Keywords

Joint distributions; joint probability function; marginal density function; mathematical expectation; moments; moment generation function; skewness and kurtosis; discrete distributions and uniform distribution

4.1 Introduction

In this chapter we introduce random variables that are defined on sample spaces associated with experiments in which the outcomes are uncertain. The term random variable is one of the basic concepts of the probability theory. The values of a random variable are real numbers associated with the outcome of an experiment and the concept of the random variables is the one of the most significant in the theory of probability. Random variable is a variable whose value is subject to variations due to chance. Random variables vary from trial to trial as the experiment is repeated. It is also called stochastic variable and is defined as follows:

We shall denote random variable by the capital letters X, Y, Z, etc., and the corresponding small (lower case) letters are used to denote the numerical values taken by the random variable.

Example 4.1: Consider an experiment A in which two coins are tossed simultaneously. Let S be the sample space associated with the experiment. We have,

$S=\{\text{HH},\text{HT},\text{TH},\mathrm{TT}\}$

If we define a random variable X, as the number of heads, then the values of X are 0, 1, and 2 corresponding to the outcomes.

$\text{TT}(0\text{head}),\text{HT}(1\text{head}),\text{TH}(1\text{head}),\text{HH}(2\text{heads}).$

We have the following table:

Example 4.2: The number of calls from subscribers at the telephone exchange during a definite time period is a random variable.

Example 4.3: An experiment is to fire four shots at a target. The random variable is the number of hits.

In the above experiment we have observed that each outcome is a simple event of the sample space S and the corresponding value is a real number.

Thus “A random variable is a rule that assigns one and only one numerical value to each simple event of an experiment” and we have the following definition:

If S is a sample space of an experiment then we can define more than one random variable on S. The set $R_X=\{X(x):x\in S\}$ is called the range of X.

In general a real-valued function, whose domain is the sample space S of an experiment and whose range set a collection of n—tuples of real numbers, is called dimensional random variable.

If X(x) ≠ 0 for all $X\in S$ then and $\frac{1}{X(x)}|X(x)|$ are also random variables on S.

If X and Y are random variables on the sample space of an experiment then X+Y, X−Y, and XY are also random variables on S.

For all real numbers a and b, a X+b Y is also a random variable on S.

The mathematical function describing the possible values of a random variable and their associated probabilities is known as a probability distribution.

Random variables can be discrete, i.e., taking any of a specified finite or countable list of values, and hence with a probability mass function (pmf) as probability distribution.

Random variables can be continuous, taking any numerical value in an interval or collection of intervals, and with a probability density function (pdf) describing the probability distribution, or a mixture of both types. The realizations of a random variable, i.e., the results of randomly choosing values according to the variable’s probability distribution, are called random variables.

4.2 Discrete Random Variable

In statistics we study variables such as heights of students, number of defective bolts, number of accidents on a road, number of male children in a family, number of printing mistakes in each page of a book, and so on. Some of these quantities can vary only by finite increments, by observable jumps in value. Such values are called discrete random variables. The sample space S of a discrete random variable contains either finite number of outcomes, which are countably infinite.

Example 4.1: In an experiment of drawing four cards from a pack of cards, the random variable “The number of kings drawn” is a discrete random variable.

Example 4.2: Consider an experiment in which a coin is tossed four times. If X denotes the number of heads obtained then X is a discrete random variable.

Discrete random variables are the random variables whose range is finite or continuously infinite.

4.3 Probability Distribution for a Discrete Random Variable

The simplest distribution for a discrete random variable X is an ordered series, which is a table whose top row contains all the values of the random variables and the bottom row contains the corresponding probabilities as shown in the following table:

x₁, x₂, …, x_n are the values of the random variable X and p_i=P(X=X_i) are the corresponding probabilities where ${\sum }_1^n{p}_i=1$. The above distribution is an ordered series. It can also be represented graphically as a frequency polygon or a histogram.

4.3.1 Probability Mass Function

Definition 4.4

Let X be a discrete random variable assuming the values x=x₁, x=x₂, …, x=x_n corresponding to the various outcomes of a random experiment. If the probability of occurrence of X=x_i, (I=1, 2, …, n) is P(X=x_i)=p_i

Such that

1. P(X=x_i)=p_i≥0 for all i $(1\le i\le n)$

2. ${\sum }_{i=1}^{n}P(X=x_i)={\sum }_{i=1}^n{p}_i=1$

Then the function P(X) is called the probability function of the random variable X and the set

$\begin{array}{l}\{(x_1,P(X=x_1)),(x_2,P(X=x_2)),\dots ,(x_n,P(X=x_n))\}\mathrm{or}\mathrm{simply}\hfill \\ \{(x,p(x_1)),(x_2,P(x_2)),\dots ,(x_n,P(x_n))\}\hfill \end{array}$

is called the probability distribution of the random variable X.

