11

Discrete Distributions

Ch3 Ch3

22

Def31-1 A function X that assigns to each element s in S exactly one real number X(s)=x is called a random variable The space of X is the set of real numbers x X(s)=x s Sisin If X(s)=s then the space of X is also S Ex31-2 S=123456 in casting a dice Let X(s)=s

P(X=5)=16 P(2leXle5)=P(X=2345=46 P(Xle2)=P(X=12)=26

Two major difficulties In many practical situations the probabilities assigned to the even

ts are unknown Repeated observations (Sampling) to estimate them

There are many ways to define X Which to use Measurement materialization of outcomes

To draw conclusions or make predictions

33

Random Variables of the Discrete Type When S contains a countable number of points

X can be defined to correspond each point in S to a positive integer S is a set of discrete points or a discrete outcome space X is a random variable of discrete type The probability mass function (pmf) f(x) denotes P(X=x)

f(x) is also called probability function probability density function frequency function

Def31-2 f(x) of a discrete random variable X is a function f(x)gt0 x Sisin sumx Sisin f(x) = 1 P(X A) = sumisin x Aisin f(x) where A Ssub

f(x)=0 if x S S is referred to as the notin support of X and the space of X A distribution is uniform if its pmf is constant over the spa

ce For instance f(x)=16 in rolling a fair 6-sided dice Ex31-3 Roll a 4-sided die twice and let X be the largest of two outc

omes S=(i j) i j=14 P(X=1)=P((11))=116 P(X=2)=P((12) (21) (22))=316 hellip f(x)=P(X=x)=(2x-1)16 for x=14 f(x)=0 otherwise

44

Graphing the Distribution (a) For a discrete probability distribution we

simply plot the points (x f(x)) for all x R (b) To get a better picture of the distribution we

use bar graphs and histograms (c) A bar graph is simply a set of lines connecting

(x 0) and (x f(x)) (d) If X takes on only integer values then a

histogram can be used by using rectangles centered at each x R of height f(x) and width one

55

Graphic Representation of f(x) The graph consists of the set of points (x f(x) x Sisin

Better visual appreciation a bar graph a probability histogram P(X=x)=f(x)=(2x-1)16 x=14

Hyper-geometric distribution a collection has N1 objects of type 1 and N2 objects of type 2 X is the number of type-1 objects in the n objects taken from the collection wo replacement

66

Examples Ex31-5 In a pond of 50 fish with 10 tagged 7 fish are caught at random wo

replacement The probability of 2 tagged fish caught is

Ex31-7 In a lot of 25 items with unknown defective 5 items are selected at random wo replacement for exam If no defective found the lot is accepted otherwise rejected Given N1 defective the acceptance probability operating characteristic curve is

Ex31-8 Roll a 4-sided die twice X is the sum 28 f(x)=(4-|x-5|)16 1000 experiments simulated on a computer

77

Mathematical Expectation Def32-1 For a random variable X with pmf f(x) if sumx Sisin u(x)f(x) exists it

is called the mathematical expectation or the expected value of the function u(X) denoted by E[u(X)] It is the weighted mean of u(x) x Sisin The function u(X) is also a random variable say Y with pmf g(y)

Ex32-2 For a random variable X with f(x)=13 x S=-1 0 1isin Let u(X)=X2 Then E[u(X)]= E[X2]=23 The support of the random variable Y=X2 is S1=0 1

and P(Y=0)=P(X=0) P(Y=1)=P(X=-1)+P(X=1) so its pmf Hence sumy S1isin yg(y) = 23 too

Thm32-1 mathematical expectation E if exists satisfies For a constant c E(c)=c For a constant c amp a function u E[c u(X)]=c E[u(X)] For constants a amp b and functions u amp v E[a u(X) + b v(X)]=a E[u(X)] + b

E[v(X)] It can be applied for 2+terms since E is a linear or distributive operator

88

Examples Ex32-3 f(x)=x10 x=1234

Ex32-4 u(x)=(x-b)2 where b is a constant Suppose E[(X-b)2] exists What is the value of b to minimize it

