Chapter 3 Discrete Random Variables

Chapter 3Discrete Random Variables:

ContentRandom VariablesThe Probability Mass FunctionsDistribution FunctionsBernoulli TrialsBernoulli DistributionsBinomial DistributionsGeometric DistributionsNegative Binomial DistributionsPoisson DistributionsHypergeometric DistributionsDiscrete Uniform Distributions

Chapter 3Discrete Random VariablesRandom Variables

Definition Random VariablesA random variable X of a probability space (, A, P) is a real-valued function defined on , i.e.,

Example 1

Example 11

Example 12

Example 1

Example 2

Notations

Example 1

Definition Discrete Random VariablesA discrete random variable X is a random variable with range being a finite or countable infinite subset {x1, x2, . . .} of real numbers R.

Definition Discrete Random VariablesA discrete random variable X is a random variable with range being a finite or countably infinite subset {x1, x2, . . .} of real numbers R.What is countablity?The set of all integersThe set of all real numbers

Example 1X, Y and Z are discrete random variables. All are finite

Example 2X, Y and Z are not discrete random variables. All are uncountable

Example 2X, Y and Z are not discrete random variables. All are uncountableIn fact, they are continuous random variables.

Chapter 3Discrete Random VariablesThe Probability Mass Functions

Definition The Probability Mass Function (pmf)The probability mass function (pmf) of r.v. X, denoted by pX(x), is defined as

Example 4

Properties of pmfsx

Chapter 3Discrete Random VariablesDistribution Functions

Cumulative Distribution Function (cdf)cdf

Cumulative Distribution Function (cdf)cdfpmf

Cumulative Distribution Function (cdf)cdfpmfxpX(x)x1x2x3x4

Cumulative Distribution Function (cdf)cdfpmfxpX(x)x1x2x3x41pX(x1)pX(x2)pX(x3)pX(x4)

Example 5

Properties of cdfs xMonotonically nondecreasing.

Properties of cdfs xMonotonically nondecreasing.

F(b)

F(a)

Chapter 3Discrete Random VariablesBernoulli Trials

Bernoulli TrialsSuppose an experiment consists of n trials, n > 0.The trials are called Bernoulli trials if three conditions are satisfied:Each trial has a sample space {S =1, F =0} (two outcomes), S to be called success and F to be called failure.For each trial P(S) = p and P(F) = q, where 0 p 1 and q = 1 p.The trials are independent.

Example 6Tossing a die ten times, the actual face number in each toss is unnoted. Instead, the outcome of 1 or 2 will be considered a success, and the outcome of 3, 4, 5, or 6 will be considered a failure. What is the sample space of the experiment?Is this experiment to performing Bernoulli Trials?Why?

DiscussionWhat probabilities may interest us on performing Bernoulli Trials?

Chapter 3Discrete Random VariablesBernoulli Distributions

Bernoulli DistributionsLet r.v. X denote the outcome of a Bernoulli trial, and let the probability of success equal to p.

Then, we havecdfpmf

Bernoulli Distributionscdfpmf

Chapter 3Discrete Random VariablesBinomial Distributions

pnXI(X)=?P(X=x)=?

Binomial DistributionsConsider n Bernoulli trials with the probability of success on each trial being p.Let r.v. X denote #successes in the experiment.Then,cdfpmfn trials

x successes

nx fails

ppp(1p)(1p)(1p)......

Binomial DistributionsConsider n Bernoulli trials with the probability of success on each trial being p.Let r.v. X denote #successes in the experiment.Then,cdfpmf

Binomial DistributionsConsider n Bernoulli trials with the probability of success on each trial being p.Let r.v. X denote #successes in the experiment.Then,cdfpmfwithout closed-form

Binomial DistributionsConsider n Bernoulli trials with the probability of success on each trial being p.Let r.v. X denote #successes in the experiment.Then,cdfpmf

Binomial Distributionscdfpmf

Binomial Distributions

Example 7Verify that b(x; n, p) is a valid pmf.

Example 810% of IC chips from an IC manufacturer are known to be defective. Taking a sample of ten IC chips from the manufacture. Find the probabilities of1. no IC chip is defective;2. at least 2 IC chips are defective.