The probability function P(X=x) is also denoted by f(x) and is called the pmf of the discrete random variable X.

Remark

Some authors refer f(x) as the frequency function or a probability function.

We have

1. f(x)=P(X=x)

2. f(x)≥0

3. ${\sum }_{x}f(x)=1$

Definition 4.5

(Finite equiprobable space) A finite probability distribution where each point X=x_i has the same probability for all i, is called a finite equiprobable space or uniform space.

4.3.2 Distribution Function

The probability $P(X\le x)$ is the probability of the event$(X\le x)$. It is a function of x. The function ${\mu }_r$ is denoted by F(x) and is called the cumulative probability distribution function of the random variable x.

Thus $F(x)=P(X\le x),(-\infty <x<\infty )$

F(x) is often called the distribution function of X.

F(x) possesses the following properties:

1. $F(-\infty )=0$

2. $F(\infty )=1$

3. $0\le F(x)\le 1$

4. $F(x_i)\le F(x_2)$ if $x_1<x_2$

5. $P(x_1<x<x_2)=F(x_2)-F(x_i)$

6. $F(x^{+})=F(x)$

F(X) can also be defined as follows:

Definition 4.6

Let X be a discrete random variable, then the function

$F(X)=P(X\le x)=\sum _{1\le x}f(t),-\infty <x<\infty$

where f(t) is the value of the probability distribution is called distribution function of X.

4.3.3 Additional Properties of Distribution Function

Property 1: (Interval property) If X is a random variable and F(x) is the distribution function of X, then $P(a<X\le b)=F(b)-F(a)$

Proof: The events $a<X\le b$ and $X\le a$ are disjoint and we have

$P(a<x\le b)\cup (x\le a)=P(x\le b)$

Hence

$P(a<x\le b)\cup (x\le a)=P(x\le b)$

or

$\begin{array}{l}P(a<x\le b)=(x\le b)-P(x\le a)\hfill \\ F(b)-F(a)\hfill \end{array}$

Property 2: If F(x) is the distribution function of a random variable x, then

$P(a\le x\le b)=P(x=a)+F(b)-F(a).$

Proof: We have

$\begin{array}{l}P(a\le x\le b)=P(x=a)+P(a<x\le b)\hfill \\ P(x=a)+F(b)–F(a)(\mathrm{by}\mathrm{property}(1)).\hfill \end{array}$

Property 3: If F(x) is the distribution function of a random variable X, then

$P(a<x<b)=F(b)–F(a)–P(x=b).$

Proof:

$\begin{array}{l}P(a<x<b)=P(a<x\le b)-P(x=b)\hfill \\ F(b)–F(a)–P(x=b)(\mathrm{by}\mathrm{property}(1)).\hfill \end{array}$

Property 4: $P(a\le x<b)=F(b)–F(a)–P(x=b)+P(x=a)$

Proof: We have

$\begin{array}{l}P(a\le x<b)=P(a<x<b)+P(x=a)\hfill \\ F(b)–F(a)–P(x=b)+P(x=a).\hfill \end{array}$

Property 5: (Monotone increasing property) If F(x) is the distribution of a random variable X and a<b, then F(a)≤F(b).

Since

$P(a<X\le b)=0$

We have

$F(b)–F(a)\ge 0$

or

$F(b)\ge F(a)$

i.e.,

$F(a)\le F(b),$

hence proved.

If the random variable X takes the values x₁, x₂, …, x_n with probabilities p₁, p₂, …, p_n respectively, then we have

$\begin{array}{l}P[X<x_1]=0\hfill \\ P[X\le x_1]=P[X<x_1]+P[X=x_1]=p_1\hfill \end{array}$

Therefore

$\begin{array}{l}P[X<x_2]=p_1\hfill \\ P[X\le x_2]=P[X<x_2]=P[X=x_2]=p_1+p_2,\hfill \\ \cdots \hfill \\ P[X\le x_n]=p_1+p_2+\cdots +p_n\hfill \end{array}$

Example: Consider an experiment in which four coins are tossed. If we define a random variable as the number of heads obtained, then we have

$\begin{array}{l}P(X=0)=\frac{1}{16},P(X=1)=\frac{4}{16}=\frac{1}{4},P(X=2)=\frac{6}{16}=\frac{3}{8}\hfill \\ P(X=3)=\frac{4}{16}=\frac{1}{4}=\frac{1}{4},P(X=4)=\frac{1}{16}\hfill \end{array}$

