CHAPTER IV RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONSsciences.kau.edu.sa/GetFile.aspx?id=262189&Lng=AR&fn=Ch(4)_210.pdf · chapter iv discrete random variables discrete probability

CHAPTER IVDISCRETE RANDOM VARIABLES

DISCRETE PROBABILITY DISTRIBUTIONS

DISCRETE RANDOM VARIABLES

• Illustrative Example:

• Consider the experiment of tossing a coin 3times. It is known that the sample space is

• S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

• Define X = the number of heads appeared

• The possible values of X are 0, 1, 2, 3.

• X is called a discrete random variable

2Statistics Department - Faculty of Science

King Abdulaziz University


• Remarks:

• The discrete random variable X is a real-valuedfunction defined on the sample space S, suchthat it assigns to each sample point in S, a realnumber r in R.

• The function X is multi – valued.

• The values of the function X are discrete.

• X is random because its values are related tothe outcomes of a random experiment




Examples of random variables

(1) The number of boys in a randomly chosen family of a certain community

(2) The number of traffic accidents at a certain time and in a certain point on a given road

(3) The number of smokers in a randomly selected sample from a group of adults




(4) The number of misprints in a given book

(5) The number of errors in a randomly selected computer program

(6) The height of a person of a certain age

(7) The weight of a person of a certain age

(8) The temperature of a living sick man

(9) The air pressure at different heights




• Based on the number and nature of values,the random variables are classified into:

(i) Discrete random variables

(ii) Continuous random variables

• The discrete random variables assumesvalues X = x1 , x2 , x3 , …

• The continuous random variables assumesvalues: x > a, x <b, a < x < b



Discrete Random Variables

• For a discrete random variables X we define:

• The Probability Mass Function (PMF), as

f (x) = P (X = x)

such that

(i) f (x) ≥ 0 for every value of X

(ii)

(iii) P (X = k) = f (k)



1 1( ) ( ) 1

n n

i ii if x P X x

Dr. Mahmoud Abd Almo'menDr. Mahmoud Abd Almo'men Atallah


• Example (1)

• Consider the PMF of the discrete r. v. X

f (x) = P (X = x) = k (5 x + 7) for x = 0, 1, 3, 4

Find the value of the constant k.

• Solution.

• 7k +12k + 22K + 27k = 1 => 68k = 1 => K = 1/68

X 0 1 3 4

f (x) 7 k 12 k 22 k 27 k




• Example (2)The discrete r. v. X has the following PMFf (x) = P (X = x) = k 5Px , x = 1, 3, 5Find the value of the constant k

• Solution

• 5k + 60k + 120k = 1 => 185k = 1 = > k = 1/185

X 1 3 5

f (x) 5P1 k 5P3 k5P5 k

5 k 60 k 120 k




• Example (3)

The discrete r. v. X has the following PMF

f (x) = P (X = x) = k 5Cx , x = 1, 2, 3, 4

Find the value of the constant k

Solution.

5k + 10k + 10k + 5k = 30 k = 1 => k = 1/30

X 1 2 3 4

f (x) 5C1 k 5C2 k 5C3 k 5C4 k




• Example (4)

The discrete r. v. X has the following PMF

f (x) = P (X = x) = k 5Cx8C 4 – x , x = 1, 2, 3, 4

Find the value of the constant k

Solution Form the table

• k [280 + 280 + 80 + 5 ] = 645 k = 1 => k = 1/645

X 1 2 3 4

f (x) 5C18C 3 k 5C2

8C 2 k 5C38C 1 k 5C4

8C 0 k




Expectation of the discrete r. v. X

(Expected value, Mean value, Mean)

The expected value of the r. v. X determines the value about which the values of X tend to accumulate.

( ) ( ) ( )all x all x

E X x P X x x f x




Properties of E ( ) :

(1) E(X) can be positive, negative, or zero

(2) For a constant k: E(k) = k

(3) For any constants a and b:

E (a X + b) = a E(X) + b

(4) For the r. v. g (X):

[ ( )] ( ) ( )all x

E g X g x f x




Higher moments of X

• The r-th non-central moment of the discreterandom variable X is denoted by

( ) ( ) ( ), 1, 2, 3,r r r

r

x x

E X x P X x x f x r




• Example (5) Consider the probabilitydistribution

• E(X)= 2.5

• E(X2 ) = 7.17

• E(5X + 2) = 5 E(X) + 2 = 5*2.5 + 2 = 14.5

X 1 2 3 4

f (x) 1/6 1/3 1/3 1/6




• E(2X2 + 5) = 2 E(X2 ) + 5 = 19.34

• E(3 X2 + 4 X +5) = 3 E(X2 ) + 4 E(X) + 5

= 36.51

• E[(4 X + 3)2 ] = E [16 X2 + 24 X + 9]

= 16 E(X2 ) + 24 E(X) + 9

= 183.72




Variance of a random variable X

• It measures the degree of scatter (measuredin squared units of X) of the values of X about its mean µ.

