f_2(y)\mathrm{are}\mathrm{marginal}\mathrm{density}\mathrm{functions}\mathrm{of}X\mathrm{and}Y,\mathrm{respectively})\hfill \\ \hfill & =E(X)+E(Y)\hfill \end{array}$ (4.2)

(4.2)

From Eqs. (4.1) and (4.2), we conclude that

$E(X+Y)=E(X)+E(Y)$

The above theorem can be generalized and stated as follows:

Cor 1: If X, Y, Z, …, are random variables then?

$E(X+Y+Z+\cdots )=E(X)+E(Y)+E(Z)+\cdots$

Definition 4.18

Two jointly distributed random variables X and Y are statistically independent for each other if and only if the joint pdf equals the product of the two marginal pdfs,

$\mathrm{i}\mathrm{.e}.,f(x,y)=f_1(x)f_2(y)$

where f₁(x) and f₂(y) are marginal pdfs of X and Y, respectively.

Theorem 3

If X and Y are two independent random variables, then

$E(XY)=E(X)E(Y)$

Proof

Let X and Y be two independent random variables and let X, Y be discrete.

Let X assume the m values x₁, x₂, …, x_m. Let p₁, p₂, …, p_m denote the corresponding probabilities.

Let Y assume the n values y₁, y₂, …, y_n and corresponding probabilities be denoted as ${p\prime }_1,{p\prime }_2,\dots ,{p\prime }_n$.

Then

$\begin{array}{ll}E(XY)\hfill & =\sum _{i=1}^m\sum _{j=1}^n{p}_i{p\prime }_j{x}_i{x}_j\hfill \\ \hfill & =\sum _{i=1}^m{x}_i{p}_i+\sum _{j=1}^n{y}_i{p\prime }_j\hfill \\ \hfill & =E(X)E(Y)\hfill \end{array}$

When X and Y are continuous random variables that are independent, the theorem holds and is left as an exercise to the student.

Generalization: The above theorem can be extended to the case of several variables and stated as follows:

If X, Y, Z, … are independent random variables

$E(XYZ\dots )=E(X)E(Y)E(Z)\dots$

4.16.2 Variance

Definition 4.19

Let X be a discrete random variable having probability function P(X=x), then the variance of X is defined as variance of X=E[(X−μ)²] where μ is the mean of X.

Since μ=E(X) the variance of X can be written as

Variance of X=E[(X−E(X))²]

Variance of X is denoted by Var[X] [or by V[X]] or σ²

If x₁, x₂, …, x_n are the values assumed by X, and p₁, p₂, …, p_n are the corresponding probabilities then

${\sigma }^2=\mathrm{Var}\left[X\right]=\sum _{i=1}^n{(x_i-\mu )}^2{p}_i$

If X is continuous random variable then the variance of X defined as follows:

Definition 4.20

Let X be a continuous random variable, having pdf f(x) then

$\mathrm{Var}\left[X\right]={\sigma }^2={\int }_{-\infty }^{\infty }{(x-\mu )}^{2}f(x)\mathrm{d}x$

Remarks

Variance gives an idea of how widely spread the values of random variables likely to be. If the value of variance is larger, then the observations are more scattered on average.

4.16.3 Properties of Variance

Theorem 4

If X is a random variable, then V[X]=E[x²]−[E(X)]²

Proof

Let X be discrete random variable.

By definition we have

$\begin{array}{ll}\mathrm{Var}\left[X\right]\hfill & =V\left[X\right]–E\left[{(X–\mu )}^2\right]\hfill \\ \hfill & =E[X^2–2X\mu +{\mu }^2]\hfill \\ \hfill & =E\left[X^2\right]–2E\left[X\mu \right]+E\left[{\mu }^2\right]\hfill \\ \hfill & =E[X^2–2\mu E[X]+{\mu }^2](\mathrm{since}{\mu }^2\mathrm{is}\mathrm{a}\mathrm{constant})\hfill \\ \hfill & =E\left[X^2\right]–2E(X)E(Y)+{[E(X)]}^2(\mathrm{Since}\mu =E(X))\hfill \\ \mathrm{Thus}\hfill & V\left[X\right]=E\left[x^2\right]–{[E(X)]}^2\hfill \end{array}$

The above property holds good for continuous variables also.