Therefore we get

$\begin{array}{l}F(0)=P(X=0)=\frac{1}{16}\hfill \\ F(1)=P(X=0)=P(X=1)=\frac{1}{16}+\frac{1}{4}=\frac{5}{16}\hfill \\ F(2)=P(X=0)=P(X=1)+P(X=2)=\frac{1}{16}+\frac{1}{4}=\frac{3}{8}=\frac{11}{16}\hfill \end{array}$

Hence the distribution function F(x) is given by

$F(x)=\{\begin{array}{ccc}0,& \mathrm{for}& X<0\\ \frac{1}{16},& \mathrm{for}& 0\le X<1\\ \frac{5}{16},& \mathrm{for}& 1\le X<2\\ \frac{11}{16},& \mathrm{for}& 2\le X<3\\ \frac{15}{16},& \mathrm{for}& 3\le X<4\\ 1,& \mathrm{for}& X\ge 4\end{array}$

4.4 Mean and Variance of a Discrete Distribution

If X is a discrete random variable, then the mean and variance of the discrete distribution can be defined as follows:

$\mathrm{Mean}=\mu =\sum _{i=1}^n{x}_{i}P(X=x_i)=\sum _{i=1}^m{x}_i{p}_i$

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

or

${\sigma }^2=\sum _{i=1}^n{(x_i-\mu )}^2{p}_i$

where σ is the standard deviation (SD).

Since

$\sum p_i=1,\mathrm{we}\mathrm{get}$

$\begin{array}{ll}\mathrm{Var}\left[x\right]\hfill & =\sum _{i=1}^n{(x_i-\mu )}^2{p}_i\hfill \\ \hfill & =\sum _{i=1}^n(x_i^2+{\mu }^2-2x_i\mu )p_i\hfill \\ \hfill & =\sum _{i=1}^n(x_i^2)p_i+\sum _{i=1}^n({\mu }^2)p_i-\sum _{i=1}^n(2x_i\mu )p_i\hfill \\ \hfill & =\sum _{i=1}^n(x_i^2)p_i+{\mu }^2\sum _{i=1}^n{p}_i–2\mu \sum _{i=1}^n{x}_i{p}_i\hfill \\ \hfill & =\sum _{i=1}^n(x_i^2)p_i+{\mu }^2-2\mu \mu \hfill \\ \hfill & =\sum _{i=1}^n(x_i^2)p_i-{\mu }^2\hfill \end{array}$

Thus

${\sigma }^2=\sum _{i=1}^n(x_i^2)p_i-{\mu }^2$

Example 4.4: Find the probability distribution of the number of blue balls drawn when 3 balls are drawn without replacement from a bag containing 4 blue and 6 red balls?

Solution: Number of balls in the bag=4 blue+6 red=10

Let X be the random variable “number of blue balls”

Then X can take values 0, 1, 2, and 3.

$\begin{array}{ll}P(X=0)\hfill & =P(\mathrm{no}\mathrm{blue}\mathrm{ball})=P(3\mathrm{red}\mathrm{balls})=\frac{{}^{6}C_3}{{}^{10}C_3}=\frac{1}{6}\hfill \\ P(X=1)\hfill & =P(1\mathrm{blue}+2\mathrm{red}\mathrm{balls})=\frac{{}^{4}C_1\times {}^{6}C_2}{{}^{10}C_3}=\frac{1}{2}\hfill \\ P(X=2)\hfill & =P(2\mathrm{blue}+1\mathrm{red}\mathrm{ball})=\frac{{}^{4}C_2\times {}^{6}C_1}{{}^{10}C_3}=\frac{3}{10}\hfill \\ P(X=3)\hfill & =P(3\mathrm{blue}\mathrm{balls})=\frac{{}^{4}C_3}{{}^{10}C_3}=\frac{1}{30}\hfill \end{array}$

The probability distribution is

x	0	1	2	3
P(X=x)	$\frac{1}{6}$	$\frac{1}{2}$	$\frac{3}{10}$	$\frac{1}{30}$

Example 4.5: Two cards are drawn successively with replacement from a well shuffled pack of cards. Find the probability distribution of the number of kings that can be drawn?

Solution: Let X denote the random variable “number of kings.”