Ex32-5 X has a hypergeometric distribution

99

More Examples Ex32-6 f(x)=x6 x=123

E(X)=μ=1(16)+2(16)+3(16)=73 Var(X)=σ2=E[(X-μ)2]=E(X2)-μ2=hellip=59 σ=0745rArr

Ex32-7 f(x)=13 x=-101 E(X)=μ=-1(13)+0(13)+1(13)=0 Var(X)=σ2=E(X2)-μ2=hellip=23 The standard deviation σ=0816rArr Comparatively g(y)=13 y=-202

Its mean is also zero but Var(Y)=83 and σY=2σ more spread outrArr

Ex32-8 uniform f(x)=1m x=1m E(X)=μ=1(1m)+hellip+m(1m)=(m+1)2 Var(X)=σ2=E(X2)-μ2=hellip=(m2-1)12

For instance m=6 when rolling a 6-sided die

1010

Derived Random Variables Linear Combination X has a mean μX and variance σX

2 Y=aX+b μrArr Y = aμX+b Var(Y)=E[(Y-μY)2] =hellip=a2σX

2 σX=|a|σX a=2 b=0 mean2 variance4 standard deviation2rArr a=1 b=-1 mean-1 variance1 standard deviation1 rArr Var(X-1) = Var(X)

The rth moment of the distribution about b E[(X-b)r] The rth factorial moment E[(X)r]=E[X(X-1)(X-2)hellip(X-r+1)]

E[(X)2] = E[X(X-1)] = E(X2)-E(X) = E(X2)-μE[(X)2]+μ-μ2= E(X2)-μ+μ-μ2= E(X2)-μ2= Var(X)=σ2

Ex32-9 X has a hypergeometric distribution (ref Ex32-5)

1111

Bernoulli Trials A Bernoulli experiment is a random experiment whose outcome can

be classified in one of two mutually exclusive and exhaustive ways success or failure

A series of Bernoulli trials occurs after independent experiments Probabilities of success p and failure q remain the same (p+q=1)

Random variable X follows a Bernoulli distribution X(success)=1 and X(failure)=0 The pmf of X is f(x)=pxq(1-x) x=01 (μ σ2)=(p pq)

A series of n Bernoulli trials a random sample will be an n-tuple of 01rsquos Ex33-4 Plant 5 seeds and observe the outcome (10101) 1st 3rd 5th seed

s germinated If the germination probability is 8 the probability of this outcome is (8)(2)(8)(2)(8) assuming independence

Let X be the number of successes in n trials X follows a binomial distribution denoted as b(n p) The pmf of X is

1212

Example Ex33-5 For lottery with 2 winning if X equals the number of winning

tickets among n=8 purchases The probability of having 2 winning tickets is

Ex33-6 The effect of n and p is illustrated as follows

1313

Cumulative Distribution Function The cumulative probability F(x) defined as P(Xle x) is called the cumulat

ive distribution function or the distribution function

Ex33-7 Assume the distribution of X is b(10 08) F(8) = P(Xle8) = 1-P(X=9)-P(X=10) = 1-10(8)9(2)-(8)10=6242 F(6) = P(Xle6) = sumx=06Cx

10 (8)x(2)10-x

Ex33-9 Y follows b(8 065) If X=8-Y X has b(8 35) whose distribution function is in Table II (p647)

Eg P(Yge6) = P(8-Yle8-6) = P(Xle2)=04278 from table lookup Likewise P(Yle5) = P(8-Yge8-5) = P(Xge3) = 1-P(Xle2) = 1-04278 =0 5722 P(Y=5) = P(X=3) = P(Xle3)-P(Xle2) = 07064-04278 = 02786

The mean and variance of the binomial distribution is (μ σ2)=(np npq) Ex34-2 details the computations

1414

Comparisons Empirical Data vs Analytical Formula

Ex33-11 b(5 5) has μ= np=

25 σ2= npq= 125 Simulate the model for 100 times

2 3 2 hellip = 247 srArr 2=15243

Suppose an urn has N1success balls and N2 failure balls N = N1+ N2

Let p = N1N and X be the number of success balls in a random sample of size n taken from this urn