Theorem 5

If X is a random variable and k is a real number, then

1. V[k x]=k² V[X]

2. V[x+k]=V[x]

Proof

1. $\begin{array}{ll}V\left[kx\right]\hfill & =E\left[{(kx)}^2\right]-{[E(kX)]}^2\hfill \\ \hfill & =E\left[k^2{x}^2\right]-{[kE(X)]}^2\hfill \\ \hfill & =k^{2}E\left[x^2\right]-k^2{[E(X)]}^2\hfill \\ \hfill & =k^2\left[E\right[x^2]-{[E(X)]}^2]\hfill \\ V\left[kx\right]\hfill & =k^{2}V\left[x\right]\hfill \end{array}$

2. $\begin{array}{ll}V(X+k)\hfill & =E[(X+k^2)]-{[E(X+k)]}^2\hfill \\ \hfill & =E[X^2+2kX+k^2]-{[E(X)+k]}^2\hfill \\ \hfill & =E\left[\right[X^2]+2kE(X)+E[k^2\left]\right]-E{(X)}^2+2kE(X)-k^2\hfill \\ \hfill & =E(X^2)+k^2-{[E(X)]}^2-k^2(\mathrm{since}E(k^2)=k^2)\hfill \\ \hfill & =E(X^2)-{[E(X)]}^2\hfill \\ \hfill & =V\left[x\right]\hfill \end{array}$

Hence proved.

Remarks

The positive square root of variance is called the SD of X.

1. $\begin{array}{ll}\mathrm{Standard}\mathrm{deviation}\hfill & =\sigma =\sqrt{V(X)}\hfill \\ \hfill & =\sqrt{E(X^2)-{[E(X)]}^2}\hfill \end{array}$

2. Since $[X=E(X^2)]\ge 0$, variance $X\ge 0$

3. V[x]=0 if and only if X takes only one value with probability 1.

4. Variance of a constant is zero.

5. V[x] is invariant to the change of origin but variance is not invariant to the change of scale.

6. Let X, Y be two random variables and $Y=\frac{X-k}{h}$ then ${\sigma }_y^2=\frac{1}{h^2}{\sigma }_x^2$ where ${\sigma }_x^2=V\left[x\right]$ and ${\sigma }_y^2=V\left[y\right]$.

Example 4.21: X and Y are two random variables such that Y≤X. If E(X) and E(Y) exist, show that E[Y]≤E[X].

Solution: We have Y≤X given,

i.e., Y−X≤0

Therefore we get E[Y−X]≤0 (applying expectation on both sides),

i.e., E[Y]−E[X]≤0

or E[Y]≤E[X].

Example 4.22: If X is a random variable and E[X] exists, then show that $\left|E\right[X\left]\right|\le E\left|X\right|$?

Solution: Since $X\le \left|X\right|$

We have

$E\left[X\right]\le E\left|X\right|$ (4.3)

(4.3)

Also we have

$-X\le \left|X\right|$

Therefore

$\begin{array}{c}E[-X]\le E\left|X\right|\\ -E\left[X\right]\le E\left|X\right|\end{array}$ (4.4)

(4.4)

From Eqs. (4.3) and (4.4), we get $\left|E\right[X\left]\right|\le E\left|X\right|$

Example 4.23: If X and Y are independent random variables with density functions

$\begin{array}{l}f(x)=\frac{8}{x^3}\mathrm{for}x>2\mathrm{and}\hfill \\ f(y)=2y\mathrm{for}0<y<1\hfill \end{array}$

Find $E\left[XY\right]$?