Since the number of cards drawn is 2, the random variable X can take values 0, 1, and 2

When one card is drawn from the pack, the probability of getting a king is

$=\frac{{}^{4}C_1}{{}^{52}C_1}=\frac{4}{52}=\frac{1}{13}$

Probability of failure, i.e., the probability of not getting a king is

$1-\frac{1}{13}=\frac{12}{13}$

Hence we get

$\begin{array}{ll}P(X=0)\hfill & =P(\text{no}\text{king})=\frac{12}{13}\cdot \frac{12}{13}=\frac{144}{169}\hfill \\ P(X=1)\hfill & =P(\text{one}\text{king})=\frac{1}{13}\cdot \frac{12}{13}+\frac{12}{13}\cdot \frac{1}{13}=\frac{24}{169}\hfill \\ P(X=2)\hfill & =P(2\text{kings})=\frac{1}{13}\cdot \frac{1}{13}+\frac{1}{169}\hfill \end{array}$

The required probability distribution is

x	0	1	2
P(X=x)	$\frac{144}{169}$	$\frac{24}{169}$	$\frac{1}{169}$

Example 4.6: A random variable X has the following probability distribution:

Find

Solution:

1. We have

$\sum p_i=1$

Therefore
3k+3k+3+2k+6k=1
or
15k=1
or

$k=\frac{1}{15}$

2. $\begin{array}{ll}\mathrm{Mean}\mu \hfill & =\sum _i{x}_i{p}_i\hfill \\ \hfill & =(0)(3k)+(1)(3k)+(2)(k)+(3)(2k)+(4)(6k)\hfill \\ \hfill & =35k=35\left(\frac{1}{15}\right)=\frac{7}{3}\hfill \end{array}$

3. $\begin{array}{ll}P(X>2)\hfill & =P(X>3)+P(X=4)\hfill \\ \hfill & =2k+6k=8k=8\left(\frac{1}{15}\right)=\frac{8}{15}\hfill \end{array}$

Example 4.7: A random variable X has the following probability distribution:

Find

Solution:

1. We have

$\sum p_i=1$

i.e., 0.1+k+0.2+2k+0.3+k=1
or 4k+0.6=1
or 4k=1−0.6=0.4
we get k=0.1

2. $\begin{array}{ll}\mathrm{Mean}\hfill & =\mu =\sum _i{x}_i{p}_i\hfill \\ \hfill & =(–2)(0.1)+{(–1)}^{2}k+(0)(0.2)+(1)(2k)+(2)(0.3)+(3)(k)\hfill \\ \hfill & =4k+0.4=4(0.1)+0.4=0.8\hfill \end{array}$

3. $\begin{array}{ll}\mathrm{Variance}{\sigma }^2\hfill & =\sum _{i=1}^n(x_i^2)p_i={\mu }^2\hfill \\ \hfill & =(0.1){(–2)}^2+{(–1)}^{2}k+(0.2){(0)}^2+(2k){(1)}^2+(0.3){(2)}^2\hfill \\ \hfill & +(k){(3)}^2–{(0.8)}^2\hfill \\ \hfill & =0.4+k+0+2k+1.2+9k–0.64\hfill \\ \hfill & =12k+1.6–0.64\hfill \\ \hfill & =12(0.1)+1.6–0.64\hfill \\ \hfill & =2.16\hfill \end{array}$

Exercise 4.1

1. Define

a. Random variable

b. Discrete random variable

c. Density function

d. Probability distribution

e. Spectrum

2. Obtain the probability distribution of the total number of heads in three tosses of a coin.

Ans:

x	0	1	2	3
P(x)	$\frac{1}{8}$	$\frac{3}{8}$	$\frac{3}{8}$	$\frac{1}{8}$

3. A fair coin is tossed 4 times. Find the probability distribution of the number of heads?

Ans:

x	0	1	2	3	4
P(x)	$\frac{1}{16}$	$\frac{1}{4}$	$\frac{3}{8}$	$\frac{1}{4}$	$\frac{1}{16}$

4. A random variable X has the following distribution:

X	0	1	2	3	4
P(X)	2λ	3λ	λ	2λ	6λ

Find

a. P(0<x<2)

b. P(x>2)

Ans: (a) $\frac{3}{14}$; (b) $\frac{4}{7}$

5. A bag contains 2 white, 3 red, and 4 blue balls. Two balls are drawn at random from the bag. If the random variable X denotes the “number of white balls” among the balls drawn, describe the probability distribution of X.

Ans:

x	0	1	2
P(X=x)	$\frac{7}{12}$	$\frac{7}{18}$	$\frac{7}{36}$

6. Three balls are drawn without replacement from a bag containing 5 white and 4 red balls. Find the probability distribution of the number of red balls drawn?