If the sampling is done one at a time with replacement X follows b(n p) If the sampling is done without replacement X has a hypergeometric distrib

ution with pmf

If N is large and n is relative small it makes little difference if the sampling is done with or without replacement (See Fig33-4)

x

1515

Moment-Generating Function (mgf )

Def34-1 X is a random variable of the discrete type with pmf f(x) and space S If there is a positive integer h st

E(etX) = sumx Sisin etxf(x) exists and is finite for ndashhlttlth thenM(t) = E(etX) is called the moment-generating function of X ndashE(etX) exists and is finite for ndashhlttlth hArrM(r)(t) exist at t=0 r=123hellip ndashUnique association pmf hArrmgf

Sharing the same mgf two random variables have the same distribution of probability

Ex34-1 X has mgf M(t) = et(36)+ e2t(26)+ e3t(16) From eldquoxrdquot Its pmf has f(0)=0 f(1)=36 f(2)=26 f(3)=16 f(4)=0 hellip Therefore f(x)=(4-x)6 x=123

Ex34-2 X has mgf M(t) = et2(1-et2) tltln2 (1-z)-1= 1 + z + z2+ z3+ hellip |z|lt1

1616

Application of Application of mgf

M(t) = E(etX) = sumx Sisin etxf(x) exists and is finite for ndashhlttlthM(t) = sumx Sisin xetxf(x) M(0) = sumx Sisin xf(x) = E(X) M(t) = sumx Sisin x2etxf(x) hellip M(0) = sumx Sisin x2f(x) = E(X2) hellip M(r)(t) = sumx Sisin xretxf(x) M(r)(0) = sumx Sisin xrf(x) = E(Xr) as t=0

M(t) must be formulated (in closed form) to get its derivatives of higher order Ex34-3 X has a binomial distribution b(n p)

Thus its mgf is

When n=1 X has a Bernoulli distribution

1717

Negative Binomial Distribution Let X be the number of Bernoulli trials to observe the rth succ

ess X has a negative binomial distribution

Its pmf g(x) is

If r=1 X has a geometric distribution with

1818

Geometric Distribution X has a geometric distribution with the pmf

P(X gt k) P(X lek)

Memory-less (EX3412)

Ex34-4 Fruit fliesrsquo eyes with frac14white and frac34red The probability of checking at least 4 flies to observe a white eye is

P(Xge4) = P(Xgt3) = (frac34)3= 04219 The probability of checking at most 4 flies to observe a white eye

isP(Xle4) = 1-(frac34)3=06836 The probability of finding the first white eye on the 4th fly checked is

P(X=4) = pq4-1= 01055 lt= P(Xle4) -P(Xle3)

Ex34-4 For a basketball player with 80 free throw X is the minimum number of throws for a total of 10 free throws Its pmf is μ= rp= 1008 = 125 σ2= rqp2= 10(02)(08)2= 3125

1919

mgf rArr pdf By Maclaurinrsquosseries expansion (ref p632)

If the moments of X E(Xr) = M(r)(0) are known M(t) is thus determined

rArr pdf can be obtained by rewriting M(t) as the weighted sum of eldquoxrdquot

Ex34-7 If the moments of X are E(Xr) = 08 r=123hellip Then M(t) can be determined as

Therefore P(X=0)=02 and P(X=1)=08

2020

Poisson Process Def35-1 An approximate Poisson process with parameter λgt0

The numbers of changes occurring in non-overlapping intervals are independent

The probability of exactly one change in a sufficiently short interval of length h is approximately λh

The probability of two or more changes in a sufficiently short interval is essentially zero

Determine pmf During the unit interval of length 1 there are x changes For nraquox we partition the unit interval into n subintervals of length 1n The probability of x changes in the unit interval equivThe probability of one

change in each of exactly x of these n subintervals The probability of one change in each subinterval is roughly λ(1n) The probability of two or more changes in each subinterval is essentially 0 The change occurrence or not in each subinterval becomes a Bernoulli trial