Solution: Since X and Y are independent random variables

$f(x,y)=f(x)f(y)$

Therefore

$\begin{array}{ll}E\left[XY\right]\hfill & ={\int }_2^{\infty }{\int }_0^{1}xyf(x,y)\mathrm{d}x\mathrm{d}y\hfill \\ \hfill & ={\int }_2^{\infty }{\int }_0^{1}xy\left(\frac{8}{x^3}\right)(2y)\mathrm{d}x\mathrm{d}y\hfill \\ \hfill & =16{\int }_2^{\infty }\left(\frac{1}{x^2}{\int }_0^1{y}^2\mathrm{d}y\right)\mathrm{d}x\hfill \\ \hfill & =16{\int }_2^{\infty }\frac{1}{x^2}{\left[\frac{y^3}{3}\right]}_0^1\mathrm{d}x\hfill \\ \hfill & =\frac{16}{3}{\left[\frac{x^{-1}}{-1}\right]}_2^{\infty }\hfill \\ \hfill & =\frac{16}{3}{\left[\frac{1}{x}\right]}_2^{\infty }=\frac{16}{3}\left[0-\frac{1}{2}\right]=\frac{8}{3}\hfill \end{array}$

Example 4.24: If X takes the values x_n=(−1)ⁿ 2ⁿ n⁻¹ for n=1, 2, … with probabilities p_n=2⁻ⁿ, then show that E[X]=−log 2

Solution: Using the definition $E\left[X\right]=\sum x_i{p}_i$ we get

$\begin{array}{ll}\sum \left[X\right]\hfill & =\sum _{i=1}^{\infty }{2}^{-n}{(-1)}^n{2}^n{n}^{-1}\hfill \\ \hfill & =\sum _{i=1}^{\infty }\frac{{(-1)}^n}{n}\hfill \\ \hfill & -1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\cdots \hfill \\ \hfill & =-\left[1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\cdots \right]\hfill \\ \hfill & =-\log 2\hfill \end{array}$

Example 4.25: Find variance for the following probability distribution:

x	8	12	16	20	24
P(X=x)	$\frac{1}{8}$	$\frac{1}{6}$	$\frac{3}{8}$	$\frac{1}{4}$	$\frac{1}{12}$

Solution: We have

$\begin{array}{ll}E\left[X\right]\hfill & =\sum _i{x}_i{p}_i\hfill \\ \hfill & =8\cdot \frac{1}{8}+12\cdot \frac{1}{6}+16\frac{3}{8}+20\cdot \frac{1}{4}+24\cdot \frac{1}{12}\hfill \\ \hfill & =1+2+6+5+2=16\hfill \end{array}$

$\begin{array}{cc}\mathrm{Mean}& =E\left[X\right]=16\end{array}$

$\begin{array}{ll}\mathrm{Variance}\hfill & =E(X^2)-{[E(X)]}^2\hfill \\ \hfill & =\sum _i{x}_i^2{p}_i-{[E(X)]}^2\hfill \\ \hfill & =8^2\cdot \frac{1}{8}+{12}^2\cdot \frac{1}{6}+{16}^2\cdot \frac{3}{8}+{20}^2\cdot \frac{1}{4}+{24}^2\cdot \frac{1}{12}-{16}^2\hfill \\ \hfill & =8+24+96+100+48–256\hfill \\ \hfill & =276–256=20\hfill \end{array}$

Therefore $V\left[X\right]=20$.

Example 4.26: Let the variable X have the distribution P(X=0)=P(X=2)=p, P(X=1)=1–2p, for $0\le p\le 1/2$. For what values of p is the Var[X] maximum?

Solution: The probability distribution of X is

$\begin{array}{llll}X:\hfill & 0\hfill & 1\hfill & 1\hfill \\ P(X=x):\hfill & p\hfill & 1-2p\hfill & p\hfill \end{array}$

$\begin{array}{ll}\mathrm{Therefore}\mathrm{expectation}\hfill & =E(X)=0\cdot p+1\cdot (1–2p)+2\cdot p\hfill \\ \hfill & =0+1–2p+2p=1\hfill \\ \hfill \mathrm{Variance}& =\mathrm{Var}\left[X\right]=E(x^2)-{[E(x)]}^2\hfill \\ \hfill & =0^2\cdot p+1^2\cdot (1–2p)+2^2\cdot p–1^2\hfill \\ \hfill & =0+1–2p+4p–1\hfill \\ \hfill & =2p\hfill \end{array}$

Clearly Var[X] is maximum when $p=1/2$.