Ans:

x	0	1	2	3
P(x)	$\frac{5}{42}$	$\frac{10}{21}$	$\frac{5}{14}$	$\frac{1}{21}$

7. The probability distribution of a random variable X is given below:

X	0	1	2
P(X)	3 λ2	4 λ–10 λ2	5 λ−1

where>0.
Find

a. λ

b. P(X≤1)

c. P(X>0)

Ans: (a) $\lambda =\frac{1}{3}$; (b) $P(X\le 1)=\frac{1}{3}$; (c) $P(X>0)=\frac{8}{9}$

8. A random variable X has the following probability distribution:

X	1	2	3	4
P(X)	k	2k	3k	4k

Find

a. k

b. P(X<3)

c. P(X≥3)

d. Mean

Ans: (a) $k=\frac{1}{10}$; (b) $\frac{3}{10}$; (c) $\frac{7}{10}$; (d) $\mu =3$

9. A random variable X has the following probability distribution:

X=xi	−2	−1	0	1	2	3
P(X=xi)	0.1	k	0.2	2k	0.3	k

Find

a. k

b. Mean

c. Variance

Ans: (a) k=0; (b) μ=0.8; (c) σ²=2.8

10. A box contains 6 tickets. Two of the tickets carry a prize of Rs. 5/- each, the other four tickets carry a prize of Rs. 1/- each. If one ticket is drawn what is the mean value of the prize?

Ans: $\frac{7}{3}$

11. A die is tossed twice. Getting a “number greater than 4” is considered a success. For the probability distribution of the number successes. Show that the mean is 2/3 and the variance is 4/9.

12. Obtain the probability distribution of the number of sixes in two tosses of a cubical die.

Ans:

x	0	1	2
P(X=x)	$\frac{25}{36}$	$\frac{10}{36}$	$\frac{1}{36}$

4.5 Continuous Random Variable

Let X be a random variable. If X takes noncountable infinite number of values, then X is called a continuous random variable. If X is a continuous random variable then the range of X is an interval on real line.

Example: The length of an electric bulb, the detection range of a radar, etc., are examples of continuous random variables.

4.6 Probability Density Function

The function f(x) is called a pdf of X. It is also referred to as density function.

If f(x) is a pdf of X then we have

$P(a\le x\le b)={\int }_a^{b}f(x)\mathrm{d}x$, for any real constants a and b with $a\le b$.

The set of values obtained from ${\int }_a^{b}f(x)\mathrm{d}x$ for various possible intervals is called a continuous probability distribution for X.

4.7 Cumulative Distribution Function

Definition 4.8

If X is a continuous random variable having f(x) as its pdf, then the function given by

$F(X)=P(X\le x)={\int }_{-\infty }^{\infty }f(x)\mathrm{d}x,\mathrm{for}-\infty <x<\infty$

is called cumulative distribution function of X.

F(x) is also referred to as the distribution function or the cumulative distribution of X.

Examples of Continuous and Mixed Random Variables

Example 4.8: A random variable X has a Simpson distribution on the interval −c to c. Find the expression for the pdf?

Solution: The random variable X obeys “the law of an isosceles triangle.” The pdf is

$\begin{array}{lll}f(x)\hfill & =\frac{1}{c}\left(1-\frac{x}{c}\right)\hfill & \mathrm{for}0<x<c\hfill \\ \hfill & =\frac{1}{c}\left(1+\frac{x}{c}\right)\hfill & \mathrm{for}–c<x<0\hfill \\ \hfill & =0\hfill & \mathrm{otherwise}\hfill \\ \mathrm{i}\mathrm{.e}.,\hfill & x<c\mathrm{or}x>c\hfill & \hfill \end{array}$

Example 4.9: A random variable X has Laplace transform defined by $f(x)=a{c}^{-\lambda \left|x\right|}$, where λ is a positive parameter. Find a?

Solution: Since the pdf is $f(x)=a{e}^{-\lambda \left|x\right|}$,

$a{\int }_{-\infty }^{\infty }{e}^{-\lambda \left|x\right|}\mathrm{d}x=1$

We have

$8^2\cdot \frac{1}{8}+{12}^2\cdot \frac{1}{6}+{16}^2\cdot \frac{3}{8}+{20}^2\cdot \frac{1}{4}+{24}^2\frac{1}{12}-{16}^2$

∴

$\frac{16}{3}{\left[\frac{1}{x}\right]}_2^{\infty }=\frac{16}{3}\left[0-\frac{1}{2}\right]=\frac{8}{3}$

$x_n={(-1)}^n{2}^n{n}^{-1}\sum x_i{p}_{i}E\left[x\right]=\sum _{i=1}^{\infty }{2}^{-n}{(-1)}^n{2}^n{n}^{-1}=\sum _{i=1}^{\infty }\frac{{(-1)}^n}{n}$

$\Rightarrow a=\frac{\lambda }{2}$

4.8 Mean and Variance of a Continuous Random Variable

Let X be a continuous random variable. If f(x) is the pdf of X, then the mean and variance of X are given as

$\mathrm{Arithmetic}\mathrm{mean}=\mu ={\int }_{-\infty }^{\infty }xf(x)\mathrm{d}x$

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

4.8.1 Solved Examples

Example 4.10: $\mathrm{If}f(x)=\{\begin{array}{ll}\frac{x}{6}+k,\hfill & 0\le x\le 3\hfill \\ 0,\hfill & \mathrm{elsewhere}\hfill \end{array}$

is a pdf, find the value of k. Also find P(1≤x≤2)?