Thus for a sequence of n Bernoulli trials with probability p = λnP(X=x) can be approximated by (binomial)

2121

Poisson DistributionPoisson Distribution

2222

ExamplesExamples Ex35-1 X has a Poisson distribution with a mean of λ=5Ex35-1 X has a Poisson distribution with a mean of λ=5

Table III on p 652 lists selected values of the distributionTable III on p 652 lists selected values of the distribution P(Xle6) = 0762P(Xle6) = 0762 P(Xgt5) = 1-P(Xle5) = 1-0616 =0384P(Xgt5) = 1-P(Xle5) = 1-0616 =0384 P(X=6) = P(Xle6)-P(Xle5) = 0762-0616 = 0146P(X=6) = P(Xle6)-P(Xle5) = 0762-0616 = 0146

Ex35-2 The Poisson probability histogramsEx35-2 The Poisson probability histograms

232310

)()(

)(

xx

etxf

tx

9900559

9936192

1)1(

1

1

21

12100559 2

)(

n

iinn

n

iin xxsx

More ExamplesMore Examples Empirical data vs Theoretical formula (Ex35-3)Empirical data vs Theoretical formula (Ex35-3)

X is the number of αparticles emittedX is the number of αparticles emitted by barium-133 in 1 sec and counted by barium-133 in 1 sec and counted by a Geiger counterby a Geiger counter

100 observations are made100 observations are made

Generally with unit interval of Generally with unit interval of length t the Poisson pmf islength t the Poisson pmf is

10486101048610 Ex35-4 Assume a tape flaw is a Poisson distribution with a Ex35-4 Assume a tape flaw is a Poisson distribution with a

mean of λ= 11200 flaw per feetmean of λ= 11200 flaw per feet What is the distribution of X the number of flaws in a 4800-foot rollWhat is the distribution of X the number of flaws in a 4800-foot roll E(X)=48001200=4E(X)=48001200=4 P(X=0) = eP(X=0) = e-4-4= 0018= 0018 By Table III on p 652 P(Xle4) By Table III on p 652 P(Xle4) =0 629=0 629

10

)4()(

)4(

xx

exf

x

2424

8570

)2()3( 210020

3

0

)2(

x

x

x

eXP

Poisson with λ=np can simulate Binomial for large n aPoisson with λ=np can simulate Binomial for large n and small pnd small p (μ σ(μ σ22)=(λ λ) asymp(np npq))=(λ λ) asymp(np npq)

Ex35-6 Bulbs with 2 defective rate Ex35-6 Bulbs with 2 defective rate The probability that a box of 100 bulbs contains at most 3 The probability that a box of 100 bulbs contains at most 3 defdef

ective bulbs isective bulbs is

By Binomial the tedious computation will result inBy Binomial the tedious computation will result in

Ex35-7 p160 the comparisons of Binomial and PoisEx35-7 p160 the comparisons of Binomial and Poisson distributionsson distributions

When Poisson asymp BinomialWhen Poisson asymp Binomial

3

0

100100 8590)980()020()3(x

xxxCXP

2525

xnxqpx

nxXPxf

)()(

Ex35-8 Among a lot of 1000 parts n=100 parts are taken at randoEx35-8 Among a lot of 1000 parts n=100 parts are taken at random wo replacement The lot is accepted if no more than 2 of100 parts m wo replacement The lot is accepted if no more than 2 of100 parts taken are defective Assume p is the defective ratetaken are defective Assume p is the defective rate Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=

2)2)HypergeometricHypergeometric

NN11=1000p=1000pN=1000 N=1000 NN22=N-N=N-N11

Since N is large it makes little difference if sampling is done with or withSince N is large it makes little difference if sampling is done with or without replacement simulated by BinomrArrout replacement simulated by BinomrArr ialial

n Bernoulli trialsn Bernoulli trials For small p p=0~1 simulated by PoissorArrFor small p p=0~1 simulated by PoissorArr nn