Example 810% of IC chips from an IC manufacturer are known to be defective. Taking a sample of ten IC chips from the manufacture. Find the probabilities of1. no IC chip is defective;2. at least 2 IC chips are defective.Let X denote #defectives

Chapter 3Discrete Random VariablesGeometric Distributions

pI(X)=?P(X=x)=?!!!X

Geometric DistributionsConsider a sequence of Bernoulli trials with the probability of success on each trial being p.Let r.v. X denote the number of trials up to and including the first success.Then,cdfpmfx trials

x1 fails

(1p)(1p)(1p). . .First successp

Geometric DistributionsConsider a sequence of Bernoulli trials with the probability of success on each trial being p.Let r.v. X denote the number of trials up to and including the first success.Then,cdfpmf

Geometric DistributionsConsider a sequence of Bernoulli trials with the probability of success on each trial being p.Let r.v. X denote the number of trials up to and including the first success.Then,cdfpmfThe parameter of the experiment.

Geometric Distributionscdfpmf

Chart4

0.1

0.09

0.081

0.0729

0.06561

0.059049

0.0531441

0.04782969

0.043046721

0.0387420489

0.034867844

0.0313810596

0.0282429536

0.0254186583

0.0228767925

0.0205891132

0.0185302019

0.0166771817

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0.0135085172

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0.0034336838

0.0030903154

0.0027812839

0.0025031555

0.00225284

0.002027556

0.0018248004

0.0016423203

x

pX(x)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

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60.13888888890.27777777780.4166666667

70.16666666670.41666666670.5833333333

80.13888888890.58333333330.7222222222

90.11111111110.72222222220.8333333333

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1811

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0.0555555556

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20.09

30.081

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110.034867844

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380.002027556

390.0018248004

400.0016423203

x

pX(x)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

200.50.250.75

00.00000095370.00317121190

10.00001907350.02114141290.0000000001

20.00018119810.06694780760.0000000016

30.00108718870.13389561520.000000028

40.00462055210.18968545490.0000003569

50.01478576660.20233115190.0000034265

60.03696441650.16860929320.0000256987

70.0739288330.11240619550.0001541923

80.12013435360.06088668920.0007516875

90.16017913820.02706075080.0030067501

100.1761970520.00992227530.0099222753

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150.01478576660.00000342650.2023311519

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170.00108718870.0000000280.1338956152

180.00018119810.00000000160.0669478076

190.00001907350.00000000010.0211414129

200.000000953700.0031712119

111

x

b(x;20,0.5)

b(x;20,0.25)

x

x

b(x;20,0.75)

Example 9Tossing a fair die, find:the probability of the first appearing of 1 is in the 5th toss;the probability of the first appearing of 1 is in the first five tosses.

Example 9Tossing a fair die, find:the probability of the first appearing of 1 is in the 5th toss;the probability of the first appearing of 1 is in the first five tosses.Let X denote #tossing to reach the 1st 1

Memoryless or Markov Property12m?

Memoryless or Markov Property12mm+1m+2m+n12n

Memoryless or Markov PropertyX

Memoryless or Markov Property

Memoryless or Markov PropertyA r.v. X is said to have memoryless or Markov property if it satisfies

Theorem 1Let r.v. X have image 1, 2,. Then, XG(p) P(X > m + n|X > m) = P(X > n)where m, n be any positive integers.

Theorem 1PF)

Theorem 1PF)Define

Theorem 1Let r.v. X have image 1, 2,. Then, XG(p) P(X > m + n|X > m) = P(X > n)where m, n be any positive integers.

Geometric DistributionsConsider a sequence of Bernoulli trials with the probability of success on each trial being p.Let r.v. X denote the number of trials up to and including the first success.Then,cdfpmfModifiedYY0,1,2,Yyy = 0,1,y+1Yyy < 00 yyfailures

Modified Geometric DistributionsConsider a sequence of Bernoulli trials with the probability of success on each trial being p.Let r.v. Y denote the number of failures up to the first success.Then,cdfpmf

