Chapter 5: Special Distributions

5.2 The Bernoulli and Binomial Distributions

Definition: A random variable X has the Bernoulli

distribution with parameter p(0 p 1) if X can takeonly the values 0 and 1 and the probabilities are

P (X = 1) = p, P (X = 0) = 1 p

E(X) = 1 p + 0 (1 p) = p,E(X2) = 12 p + 02 (1 p) = p,V ar(X) = E(X2) [E(X)]2 = p p2 = p(1 p).Definition: If the random variables in a finite or in-

finite sequence X1, X2, ... are i.i.d., and if each random

variable Xi has the Bernoulli distribution with parameter

p, then it is said that X1, X2, ... are Bernoulli trials with

parameter p.

Theorem: If the random variables X1, ..., Xn form n

Bernoulli trials with parameter p, and if X = X1 + ... +

Xn, then X has the binomial distribution with parameters

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n and p. The p.f. is as follows:

f (x) = P (X = x) =

{ (nx

)px(1 p)nx x = 0, 1, ..., n

0 otherwise

E(X) =n

i=1E(Xi) = np

V ar(X) =n

i=1 V ar(Xi) = np(1 p)Note: not every sum of Bernoulli random variables has

a binomial distribution. The Bernoulli random variables

must be mutually independent, and they must all have

the same parameter.

5.3 The Hypergeometric Distributions

In this section, we consider dependent Bernoulli ran-

dom variables.

Example: Sampling without Replacement. Suppose

that a box contains A red balls and B blue balls. Sup-

pose also that n balls are selected at random from the

box without replacement, and let X denote the number

of red balls that are obtained.

let Xi = 1 if the ith ball drawn is red and Xi = 0 if not.

Then each Xi has a Bernoulli distribution, but X1, ..., Xn

are not independent in general. X is not Binomial.

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Theorem: The distribution of X has the p.f.

f (x) = P (X = x) =

(Ax

)(Bnx)(

A+Bn

)for max{0, nB} x min{n,A}.

It is said that X has the hypergeometric distribution

with parameters A, B, and n.

Theorem: Let X have a hypergeometric distribution

with strictly positive parameters A, B, and n. Then

E(X) =nA

A + B

V ar(X) =nAB

(A + B)2 A + B nA + B 1

5.4 The Poisson Distributions

Many experiments consist of observing the occurrence

times of random arrivals. Examples include arrivals of

customers for service, arrivals of calls at a switchboard,

occurrences of floods and other natural and man-made

disasters, and so forth.

Definition: Let > 0. A random variable X has the

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Poisson distribution with mean if the p.f. of X is as

follows:

f (x) = P (X = x) =ex

x!for x = 0, 1, 2, ...

Theorem: If X Poisson(), then

E(X) = V ar(X) = .

The Poisson Approximation to Binomial Distributions

Theorem: X Binomial(n, pn), If limn npn = ,then X can be approximated by X Poisson(). Thatis, (

n

x

)pxn(1 pn)nx

ex

x!

for all x = 0, 1, 2...

Example: Approximating a Probability. Suppose that

in a large population the proportion of people who have a

certain disease is 0.01. We shall determine the probability

that in a random group of 200 people at least four people

will have the disease.

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X=the number of people having the disease among the

200 people. X Binomial(n = 200, p = 0.01).This distribution can be approximated by the Poisson

distribution for which the mean is = np = 2. X Poisson( = 2).

P (X 4) = 1 P (X = 0) P (X = 1)P (X = 2) P (X = 3)

= 1 e220

0! e

221

1! e

222

2! e

223

3!= 0.1428

5.5 The Negative Binomial Distributions

Suppose that an infinite sequence of Bernoulli trials

with probability of success p are available. The number

X of failures that occur before the rth success has the

following p.d.f.:

f (x) = P (X = x) =

(r + x 1

x

)pr(1 p)x,

for x = 0, 1, 2, ....A random variable X has the negative binomial distri-

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bution with parameters r and p. X NB(r, p).

If r = 1, X NB(1, p), f (x) = p(1 p)x, x =0, 1, 2, ... This would be the number of failures until the

first success. X has the geometric distribution with pa-

rameter p.

Example: Suppose that a machine produces parts that

can be either good or defective. Assume that the parts

are good or defective independently of each other with p

for all parts. An inspector observes the parts produced

by this machine until she sees four defectives. Let X be

the number of good parts observed by the time that the

fourth defective is observed. What is the distribution of

X?

X NB(4, p)

Theorem: IfX has the negative binomial distribution

with parameters r and p, the mean and the variance of X

must be

E(X) =r(1 p)

p, V ar(X) =

r(1 p)p2

Discrete distributions:

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The Bernoulli and Binomial Distributions

The Hypergeometric Distributions

The Poisson Distributions

The Negative Binomial Distributions and Geometricdistributions

...