(since the maximum value Var[X] can take is 1 in $0\le p\le \frac{1}{2}$ occurs is at $p=\frac{1}{2}$)

Example 4.27: A coin is tossed until a head appears. What is the expectation of the number of tosses required?

Solution: Head can appear in the first toss, or in the second toss, or in the third toss, and so on.

The favorable cases are $\mathrm{H},\text{TH},\text{TTH},\mathrm{TTTH},\dots$

The probability in which head appears in the first toss=$\frac{1}{2}$

The probability in which head appears in the second toss=${\left(\frac{1}{2}\right)}^2$

The probability of getting head in the nth toss=${\left(\frac{1}{2}\right)}^n$

Let X denote the number of tosses required to get the first head. The probability distribution of X is

X=x	1	2	…	n	…
P(X=x)	$\frac{1}{2}$	${\left(\frac{1}{2}\right)}^2$	…	${\left(\frac{1}{2}\right)}^n$	…

The expectation of number of tosses required is $E(X)=\sum _i{x}_i{p}_i$

i.e.,

$\begin{array}{ll}E(X)\hfill & =1\cdot \frac{1}{2}+2\cdot {\left({\displaystyle \frac{1}{2}}\right)}^2+3\cdot {\left({\displaystyle \frac{1}{2}}\right)}^3+\cdots +n\cdot {\left({\displaystyle \frac{1}{2}}\right)}^n+\cdots \hfill \\ \hfill & =\frac{1}{2}\left[1+2+3{\left({\displaystyle \frac{1}{2}}\right)}^2+4{\left({\displaystyle \frac{1}{2}}\right)}^3+\cdots +n{\left({\displaystyle \frac{1}{2}}\right)}^{n-1}+\cdots \right]\hfill \\ \hfill & =\frac{1}{2}{\left[1-{\displaystyle \frac{1}{2}}\right]}^{-2}\hfill \\ \hfill & =\frac{1}{2}{\left({\displaystyle \frac{1}{2}}\right)}^{-2}=\frac{2^2}{2}=2\hfill \end{array}$

Example 4.28:

1. What is the expected value of the number of points obtained in a single throw with an ordinary die. Also find the variance?

2. What is the mathematical expectation of the sum of points obtained on n dice?

Solution:

1. Let X be the random variable “the number of points” obtained X assumes the values 1, 2, 3, 4, 5, and 6, with probabilities $1/6$ in each case.
Hence the expectation of X is

$\begin{array}{ll}E(X)\hfill & =1\cdot \frac{1}{6}+2\cdot \frac{1}{6}+3\cdot \frac{1}{6}+4\cdot \frac{1}{6}+5\cdot \frac{1}{6}+6\cdot \frac{1}{6}\hfill \\ \hfill & =\frac{1}{6}(1+2+3+4+5+6)=\frac{21}{6}=\frac{7}{2}\hfill \end{array}$

2. Let x_i denote the number of points ith die. Then $E(x_i)=\frac{7}{2}$, i=1, 2, 3, …n, …
The sum of points on n dice is x₁+x₂++ x_n
Therefore

$\begin{array}{ll}E(x_1+x_2+\cdots +x_n)\hfill & =E(x_1)+E(x_2)+\cdots +E(x_n)\hfill \\ \hfill & =\frac{7}{2}+\frac{7}{2}+\cdots n\mathrm{terms}\hfill \\ \hfill & =n\left({\displaystyle \frac{7}{2}}\right)\hfill \\ \hfill & =\frac{7}{2}n\hfill \end{array}$

Hence the expectation of sum of points on n dice=$\frac{7}{2}n$.