Modified Geometric Distributionscdfpmf

Chapter 3Discrete Random VariablesNegative Binomial Distributions

pI(X)=?P(X=x)=?!!!rX:r

Negative Binomial DistributionsConsider a sequence of Bernoulli trials with the probability of success on each trial being p.Let r.v. X denote #trials up to and including the rth success.Then,cdfpmfrX = x

x1r1

Negative Binomial DistributionsConsider a sequence of Bernoulli trials with the probability of success on each trial being p.Let r.v. X denote #trials up to and including the rth success.Then,cdfpmf

Negative Binomial DistributionsConsider a sequence of Bernoulli trials with the probability of success on each trial being p.Let r.v. X denote #trials up to and including the rth success.Then,cdfpmfwithout closed-form

Negative Binomial DistributionscdfpmfFact:

Negative Binomial Distributionscdfpmf

Chart1

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0.00917504

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x

pX(x)

X~NB(5,0.2)

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b(x;20,0.5)

b(x;20,0.25)

x

x

b(x;20,0.75)

10.1

20.09

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80.04782969

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340.0030903154

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360.0025031555

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380.002027556

390.0018248004

400.0016423203

x

pX(x)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

50.00032

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x

pX(x)

X~NB(5,0.2)

Negative Binomial Distributionspmf

Negative Binomial DistributionspmfVerify that nb(x;r,p) is a valid pmf.Exercise

pI(X)=?P(X=x)=?!!!rX:r

pI(Y)=?P(Y=y)=?!!!rY:rFact: Y = X r

Negative Binomial DistributionspmfModifiedFact: Y = X rYYyyy =0,1,2, yyy

Modified Negative Binomial DistributionspmfFact:

Chapter 3Discrete Random VariablesPoisson Distributions

A Toll Station

Arriving/Failure Rate:

Poisson DistributionsConsider a highway toll station.Assume that, on average, vehicles pass the station per unit time interval (e.g., an hour). Let X denote #vehicles passing in a time interval of duration t, i.e., (0, t].: Arriving rateX: #vehicles passing in (0, t]I(X) = ?P(X=x) = ?{0, 1, 2, }

Poisson Distributions: Arriving rateX: #vehicles passing in (0, t]P(X=x) = ? ?

t

n

n


t

n

n()?010101001000110001010


t

n

nn

p = ?


t

n

nn

p = ?np


t

n

nn

p = ?

Poisson Distributions: Arriving rateX: #vehicles passing in (0, t]

=1, n

=1, n

Poisson Distributions: Arriving rateX: #vehicles passing in (0, t]Let

Poisson Distributions: Arriving rateX: #vehicles passing in (0, t]

Poisson DistributionsConsider a highway toll station.Assume that, on average, vehicles pass the station per unit time interval (e.g., an hour). Let X denote #vehicles passing in a time interval of duration t, i.e., (0, t].: Arriving rateX: #vehicles passing in (0, t]cdfpmfwithout closed-form

Poisson DistributionsConsider a highway toll station.Assume that, on average, vehicles pass the station per unit time interval (e.g., an hour). Let X denote #vehicles passing in a time interval of duration t, i.e., (0, t].cdfpmfwithout closed-formin an interval the same interval

Poisson DistributionscdfpmfVerify that p(x;) is a valid pmf.Exercise

Poisson DistributionscdfpmfVerify that p(x;) is a valid pmf.Exercise

Chart1

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x

b(x;20,0.75)

