Chapter 4. Random Variables - 3

Outline

• Bernoulli and Binomial random variables– Properties of Binomial Random Variables

• Poisson Random Variables

• Geometric Random Variables

Properties of Binomial RV – Expected Value

0

1

1 1

1

[ ] (1 )

(1 )

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we get1

[ ] (1 )1

nk k i n i

i

nk i n i

i

nk k i n i

i

nE X i p p

i

ni p p

i

n ni n

i i

nE X np i p p

i

1 1

1

11 1

0

1

1[ ] (1 )

1by letting -1

1[ ] ( 1) (1 )

[( 1) ] is a binomial random variable with parameters -1, .

When 1, [ ]

nk k i n i

i

nk k j n j

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k

nE X np i p p

ij i

nE X np j p p

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npE YY n p

k E X np

Properties of Binomial RV – Variance

)1( )(]1)1[(

])[(][)(Var

have we,][ Since]1)1[(]1[][

2

22

2

pnpnppnnp

XEXEX

npXEpnnpYnpEXE

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pnpXnpXE

Binomial Distribution

Poisson Random Variable - Examples

• Studying processes that generate rare, discrete number of events – The number of wrong telephone numbers that are dialed in

a day– The number of customers entering a post office on a given

day– The number of vacancies occurring during a year in the

federal judicial system– The number of misprints on a page of a book– The number of bacteria in a particular plate– The number of people in a community living to 100 years

of age

Poisson Random Variable• The probability of i events in a time period t for a Poisson

random variable with parameter ( is also commonly used) is

• Parameter r represents expected number of events per unit time. is the expected number of events over time period t.

• Difference between Binomial and Poisson distribution– There are a finite number of trials n in Binomial distribution– The number of events can be infinite for Poisson distribution

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

where

i

P X i e ii

rt

Probability mass function of Poisson RV

Poisson Distribution – Expected Value

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since

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Poisson Distribution – Variance

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• Ex 7a. Suppose that the number of typographical errors on a single page of a book has a Poisson distribution with parameter λ=1/2. Calculate the probability that there is at least one error on one page.X: the number of errors on this pageP(X ≥ 1) = 1 - P(X = 0)

P(X ≥ 1) = 1 - P(X = 0) = 1- e-1/2((1/2)0/0!) = 1 - e-1/2

,...2 ,1 ,0 !

)2/1()( 2/1 ii

eiXPi

• Ex 7b. Suppose that the probability that an item produced by a certain machine will be defective is .1. Find the probability that a sample of 10 items will contain at most 1 defective item.

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10)9(.)1(.

010

: (10,0.1)parameter on with Distributi Binomial

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Ex 7c. Consider an experiment that consists of counting the number of α-particles given off in a 1-second interval by 1 gram of radioactive material. If we know from past experience that, on the average, 3.2 such α-particle are given off, what is a good approximation to the probability that no more than 2 α-particles will appear?The number of α-particles given off can be modeled by a Poisson random variable with parameter λ = 3.2.

2.32

2.32.3

22.32.3

)2()1()0()2(

eee

XPXPXPXP

• Ex. If the area of a plate, A, is 100cm2 and there are r =.02 colonies per cm2, calculate the probability of at least 2 bacterial colonies on this plate.

• We have = rA = 100(0.2) = 2. Let X = number of colonies.

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

k

P X k e kk

22 21

)1()0(1)2(

ee

XPXPXP

• Ex. The number of deaths attributable to polio during the years 1968-1976 is given in the following table. Based on this data set, can we use Poisson distribution to model the number of deaths from polio?

-----------------------------------------------------------------------------Year (19-) 68 69 70 71 72 73 74 75 76---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

# deaths 24 13 7 18 2 10 3 9 16

-----------------------------------------------------------------------------The sample mean is 11.3 and the variance is 51.5.The Poisson distribution will not fit here because the mean and variance are too different.

Poisson Distribution – Weakly Correlated Events

• Poisson distribution remains a good approximation even when the trials are not independent, provided that their dependence is weak.

• Hat-matching problem, where n men randomly select hats from a set consisting of one hat from each person. What is the probability that there are k matches?

• We say that trial i is a success if person i selects his own hat.Ei = {trial i is a success}P(Ei) = 1/n

• The number of successes will approximately have a Poisson distribution with parameter n×1/n=1.

• The probability that there are k matches is e-1/k!.– This is the same as obtained in Chapter 2, Ex 5m.

Geometric Random Variable

• Independent trials, each having a probability p of being a success, are performed until a success occurs. If X is the number of trials required,

• X is a geometric variable with parameter p.

,...2 ,1 ,)1()( 1 nppnXP n

Ex 8a. An urn contains N white and M black balls. Balls are randomly selected, one at a time, until a black one is obtained. If we assume that each selected ball is replaced before the next one is drawn, what is the probability that(a) exactly n draws are needed;(b) at least k draws are needed?If we let X denote the number of draws needed to select a black ball, then X is a geometric random variable with parameter M/(M+N).

n

nn

NMMN

NMM

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)()( )(

11

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Ex 8b. Find the expected value of a geometric random variable.

1

1

1

1

1 1

1 1

0

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1

Let 1 ,

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( 1 1)

( 1)

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