Example 4.29: If X and Y are random variables having joint density function

$\begin{array}{lll}f(x,y)\hfill & =4xy,\hfill & 0\le x\le 1,0\le y\le 1\hfill \\ \hfill & =0,\hfill & \mathrm{otherwise}\hfill \end{array}$

Verify that

1. E(X+Y)=E(X)+E(Y)

2. E(XY)=E(X) E(Y)

Solution: We have

$\begin{array}{ll}E(X)\hfill & ={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }xf(x,y)\mathrm{d}x\mathrm{d}y\hfill \\ \hfill & ={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }x(4xy)\mathrm{d}x\mathrm{d}y\hfill \\ \hfill & =4{\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }{x}^{2}y\mathrm{d}x\mathrm{d}y\hfill \\ \hfill & =4{\int }_{x=0}^{\infty }{\int }_{y=0}^{\infty }{x}^{2}y\mathrm{d}x\mathrm{d}y\hfill \\ \hfill & =4{\int }_{y=0}^{\infty }y\left({\int }_{x=0}^1{x}^2\mathrm{d}x\right)\mathrm{d}y\hfill \\ \hfill & =4{\int }_{y=0}^{\infty }y{\left[\frac{x^3}{3}\right]}_0^1\mathrm{d}y\hfill \\ \hfill & =\frac{4}{3}{\int }_{y=0}^{1}y\mathrm{d}y=\frac{4}{3}{\left[\frac{y^2}{2}\right]}_0^1\hfill \\ \hfill & =\frac{4}{6}=\frac{2}{3}\hfill \end{array}$

Similarly

$\begin{array}{ll}E(Y)\hfill & ={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }yf(x,f)\mathrm{d}x\mathrm{d}y\hfill \\ \hfill & ={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }y(4xy)\mathrm{d}x\mathrm{d}y\hfill \\ \hfill & =4{\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }x{y}^2\mathrm{d}x\mathrm{d}y\hfill \\ \hfill & =4{\int }_{x=0}^1{\int }_{y=0}^{\infty }x{y}^2\mathrm{d}x\mathrm{d}y=\frac{2}{3}\hfill \end{array}$

Now

1. $\begin{array}{ll}E(X+Y)\hfill & ={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }(x+y)f(x,y)\mathrm{d}x\mathrm{d}y\hfill \\ \hfill & ={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }(x+y)(4xy)\mathrm{d}x\mathrm{d}y\hfill \\ \hfill & =4{\int }_0^1{\int }_0^1(x^{2}y+x{y}^2)\mathrm{d}x\mathrm{d}y\hfill \\ \hfill & =4{\int }_{y=0}^1\left({\int }_{x=0}^1(x^{2}y+x{y}^2)\mathrm{d}x\right)\mathrm{d}y\hfill \\ \hfill & =4{\int }_{y=0}^{1}y{\left[\frac{x^3}{3}y+\frac{x^2}{2}{y}^2\right]}_0^1\mathrm{d}y\hfill \\ \hfill & =4{\int }_{y=0}^1\left[\frac{y}{3}+\frac{y^2}{2}\right]\mathrm{d}y\hfill \\ \hfill & =4{\left[\frac{y^2}{6}+\frac{y^3}{6}\right]}_0^1=4\left(\frac{1}{6}+\frac{1}{6}\right)\hfill \\ \hfill & =4\left(\frac{2}{6}\right)=\frac{4}{3}\hfill \end{array}$
Therefore $E(X+Y)=\frac{4}{3}=\frac{2}{3}+\frac{2}{3}=E(X)+E(Y)$
Also,

2. $\begin{array}{ll}E(XY)\hfill & ={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }(xy)f(x,y)dxdy\hfill \\ \hfill & ={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }(xy)(4xy)dxdy\hfill \\ \hfill & =4{\int }_0^1{\int }_0^1{x}^2{y}^{2}dxdy\hfill \\ \hfill & =4{\int }_{y=0}^1{y}^2{\left[\frac{x^3}{3}\right]}_0^{1}dy\hfill \\ \hfill & =\frac{4}{3}{\int }_{y=0}^1{y}^{2}dy=\frac{4}{3}{\left[\frac{y^3}{3}\right]}_0^1\hfill \\ \hfill & =\frac{4}{3}\cdot \frac{1}{3}=\frac{4}{9}\hfill \end{array}$
Thus $E(XY)=\frac{4}{9}=\frac{2}{3}\cdot \frac{2}{3}=E(X)E(Y)$
Hence verified.