OC(001)=092OC(001)=092OC(002)=0677OC(002)=0677OC(003)=0423OC(003)=0423OC(005)=0125OC(005)=0125OC(010)=0003OC(010)=0003

When Poisson asymp Binomial Binomial When Poisson asymp Binomial Binomial asymp Hypergeometric Hypergeometricasymp Hypergeometric Hypergeometric

21 )()(21

21

NxnNxnxxXPxfn

NN

xn

N

x

N

2

0

)100(

)100()2( 100

x

px

x

epXPpnp

22

Def31-1 A function X that assigns to each element s in S exactly one real number X(s)=x is called a random variable The space of X is the set of real numbers x X(s)=x s Sisin If X(s)=s then the space of X is also S Ex31-2 S=123456 in casting a dice Let X(s)=s

P(X=5)=16 P(2leXle5)=P(X=2345=46 P(Xle2)=P(X=12)=26

Two major difficulties In many practical situations the probabilities assigned to the even

ts are unknown Repeated observations (Sampling) to estimate them

There are many ways to define X Which to use Measurement materialization of outcomes

To draw conclusions or make predictions

33

Random Variables of the Discrete Type When S contains a countable number of points

X can be defined to correspond each point in S to a positive integer S is a set of discrete points or a discrete outcome space X is a random variable of discrete type The probability mass function (pmf) f(x) denotes P(X=x)

f(x) is also called probability function probability density function frequency function

Def31-2 f(x) of a discrete random variable X is a function f(x)gt0 x Sisin sumx Sisin f(x) = 1 P(X A) = sumisin x Aisin f(x) where A Ssub

f(x)=0 if x S S is referred to as the notin support of X and the space of X A distribution is uniform if its pmf is constant over the spa

ce For instance f(x)=16 in rolling a fair 6-sided dice Ex31-3 Roll a 4-sided die twice and let X be the largest of two outc

omes S=(i j) i j=14 P(X=1)=P((11))=116 P(X=2)=P((12) (21) (22))=316 hellip f(x)=P(X=x)=(2x-1)16 for x=14 f(x)=0 otherwise

44

Graphing the Distribution (a) For a discrete probability distribution we

simply plot the points (x f(x)) for all x R (b) To get a better picture of the distribution we

use bar graphs and histograms (c) A bar graph is simply a set of lines connecting

(x 0) and (x f(x)) (d) If X takes on only integer values then a

histogram can be used by using rectangles centered at each x R of height f(x) and width one

55

Graphic Representation of f(x) The graph consists of the set of points (x f(x) x Sisin

Better visual appreciation a bar graph a probability histogram P(X=x)=f(x)=(2x-1)16 x=14

Hyper-geometric distribution a collection has N1 objects of type 1 and N2 objects of type 2 X is the number of type-1 objects in the n objects taken from the collection wo replacement

66

Examples Ex31-5 In a pond of 50 fish with 10 tagged 7 fish are caught at random wo

replacement The probability of 2 tagged fish caught is

Ex31-7 In a lot of 25 items with unknown defective 5 items are selected at random wo replacement for exam If no defective found the lot is accepted otherwise rejected Given N1 defective the acceptance probability operating characteristic curve is

Ex31-8 Roll a 4-sided die twice X is the sum 28 f(x)=(4-|x-5|)16 1000 experiments simulated on a computer

77

Mathematical Expectation Def32-1 For a random variable X with pmf f(x) if sumx Sisin u(x)f(x) exists it

is called the mathematical expectation or the expected value of the function u(X) denoted by E[u(X)] It is the weighted mean of u(x) x Sisin The function u(X) is also a random variable say Y with pmf g(y)

Ex32-2 For a random variable X with f(x)=13 x S=-1 0 1isin Let u(X)=X2 Then E[u(X)]= E[X2]=23 The support of the random variable Y=X2 is S1=0 1

and P(Y=0)=P(X=0) P(Y=1)=P(X=-1)+P(X=1) so its pmf Hence sumy S1isin yg(y) = 23 too