Solution: f(x) is a pdf. Therefore we have

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

or

${\int }_{-\infty }^{\infty }\left(\frac{x}{6}+k\right)\mathrm{d}x=1$

or

${\int }_{-\infty }^{\infty }\left(\frac{x}{6}+k\right)\mathrm{d}x+{\int }_0^3\left(\frac{x}{6}+k\right)\mathrm{d}x+{\int }_3^{\infty }\left(\frac{x}{6}+k\right)\mathrm{d}x=1$

or

$0+{\left[\frac{x^2}{12}+kx\right]}_1^3+0=1$

or

$\frac{9}{12}+3k=1\mathrm{or}3k=1-\frac{3}{4}=\frac{1}{4}$

or

$k=\frac{1}{12}$

Now

$\begin{array}{l}P(1\le x\le 2)={\int }_1^{2}f(x)\mathrm{d}x={\int }_1^2\left(\frac{x}{6}+\frac{1}{12}\right)\mathrm{d}x\hfill \\ {\left[\frac{x^2}{12}+\frac{x}{12}\right]}_1^2=\left(\frac{4}{12}+\frac{2}{12}\right)-\left(\frac{1}{12}+\frac{1}{12}\right)=\frac{1}{3}+\frac{1}{6}-\frac{1}{6}=\frac{1}{3}\hfill \end{array}$

Example 4.11: Given the cumulative distribution function

$f(x)=\{\begin{array}{l}0,\dots ,x>0\hfill \\ x^2,\dots ,0\le x\le 1\hfill \\ 1,\dots ,x>1\hfill \end{array}$

1. Find the pdf?

2. Find $P(0.5<x\le 0.75)$?

Solution:

1. Since $f(x)=\frac{\mathrm{d}}{\mathrm{d}x}(F(x))$
We get $f(x)=\{\begin{array}{ll}0,\hfill & x>0\hfill \\ 2x,\hfill & 0\le x\le 1\hfill \\ 1,\hfill & x>1\hfill \end{array}$
or simply $f(x)=\{\begin{array}{ll}2x,\hfill & 0\le x\le 1\hfill \\ 0,\hfill & \mathrm{otherwise}\hfill \end{array}$

2. $\begin{array}{ll}P(0.5<x\le 0.75)\hfill & =P(X\le 0.75)–P(X\le 0.5)\hfill \\ \hfill & =F(0.75)–F(0.5)\hfill \\ \hfill & ={(0.75)}^2–{(0.5)}^2\hfill \\ \hfill & =0.5625–0.25=0.3125\hfill \end{array}$

Example 4.12: Find k such that f(x) is a pdf of a continuous random variable X where f(x) is defined as follows:

$f(x)=\{\begin{array}{ll}kx{e}^{-x},\hfill & 0<x<1\hfill \\ 0,\hfill & \mathrm{otherwise}\hfill \end{array}$

Also find the mean.

Solution:

1. Since f(x) is a pdf we have ${\int }_{-\infty }^{\infty }f(x)\mathrm{d}x=1$
i.e., ${\int }_0^{1}kx{e}^{-x}dx=1$
or $k\left[\frac{x{e}^{-x}}{(-1)}-\frac{e^{-x}}{{(-1)}^2}\right]=1$
or $k[(e^{-1}-0)-(e^{-1}-e^0)]=1$
or $k\left(1-\frac{2}{e}\right)=1$
or $k=\frac{e}{e-2}$

2. $\begin{array}{ll}\mathrm{Mean}\mu \hfill & ={\int }_{-\infty }^{\infty }xf(x)\mathrm{d}x\hfill \\ \hfill & ={\int }_0^{1}x\left[\frac{e}{e-2}x{e}^{-x}\right]\mathrm{d}x\hfill \\ \hfill & =\frac{e}{e-2}{\int }_0^1{x}^2{e}^{-x}\mathrm{d}x\hfill \\ \hfill & =\frac{e}{e-2}{\left[\frac{x^2{e}^{-x}}{-1}-2x\frac{e^{-x}}{{(-1)}^2}+2\frac{e^{-x}}{{(-1)}^3}\right]}_0^1\hfill \\ \hfill & =\frac{e}{e-2}[-e^{-1}-2e^{-1}(e^{-1}-1)]\hfill \\ \hfill & =\frac{e}{e-2}\left[-\frac{1}{e}-\frac{2}{e}-\frac{2}{e}+2\right]\hfill \\ \hfill & =\frac{e}{e-2}\left[-\frac{5}{e}+2\right]=\frac{2e-5}{e-2}\hfill \end{array}$

Example 4.13: The distribution of a random variable given by

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

Find the corresponding density function of random variable X?