10.1

20.09

30.081

40.0729

50.06561

60.059049

70.0531441

80.04782969

90.043046721

100.0387420489

110.034867844

120.0313810596

130.0282429536

140.0254186583

150.0228767925

160.0205891132

170.0185302019

180.0166771817

190.0150094635

200.0135085172

210.0121576655

220.0109418989

230.009847709

240.0088629381

250.0079766443

260.0071789799

270.0064610819

280.0058149737

290.0052334763

300.0047101287

310.0042391158

320.0038152042

330.0034336838

340.0030903154

350.0027812839

360.0025031555

370.00225284

380.002027556

390.0018248004

400.0016423203

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

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0

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0

0

0

0

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0

x

pX(x)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

50.00032

60.00128

70.003072

80.0057344

90.00917504

100.0132120576

110.0176160768

120.0221459251

130.0265751101

140.0307090162

150.0343940981

160.0375208343

170.0400222233

180.0418694028

190.0430656714

200.0436398804

210.0436398804

220.04312647

230.042168104

240.0408364797

250.0392030205

260.03733621

270.0352996895

280.0331510127

290.0309409452

300.0287131971

310.0265044897

320.0243448646

330.0222581619

340.0202626026

350.0183714263

360.0165935464

370.0149341917

380.0133955174

390.0119771685

400.0106767902

410.0094904802

420.0084131824

430.0074390245

440.0065616011

450.0057742089

460.0050700371

470.0044423182

480.0038844457

490.0033900617

500.0029531204

510.0025679308

520.0022291825

530.0019319582

540.0016717352

550.0014443792

560.0012461311

570.0010735899

580.0009236924

590.0007936913

600.0006811314

610.0005838269

620.0004998378

630.0004274475

640.0003651416

650.0003115875

660.0002656156

670.0002262016

680.0001924509

690.0001635833

700.00013892

710.0001178715

720.0000999269

730.0000846439

740.0000716407

750.0000605875

760.0000512007

770.0000432362

x

pX(x)

X~NB(5,0.2)

00

10.0000000003

20.0000000043

30.0000000362

40.000000226

50.0000011302

60.0000047092

70.0000168185

80.0000525578

90.000145994

100.000364985

110.0008295113

120.0017281486

130.0033233628

140.0059345764

150.0098909606

160.0154546259

170.0227273911

180.0315658209

190.0415339749

200.0519174686

210.0618065102

220.0702346707

230.0763420334

240.0795229515

250.0795229515

260.0764643764

270.0708003485

280.0632145969

290.0544953422

300.0454127851

310.0366232138

320.0286118858

330.0216756711

340.0159379934

350.011384281

360.0079057507

370.0053417234

380.0035142917

390.0022527511

400.0014079694

410.000858518

420.0005110226

430.0002971062

440.0001688103

450.0000937835

460.0000509693

470.0000271113

480.0000141205

490.0000072043

500.0000036022

510.0000017658

520.0000008489

530.0000004004

540.0000001854

550.0000000843

560.0000000376

570.0000000165

580.0000000071

590.000000003

600.0000000013

610.0000000005

620.0000000002

630.0000000001

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x

pX(x)

Example 12On average, a job arrives for CPU service every 6 seconds. Find the probability that there will be less than or equal to 4 arrivals in a given minute? = 1/6 job/sec.Let X denote #arrivals in the minute.t = 60 secs. = t = 10 jobs.

Poisson ApproximationThe binomial distribution is important, but the probability values associated with it are hardly evaluated.

Poisson Approximationppnp

Poisson Approximationpp

Poisson Approximationpppp

Poisson Approximation

Poisson ApproximationnpE.g., n 20 and p 0.05.

Example 13A manufacture produces IC chips, 1% of which are defective. A box contains 100 chips. Find the probability that1. the box contains no defective.2. the box contains less than or equal to 2 defectives.Let X denote the number of defectives.

Chapter 3Discrete Random VariablesHypergeometric Distributions

Hypergeometric Distributions+= N

d

NdI(X)=?P(X=x)=?

Hypergeometric DistributionsI(X)=?P(X=x)=?cdfpmfwithout closed-form

Hypergeometric Distributionspmf

Example 14Compute the probability of obtaining three defectives in a sample of size ten taken without replacement from a box of twenty components containing four defectives.Let X denote the number of defectives.

Hypergeometric Distributionspmf

Example 15A box contains 200 red balls and 800 black balls. Now 10 balls are taken without replacement. Find the probability of obtaining none red ball.Let X denote #red balls taken.

Chapter 3Discrete Random VariablesDiscrete Uniform Distributions

Discrete Uniform DistributionsA r.v. X is said to possess a discrete uniform distribution if it has a finite image {x1, x2, , xN} and has the pmf

Example 16One ball is drawn from a box containing 10 balls numbered 1,2,. . . ,10. Find the probability that the ball number is less than 4.Let r.v. X denote the ball number.Is uniform distribution really that simple?

ReviewBernoulli DistributionsBinomial DistributionsGeometric DistributionsNegative Binomial DistributionsPoisson DistributionsHypergeometric DistributionsDiscrete Uniform Distributions

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Chapter 3 Discrete Random Variables