• It is denoted by Var (X) or σ2 and defined by

σ2 = Var (X) = E[(X - µ)2 ] = E(X2 ) - µ2




Properties of the variance

(1) Var (X) ≥ 0

(2) Var (k) = 0, k is a constant

(3) Var (k X) = k2 Var (X), k is a constant

(4) Var (a X + b) = a2 Var (X), a & b are constants




The standard deviation of X

• It measures the degree of scatter of the valuesof X (measured in the same units as X) aboutits mean µ.

• It is denoted by σ and defined as

( )Var X




• Example (6)• For the probability distribution

We have: E(X) = 2.5 and E(X2 ) = 7.17

• Var (X) = 7.17 – (2.5)2 = 0.92

X 1 2 3 4

f (x) 1/6 1/3 1/3 1/6

0.92 0.96X




• It follows that

• Var (2 X + 3) = 4 * 0.92 = 3.68

• Var (3 X – 5) = 9 * 0.92 = 8.28

• Var (4 X) = 16 * 0.92 = 14.72

•2 3 (2 3) 3.68 1.92X Var X




Cumulative Distribution Function of X

• It is denoted by F (x) and defined as follows:

• Some properties of F(x) :

(1) F(- ∞) = 0, F(+ ∞) = 1

(2) 0 ≤ F(x) ≤ 1

( ) ( ) ( ) ( )u x u x

F x P X x P X u f u




(3) F(x) is a stepwise function

(4) F(x) is a nondecreasing function of X, i.e.

x < y => F(x) ≤ F(y)

(5) F(x) is continuous from the right, i.e.

F(a) = F(a + )




Example (7)

Find the cumulative distribution function for

Solution

X 0 3 5

f (x) 0.25 0.5 0.25

X x < 0 0 ≤ x < 3 3 ≤ x < 5 x ≥ 5

F (x) 0 0.25 0.75 1




The Probability Generating Function

• It is denoted by GX (t) and defined by

• Such that

(1)

(2)

(3)

Statistics Department - Faculty of Science King Abdulaziz University

25

( ) ( ) ( ) ( )X x x

X

x x

G t E t t P X x t f x

(1) 1XG

( ) (1), [ ( 1)] (1)X XE X G E X X G

2var (1) (1) (1)X X Xiance G G G


• Example (8) The PGF of the distribution of the r. v. Xis given by

• Find the value of E (X), E[X(X-1)], and Var (X).

• Solution


26

24 6 4 1( )

15 15 15 15G t t t t t

( ) (1) 2.133Xmean E X G

2var (1) (1) (1) 0.782X X Xiance G G G

[ ( 1)] (1) 3.2XE X X G


27


• Remark. The PMF f(x) can be determined from thegiven PGF. We have

• Then, we can find the requirements using theobtained PMF.

24 6 4 1( )

15 15 15 15G t t t t t


28



• Find the probability mass function of X.

• Solution

24 6 4 1( )

15 15 15 15G t t t t t

X 1 2 3 4

f(x) 4/15 6/15 4/15 1/15




• Solution


29

2var (1) (1) (1) 1.64X X Xiance G G G

2 3 4( ) 0.08 0.14 0.20 0.26 0.32G t t t t t

( ) (1) 2.60Xmean E X G [ ( 1)] (1) 5.80XE X X G


30



• Find the probability mass function of X.

• Solution

2 3 4( ) 0.08 0.14 0.20 0.26 0.32G t t t t t

X 0 1 2 3 4

f (x) 0.08 0.14 0.20 0.26 0.32


• Example (12) The PGF of the distribution ofthe r. v. X is given by:


• Solution


31

0.65( )

1 0.35

tG t

t

( ) (1) 1.538Xmean E X G [ ( 1)] (1) 1.657XE X X G

2var (1) (1) (1) 0.828X X Xiance G G G


• Example (13) The PGF of the distribution of the r. v. Xis given by:


• Solution


32

5

( ) 0.25 0.75G t t

( ) (1) 3.75Xmean E X G [ ( 1)] (1) 11.25XE X X G

2var (1) (1) (1) 0.938X X Xiance G G G


• Example (14) The PGF of the distribution ofthe r. v. X is given by:


• Solution


33

5 ( 1)( ) tG t e

( ) (1) 5Xmean E X G [ ( 1)] (1) 25XE X X G

2var (1) (1) (1) 5X X Xiance G G G


The Moment Generating Function

• It is denoted by MX (t) and defined by

• Such that

(1) MX (0) = 1

(2)

( ) ( ) ( ) ( )t X t x t x

X

x x

M t E e e P X x e f x

( ) (0) ( ), 0,1,2,r r

XM E X r




Example (15)

Find the moment generating function, given that

Solution

1 3 5( ) ( ) 0.25 0.5 0.25t x t t t

X

x

M t e f x e e e

X 1 3 5

f (x) 0.25 0.5 0.25




• Example (16)

• The MGF of the r. v. X is given by

Find the corresponding PMF f(x).