Theorem 6

(Cauchy–Schwartz inequality) If X, Y are random variables taking real values, then

${\left[E\right[XY\left]\right]}^2\le E\left[X^2\right]E\left[Y^2\right]$

Proof

Let t be a real variable.

Consider the expression (X+tY)²

(X+tY)² is a nonnegative for all values of X and Y.

Hence E(X+tY)²≥0 for all i.

i.e., E(x²+2tXY+t²Y²)≥0 for all i

i.e., E(x²)+2tE(XY)+t² E(Y²)≥0 for all i

t² E(Y²)+2tE(XY)+E(x²) is of the form, at²+bt+c, a quadratic in t.

We have

$\begin{array}{l}a=E(Y^2),b=2E(XY),c=E(x^2)\hfill \\ a{t}^2+bt+c\ge 0,\mathrm{implies}{b}^2–4ac\le 0\hfill \end{array}$

Therefore

$4{\left[E\right[XY\left]\right]}^2-E\left[X^2\right]E\left[Y^2\right]\le 0$

or

${\left[E\right[XY\left]\right]}^2-E\left[X^2\right]E\left[Y^2\right]\le 0$

or

${\left[E\right[XY\left]\right]}^2\le E\left[X^2\right]E\left[Y^2\right]$

Hence proved.

Theorem 7

(Chebyshev’s inequality) If X is a random variable with mean μ and variance σ² then

$P(|x-\mu |\ge k)\le \frac{{\sigma }^2}{k^2}$

or

$P(|x-\mu |\ge k)\ge 1-\frac{{\sigma }^2}{k^2}$

where k is a real number.

Proof

Case 1: Let X be a discrete random variable and f(x) denote the probability function of X.

Then

$\begin{array}{ll}\mathrm{Var}\left[X\right]\hfill & ={\sigma }^2=E\left|{(x-\mu )}^2\right|\hfill \\ \hfill & =\sum {(x-\mu )}^{2}f(x)\hfill \end{array}$ (4.5)

(4.5)

Right hand side of the sum given by Eq. (4.5) is nonnegative.

Hence

${\sigma }^2\ge \sum _{|x-\mu |\ge k}{(x-\mu )}^{2}f(x)\ge \sum _{|x-\mu |\ge k}{k}^{2}f(x)$

i.e.,

${\sigma }^2\ge k^2\sum _{|x-\mu |\ge k}f(x)$ (4.6)

(4.6)

But

$\sum _{|x-\mu |\ge k}f(x)=P(|x-\mu |\ge k)$

Therefore from Eq. (4.6) we get

${\sigma }^2\ge k^{2}P(|x-\mu |\ge k)$

i.e.,

$\frac{{\sigma }^2}{k^2}\ge P(|x-\mu |\ge k)$

Thus we get

$P(|x-\mu |\ge k)\le \frac{{\sigma }^2}{k^2}$

Case 2: Let X be a continuous random variable and f(x) be the pdf of X then ${\sigma }^2={\int }_{-\infty }^{\infty }{(x-\mu )}^{2}f(x)\mathrm{d}x$.

Clearly the above integrand is positive. Therefore when the range of integration is reduced, the value of the integral decreases.

Thus we have ${\sigma }^2\ge {\int }_{|x-\mu |\ge k}{(x-\mu )}^{2}f(x)\mathrm{d}x\ge {\int }_{|x-\mu |\ge k}{k}^{2}f(x)\mathrm{d}x$

or

${\sigma }^2\ge k^2{\int }_{|x-\mu |\ge k}f(x)\mathrm{d}x$