Thm32-1 mathematical expectation E if exists satisfies For a constant c E(c)=c For a constant c amp a function u E[c u(X)]=c E[u(X)] For constants a amp b and functions u amp v E[a u(X) + b v(X)]=a E[u(X)] + b

E[v(X)] It can be applied for 2+terms since E is a linear or distributive operator

88

Examples Ex32-3 f(x)=x10 x=1234

Ex32-4 u(x)=(x-b)2 where b is a constant Suppose E[(X-b)2] exists What is the value of b to minimize it

Ex32-5 X has a hypergeometric distribution

99

More Examples Ex32-6 f(x)=x6 x=123

E(X)=μ=1(16)+2(16)+3(16)=73 Var(X)=σ2=E[(X-μ)2]=E(X2)-μ2=hellip=59 σ=0745rArr

Ex32-7 f(x)=13 x=-101 E(X)=μ=-1(13)+0(13)+1(13)=0 Var(X)=σ2=E(X2)-μ2=hellip=23 The standard deviation σ=0816rArr Comparatively g(y)=13 y=-202

Its mean is also zero but Var(Y)=83 and σY=2σ more spread outrArr

Ex32-8 uniform f(x)=1m x=1m E(X)=μ=1(1m)+hellip+m(1m)=(m+1)2 Var(X)=σ2=E(X2)-μ2=hellip=(m2-1)12

For instance m=6 when rolling a 6-sided die

1010

Derived Random Variables Linear Combination X has a mean μX and variance σX

2 Y=aX+b μrArr Y = aμX+b Var(Y)=E[(Y-μY)2] =hellip=a2σX

2 σX=|a|σX a=2 b=0 mean2 variance4 standard deviation2rArr a=1 b=-1 mean-1 variance1 standard deviation1 rArr Var(X-1) = Var(X)

The rth moment of the distribution about b E[(X-b)r] The rth factorial moment E[(X)r]=E[X(X-1)(X-2)hellip(X-r+1)]

E[(X)2] = E[X(X-1)] = E(X2)-E(X) = E(X2)-μE[(X)2]+μ-μ2= E(X2)-μ+μ-μ2= E(X2)-μ2= Var(X)=σ2

Ex32-9 X has a hypergeometric distribution (ref Ex32-5)

1111

Bernoulli Trials A Bernoulli experiment is a random experiment whose outcome can

be classified in one of two mutually exclusive and exhaustive ways success or failure

A series of Bernoulli trials occurs after independent experiments Probabilities of success p and failure q remain the same (p+q=1)

Random variable X follows a Bernoulli distribution X(success)=1 and X(failure)=0 The pmf of X is f(x)=pxq(1-x) x=01 (μ σ2)=(p pq)

A series of n Bernoulli trials a random sample will be an n-tuple of 01rsquos Ex33-4 Plant 5 seeds and observe the outcome (10101) 1st 3rd 5th seed

s germinated If the germination probability is 8 the probability of this outcome is (8)(2)(8)(2)(8) assuming independence

Let X be the number of successes in n trials X follows a binomial distribution denoted as b(n p) The pmf of X is

1212

Example Ex33-5 For lottery with 2 winning if X equals the number of winning

tickets among n=8 purchases The probability of having 2 winning tickets is

Ex33-6 The effect of n and p is illustrated as follows

1313

Cumulative Distribution Function The cumulative probability F(x) defined as P(Xle x) is called the cumulat

ive distribution function or the distribution function

Ex33-7 Assume the distribution of X is b(10 08) F(8) = P(Xle8) = 1-P(X=9)-P(X=10) = 1-10(8)9(2)-(8)10=6242 F(6) = P(Xle6) = sumx=06Cx

10 (8)x(2)10-x

Ex33-9 Y follows b(8 065) If X=8-Y X has b(8 35) whose distribution function is in Table II (p647)