• Solution

1 2 3( ) 0.125 0.375 0.375 0.125t t t

XM t e e e

X 0 1 2 3

f (x) 0.125 0.375 0.375 0.125




1. Bernoulli distribution

Definition “Bernoulli trial”

It is a random experiment, whose outcomes is either:

(a) “A = success” with probability p, 0<p<1, or

(b) “not A = failure” with probability 1 – p = q

Examples :

(i) Tossing a coin once and looking for “Head”

(ii) Rolling a die and looking for “face 1”




Now let X denote the number of successes in one Bernoulli trial.

The possible values of X are 0, 1.

It is clear that X is a discrete r. v. with

P(X = 1) = P(success) = p

P(X = 0) = P(failure) = q

Such that p + q = 1.




The probability mass function (PMF) of X is

This is called the Bernoulli distribution with parameter p.

This is referred to simply, as

1( ) ( ) , 0, 1, 0 , 1, 1x xf x P X x p q x p q p q

( )X Ber p




Characteristics of Bernoulli distribution

(a) Mean E(X) = p

(b) Variance Var(X) = p q

(c) Standard deviation

(d) The moment generating function

( ) [ ]t x t

XM t E e q pe

p q




41


Example (17): Let X ~ Ber (0.85).

This means that X has Bernoulli distribution with p = 0.85.

f(x) = P(X = x) = (0,85)x (0.15)1 – x x = 0, 1

(a) Mean E(X) = 0.85

(b) Variance Var(X) = (0.85)(0.15) = 0.1275

(c) Standard deviation

(d) MGF ( ) [ ] 0.15 0.85t x t

XM t E e e

0.1275 0.357


2. Binomial distribution

Consider of n Bernoulli trials such that:

(1) The outcome of each trial is

(i) success with probability p, or

(ii) failure with probability 1 – p = q

(2) The probability p of success is constant, i.e. it doesn’t change from trial to trial.

(3) The trials are independent.




Let X be a r. v. denoting the number of successes in these n trials

The possible values of X are X = 0, 1, 2, …, n

It is clear that X is a discrete r. v.

The probability mass function of X is

This is called binomial distribution with parameters n and p. We write

( ) ( ) , 0,1,2,...,x n xn

f x P X x p q x nx

( , )X Bin n p




Characteristics of binomial distribution

(a) Mean = E(X) = n p(b) Variance = Var(X) = n p q(c) Standard deviation(d) Moment generating function

(e) Probability generating function

n p q

( ) [ ]n

t X t

XM t E e q p e



( )n

G t q p t


Example (18)

A fair coin is tossed 6 times. Find the probability that the Head will appear:

(a) Exactly 3 times

(b) At least 2 times

(c) At most 4 times

Solution



We have

(a) P(X=3) = f(3)=0.313

(b) P(X ≥ 2) = 0.891

(c) P(X ≤ 4) = 0.891


66 1

( ) ( ) , 0,1,..., 62

f x P X x xx



Example (19)

The moment generating function of the r. v. X is

Determine: the probability distribution, the mean and variance.

Solution

X ~ Bin (6, 0.75). E(X) = 6*0.75 = 4.5

Var (X) = 6*0.75*0.25 = 1.125


6

( ) 0.25 0.75 t

XM t e




3. Poisson distribution

Let X be a r. v. denoting the number of successes in a sequence of n (n ≥ 30) Bernoulli trials with probability p (p < 0.5) of successes, satisfying the conditions of the binomial distribution such that n p = λ is finite.

The possible values of X are X = 0, 1, 2, …

It is clear that X is a discrete r. v.




The probability mass function of X is

In this case we say

X has a Poisson Distribution with parameter λand write

( ) ( ) , 0,1,2,...!

x

f x P X x e xx

( )X Poi




Characteristics of Poisson distribution

(a) Poisson distribution is a limiting distribution of the binomial distribution as n tends to ∞.

(b) Poisson distribution is the distribution of rare events. It is used when n is large and p is small, such that λ = n p < ∞.