Next, we will study some continuous distributions:

5.6 The Normal Distributions

Definition: A random variable X has the normal distri-

bution with mean and variance 2 ( < 0) if X has a continuous distribution with the following

p.d.f.:

f (x) =12

e(x)

2

22

for < x

It is seen that the curve is symmetric and bell shaped.

Theorem: If X has the normal distribution with mean

and variance 2 and if Y = aX + b, where a and b

are given constants and a 6= 0, then Y has the normaldistribution with mean a + b and variance a22.

The Standard Normal Distribution

Definition: The normal distribution with mean 0 and

variance 1 is called the standard normal distribution. Z N(0, 1).

The p.d.f. and c.d.f. of the standard normal distribu-

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tion are usually denoted by

(x) =12e

x2

2 , < x

x0

(x)

Example: Quantiles of Normal Distributions. If X N( = 1.329, 2 = 0.48442), find x0 such that P (X x0) = 0.05.

0.05 = P (X x0) = P (X 1.329

0.4844 x0 1.329

0.4844)

= P (Z x0 1.3290.4844

)

= (x0 1.329

0.4844)

1.64501.645

0.05 0.05

From the Z-table, (0.95) = 1.645,(0.05) = 1.645 =x01.329

0.4844 , x0 = 0.5322.

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Linear Combinations of Normally Distributed Vari-ables

Theorem: If the random variables X1, ..., Xk are inde-

pendent and if Xi N(i, 2i ), (i = 1, ..., k), then thesum X1 + ... + Xk N(1 + ... + k, 21 + ...2k).

Definition: Let X1, ..., Xn be random variables. The

average of these n random variables, Xn =n

i=1Xi, is

called their sample mean.

Theorem: Suppose that the random variablesX1, ..., Xn

form a random sample from the normal distributionN(, 2),

then Xn N(, 2

n ).

Example: Suppose that a random sample of size n is

to be taken from the normal distribution N(, 2 = 9),

determine the minimum value of n for which P (|X| 1) 0.95.

X N(, 9n), Z =X

9/n= n

1/2

3 (X ) N(0, 1)

P (|X | 1) = P (|Z| n1/23 ) 0.95.That is, P (Z > n

1/2

3 ) = 1 (n1/2

3 ) 0.025

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(n1/2

3) 0.975

n1/2

3 1(0.975) = 1.96

n 34.6

the sample size must be at least 35.

1.9601.96

0.025 0.0250.95

5.7 The Gamma Distributions

Definition: gamma function:

() =

0

x1exdx, > 0.

Theorem: If > 1, () = ( 1)( 1).Theorem: (n) = (n 1)!,(1) = 1,(12) =

Definition: A random variable X has the gamma dis-

tribution with parameters > 0, > 0, if X has a con-

tinuous distribution for which the p.d.f. is

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f (x) =

{

()x1ex x > 0

0 x 0Theorem: If X Gamma(, ), then

E(X) =

, V ar(X) =

2.

0 1 2 3 4 5

0.0

0.5

1.0

1.5

x

Gam

ma

f(x)

= 0.1, = 0.1 = 1, = 1 = 2, = 2 = 3, = 3

Definition: If = 1, Gamma distribution reduces to

The Exponential Distribution. X Exp().104

f (x) =

{ex x > 0

0 x 0Theorem: If X Exp(), then

E(X) =1

, V ar(X) =

1

2.

Theorem: Memoryless Property of Exponential Distri-

butions. If X Exp(), for t > 0, h > 0,

P (X t + h|X t) = P (X h)

5.10 The Bivariate Normal Distributions Definition: When

the joint p.d.f. of two random variables X1 and X2 is of

the form

f (x1, x2) =1

212(1 2)1/2exp

{ 1

2(1 2)[(x1 11

)2 2(x1 11

)(x2 22

) + (x2 22

)2]}

it is said that X1 and X2 have the bivariate normal dis-

tribution N(1, 2, 21,

22, ).

Theorem: If (X1, X2) N(1, 2, 21, 22, ), then X1and X2 are independent if and only if they are uncorre-

lated = 0.

105

Theorem: If (X1, X2) N(1, 2, 21, 22, ), thenX1 N(1, 21), X2 N(2, 22),a1X1 + a2X2 + b N(a11 + a22 + b, a2121 + a2222 +

2a1a212) .

Example: If (X1, X2) N(1 = 66.8, 2 = 70, 21 =22, 22 = 2

2, = 0.68), find P (X1 > X2).

Since X1 and X2 have a bivariate normal distribution,

it follows that the distribution X1X2 will be the normaldistribution.

E(X1 X2) = 66.8 70 = 3.2

V ar(X1X2) = V ar(X1)+V ar(X2)2 Cov(X1, X2) = 2.56

X1 X2 N(3.2, 2.56)

P (X1 > X2) = P (X1 X2 > 0)

= P

[X1 X2 (3.2)

2.56>

0 (3.2)2.56

]= P (Z > 2)

= 1 (2)= 0.0227

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