Solution: We have

$\begin{array}{ll}F(x)\hfill & =\frac{\mathrm{d}}{\mathrm{d}x}(F(x))\hfill \\ \hfill & =\frac{\mathrm{d}}{\mathrm{d}x}(1-(1+x)e^{-x})\hfill \\ \hfill & =\frac{\mathrm{d}}{\mathrm{d}x}(1-e^{-x}-x{e}^{-x})\hfill \\ \hfill & =0-(-e^{-x})-[x(-e^{-x})+e^{-x}]\hfill \\ \hfill & =e^{-x}+x{e}^{-x}-e^{-x}=x{e}^{-x}\hfill \end{array}$

Hence $f(x)=\{\begin{array}{ll}x{e}^{-x},\hfill & x\ge 0\hfill \\ 0,\hfill & \mathrm{otherwise}\hfill \end{array}$ is the required density function.

Example 4.14: Find the cumulative distribution function for the following probability function of a random variable X?

$f(x)=\{\begin{array}{ll}x{e}^{-x},\hfill & x\ge 0\hfill \\ 0,\hfill & \mathrm{otherwise}\hfill \end{array}$

Solution: Since

$f(x)=\frac{\mathrm{d}}{\mathrm{d}x}(F(x))$

We have

$\begin{array}{ll}F(x)\hfill & ={\int }_{-\infty }^{x}f(x)\mathrm{d}x={\int }_{-\infty }^{0}f(x)\mathrm{d}x+{\int }_0^{x}f(x)\mathrm{d}x\hfill \\ \hfill & =0+6{\int }_{-\infty }^x(x-x^2)\mathrm{d}x=6{\left[\frac{x^2}{2}-\frac{x^3}{3}\right]}_0^x\hfill \\ \hfill & =6\left[\frac{3x^2-2x^3}{6}\right]=3x^2-2x^3\hfill \end{array}$

Hence

$F(x)=3x^2–2x^3,0\le x\le 1$

Example 4.15: For the following density function:

$F(x)=a{e}^{-\left|x\right|}-\infty \le x\le \infty$

of a random variable X. Find 1. a, 2. Mean, 3. Variance

Solution:

1. f(x) is a pdf we have,
Therefore $\underset{-\infty }{\overset{\infty }{\int }}f(x)\mathrm{d}x=1$
or $\underset{-\infty }{\overset{\infty }{\int }}a{e}^{-\left|x\right|}\mathrm{d}x=1$
or $2a\underset{0}{\overset{\infty }{\int }}{e}^{-\left|x\right|}\mathrm{d}x=1(\mathrm{since}{e}^{–\left|x\right|}\mathrm{is}\mathrm{an}\mathrm{even}\mathrm{function}.)$
or $2a|-(0-1)|=1$
Hence $a=\frac{1}{2}$

2. $\mathrm{Mean}={\int }_{-\infty }^{\infty }f(x)\mathrm{d}x$
$\mathrm{i}\mathrm{.e}.,\mu ={\int }_{-\infty }^{\infty }x\left(\frac{1}{2}{e}^{-\left|x\right|}\right)\mathrm{d}x=\frac{1}{2}{\int }_{-\infty }^{\infty }x{e}^{-\left|x\right|}\mathrm{d}x=0(\mathrm{since}x{e}^{-\left|x\right|}\mathrm{is}\mathrm{odd})$

3. $\begin{array}{ll}\mathrm{Variance}\hfill & ={\int }_{-\infty }^{\infty }{x}^{2}f(x)\mathrm{d}x-{\mu }^2\hfill \\ \hfill & =\frac{1}{2}{\int }_{-\infty }^{\infty }{x}^2{e}^{-\left|x\right|}\mathrm{d}x-{\theta }^2\hfill \\ \hfill & =\frac{1}{2}\cdot 2{\int }_0^{\infty }{x}^2{e}^{-\left|x\right|}\mathrm{d}x-{\theta }^2(\mathrm{since}{x}^2{e}^{-\left|x\right|}\mathrm{is}\mathrm{odd})\hfill \\ \hfill & ={\int }_0^{\infty }{x}^2{e}^{-\left|x\right|}\mathrm{d}x-{\theta }^2\hfill \\ \hfill & ={[x^2(-e^{-x})-2x{e}^{-x}-2e^{-x}]}_0^{\infty }\hfill \\ \hfill & =0–0–2(0–1)=2\hfill \end{array}$

Exercise 4.2

1. A continuous random variable X has a pdf

$\begin{array}{ll}f(x)\hfill & =3x^2,0<x\le 1\hfill \\ \hfill & =0,\mathrm{otherwise}.\hfill \end{array}$

Find a and b, such that P(x≤a)=P(x>a) and P(x>4)=0.05?