(c) Mean = variance = λ

(d) The MGF is (exp 1)( ) t

XM t e




Example (20)

The number X of annual earthquakes in a certain country has a mean 4. What is the probability distribution of X.

Solution

Because of the earthquake is a rare event, then the distribution of X is Poisson distribution with parameter λ = 4.




Example (21)

Consider the case when X has a Poisson distribution with parameter 3.

In this case:

(i) The PMF of X is

(ii) E(X) = Var(X) = 3

(iii)P(X = 2) = f(2) = 4.5 e- 3 = 0.224

33( ) ( ) , 0,1,2,...

!

x

f x P X x e xx




(iv) P(X is at least 3) = P(X ≥ 3) = f(3) + f(4) + …

= 0.577

(v) P(X is at most 2) = P(X ≤ 3) = f(0) + f(1) + f(2)

= 0.423




4. Geometric distribution

Let X be a r. v. denoting the number of Bernoulli trials, required to obtain the first success.

The possible values of X are X = 1, 2, …

It is clear that X is a discrete r. v. with

P(X = 1) = p

P(X = 2) = p q, …




In general

This is called the geometric distribution with parameter p.

We denote this by writing

1( ) ( ) , 1,2,...xf x P X x p q x

( )X Geom p

2

1( ) , ( ) , ( )

1

t

X t

p eqE X Var X M t

p p q e



( )1

X

p tG t

q t


Example (22)

Consider the case

This means that X has geometric distribution with parameter p = 0.65.

Thus

We have:

(0.65)X Geom

1( ) ( ) (0.65)(0.35) , 1,2,...xf x P X x x




(i) E(X) = 1/0.65 = 1.538

(ii) Var(X) = 0.35 / (0.65)2 = 0.828

(iii) P(X = 3) = f(3) = 0.080

(iv) P(X > 2) = 1 – f(1) – f(2) = 1 – 878 = 0.122

(v) P(X < 4) = f(1) + f(2) + f(3) = 0.957




5. Negative binomial distribution

Let X be a r. v. denoting the number of Bernoulli trials required to obtain the first k successes.

The possible values of X are k, k+1, k+2, …

It is clear that X is a discrete random variable.

We can prove that the PMF of X has the form:

1

1( ) ( ) , , 1,...x k x k

kf x P X x C p q x k k




In this case, we say that X has a negative binomial distribution with parameters k, p .

This is referred to by writing

Note that

The negative binomial distribution reduces to the geometric distribution when k = 1.

Thus the geometric distribution is a special case of the negative binomial distribution.

( , )X NB k p




Characteristics of negative binomial distribution

(a) E(X) = k/p

(b) Var(X) = k q/p2

(c) The MGF is

(d) The probability generating function is

( )1

kt

X t

p eM

q e



( )1

k

X

p tG t

q t


Example (23)

Consider the case when . This means that X has the negative binomial distribution with

k = 5 and p = 0.8.

In this case

(a) E(X) = 5/0.8 = 6.25

(b) Var (X) = 5*0.2 / (0.8)2 = 1.563

(c)

(5,0.8)X NB

56

( 7) (0.8) (0.2) 0.1974

P X




6. Hyper-geometric distribution

Consider a collection of k of objects of a certain Type and N – k of another Type.

A random sample of size n is drawn without replacement.

Let X be a random variable denoting the number of objects of the first type in the selected sample.




The PMF of X is

In this case, we say that X has a hyper-geometric distribution with parameters N, k, n and write

( ) ( ) , max(0, ),...,min( , )

k N k

x n xf x P X x x n k N k n

N

n

( , , )X HG N k n




64

Discrete Probability Distributions

Characteristics of hyper-geometric distribution

X ~ HG (N, k, n)

(i) Mean

(ii) Variance

( )k

E X n p nN

( ) 11

k k N nVar X npq n

N N N


Example (24)

From a group of 6 men and 4 women, a random sample of size 5 persons is selected without replacement. Let X denotes the number of men in the sample.

It is clear that X has HG(10, 6, 5). The PMF of X is

6 4

5( ) ( ) , 1,...,5

10

5

x xf x P X x x



Dr. Mahmoud Abd Almo

men


We have : N = 10, k = 6, n = 5

(i) Mean = 3

(ii) Variance = 0.67

(iii)P(2 men are selected) = P(X=2) = f(2) = 0.238

(iv) P(selecting 2 women) = P(X=3) = f(3) = 0.476

(v) P(less than 3 men) = f(1)+f(2) = 0.262



Documents

CHAPTER IV RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONSsciences.kau.edu.sa/GetFile.aspx?id=262189&Lng=AR&fn=Ch(4)_210.pdf · chapter iv discrete random variables discrete probability