Ans: $\begin{array}{cc}a={\left({\displaystyle \frac{1}{2}}\right)}^{1/3}& b={\left({\displaystyle \frac{19}{20}}\right)}^{1/3}\end{array}$

2. A random variable X has the pdf

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

Find

a. $P\left(x<\frac{1}{2}\right)$

b. $P\left(\frac{1}{4}<x<\frac{1}{2}\right)$

Ans: (a) $\frac{1}{4}$; (b) $\frac{3}{16}$

3. Find the cumulative distribution function for the following probability distribution function of a random variable x?

$\begin{array}{lll}f(x)\hfill & =\frac{x}{4}{e}^{-x/2},\hfill & 0<x\le \infty \hfill \\ \hfill & =0,\hfill & \mathrm{otherwise}\hfill \end{array}$

Ans: $f(x)=1-e^{-x/2}-\frac{x}{2}{e}^{-x/2},0<x<\infty$

4. If the pdf of a random variable is given by

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

Find

a. k

b. The distribution function of the random variable.

Ans: (a) $\frac{3}{2}$; (b) 1

5. If X has the pdf

$\begin{array}{lll}f(x)\hfill & =k{e}^{-3x},\hfill & x>0\hfill \\ \hfill & =0,\hfill & \mathrm{otherwise}\hfill \end{array}$

Find k and P(0.5≤x≤1)

Ans: 1, 0.173

6. 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 P(x<4)?

Ans: $\frac{2}{27},\frac{16}{27}$

7. Find the pdf for the random variable whose distribution function is given by

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

Ans: $\begin{array}{lll}f(x)\hfill & =1,\hfill & 0<x<1\hfill \\ \hfill & =0,\hfill & \mathrm{otherwise}\hfill \end{array}$

8. A continuous random variable X has the following pdf:

$\begin{array}{lll}f(x)\hfill & =\frac{1}{2},\hfill & -1\le x\le 0\hfill \\ \hfill & =\frac{1}{4}(2-x),\hfill & 0<x<2\hfill \\ \hfill & =0,\hfill & \mathrm{elsewhere}\hfill \end{array}$

Obtain the distribution of X.

Ans: $\begin{array}{lll}f(x)\hfill & =\frac{x+1}{2},\hfill & -1\le x\le 0\hfill \\ \hfill & =\frac{1}{8}(4+4x+x^2),\hfill & 0<x<2\hfill \\ \hfill & =1,\hfill & \mathrm{if}x\ge 2\hfill \end{array}$

9. A random process gives measurements x between 0 and 1 with a pdf

$\begin{array}{lll}f(x)\hfill & =12x^3–21x^2+10x,\hfill & 0\le x\le 1\hfill \\ \hfill & =0,\hfill & \mathrm{elsewhere}\hfill \end{array}$

Find

a. $P\left(X\le \frac{1}{2}\right)$

b. $P\left(X>\frac{1}{2}\right)$

c. a number k such that $P(X\le k)=\frac{1}{2}$

Ans: (a) $\frac{9}{16}$; (b) $\frac{7}{16}$; (c) $\frac{1}{2}$

10. Let X be a continuous random variable with pdf

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

a. Find the value of a

b. Find P(X<1.5)

Ans: $a=\frac{1}{2},P(X<1.5)=\frac{1}{2}$

Note: If X is continuous random variable and f(x) is the pdf of X defined in the interval (a, b) then

a. The median M of the distribution is obtained by solving ${\int }_a^{M}f(x)\mathrm{d}x=\frac{1}{2}\mathrm{and}={\int }_M^{\infty }f(x)\mathrm{d}x=\frac{1}{2}$, for M

b. The mode of the distribution is the value of x for which $f\prime (x)=0$ and $f\prime (x)<0$ (a<x<b) hold, i.e., f(x) is maximum.

c. Mean deviation about the mean is given by

$\mathrm{Mean}\mathrm{deviation}={\int }_a^b|x-\mu |f(x)\mathrm{d}x$

d. $\mathrm{Harmonic}\mathrm{mean}={\int }_a^b\frac{1}{x}f(x)\mathrm{d}x$

e. $\mathrm{Geometric}\mathrm{mean}=\log =G={\int }_a^b\log xf(x)dx$, where G is the Geometric mean.