Eg P(Yge6) = P(8-Yle8-6) = P(Xle2)=04278 from table lookup Likewise P(Yle5) = P(8-Yge8-5) = P(Xge3) = 1-P(Xle2) = 1-04278 =0 5722 P(Y=5) = P(X=3) = P(Xle3)-P(Xle2) = 07064-04278 = 02786

The mean and variance of the binomial distribution is (μ σ2)=(np npq) Ex34-2 details the computations

1414

Comparisons Empirical Data vs Analytical Formula

Ex33-11 b(5 5) has μ= np=

25 σ2= npq= 125 Simulate the model for 100 times

2 3 2 hellip = 247 srArr 2=15243

Suppose an urn has N1success balls and N2 failure balls N = N1+ N2

Let p = N1N and X be the number of success balls in a random sample of size n taken from this urn

If the sampling is done one at a time with replacement X follows b(n p) If the sampling is done without replacement X has a hypergeometric distrib

ution with pmf

If N is large and n is relative small it makes little difference if the sampling is done with or without replacement (See Fig33-4)

x

1515

Moment-Generating Function (mgf )

Def34-1 X is a random variable of the discrete type with pmf f(x) and space S If there is a positive integer h st

E(etX) = sumx Sisin etxf(x) exists and is finite for ndashhlttlth thenM(t) = E(etX) is called the moment-generating function of X ndashE(etX) exists and is finite for ndashhlttlth hArrM(r)(t) exist at t=0 r=123hellip ndashUnique association pmf hArrmgf

Sharing the same mgf two random variables have the same distribution of probability

Ex34-1 X has mgf M(t) = et(36)+ e2t(26)+ e3t(16) From eldquoxrdquot Its pmf has f(0)=0 f(1)=36 f(2)=26 f(3)=16 f(4)=0 hellip Therefore f(x)=(4-x)6 x=123

Ex34-2 X has mgf M(t) = et2(1-et2) tltln2 (1-z)-1= 1 + z + z2+ z3+ hellip |z|lt1

1616

Application of Application of mgf

M(t) = E(etX) = sumx Sisin etxf(x) exists and is finite for ndashhlttlthM(t) = sumx Sisin xetxf(x) M(0) = sumx Sisin xf(x) = E(X) M(t) = sumx Sisin x2etxf(x) hellip M(0) = sumx Sisin x2f(x) = E(X2) hellip M(r)(t) = sumx Sisin xretxf(x) M(r)(0) = sumx Sisin xrf(x) = E(Xr) as t=0

M(t) must be formulated (in closed form) to get its derivatives of higher order Ex34-3 X has a binomial distribution b(n p)

Thus its mgf is

When n=1 X has a Bernoulli distribution

1717

Negative Binomial Distribution Let X be the number of Bernoulli trials to observe the rth succ

ess X has a negative binomial distribution

Its pmf g(x) is

If r=1 X has a geometric distribution with

1818

Geometric Distribution X has a geometric distribution with the pmf

P(X gt k) P(X lek)

Memory-less (EX3412)

Ex34-4 Fruit fliesrsquo eyes with frac14white and frac34red The probability of checking at least 4 flies to observe a white eye is

P(Xge4) = P(Xgt3) = (frac34)3= 04219 The probability of checking at most 4 flies to observe a white eye

isP(Xle4) = 1-(frac34)3=06836 The probability of finding the first white eye on the 4th fly checked is

P(X=4) = pq4-1= 01055 lt= P(Xle4) -P(Xle3)

Ex34-4 For a basketball player with 80 free throw X is the minimum number of throws for a total of 10 free throws Its pmf is μ= rp= 1008 = 125 σ2= rqp2= 10(02)(08)2= 3125

1919

mgf rArr pdf By Maclaurinrsquosseries expansion (ref p632)

If the moments of X E(Xr) = M(r)(0) are known M(t) is thus determined

rArr pdf can be obtained by rewriting M(t) as the weighted sum of eldquoxrdquot

Ex34-7 If the moments of X are E(Xr) = 08 r=123hellip Then M(t) can be determined as

Therefore P(X=0)=02 and P(X=1)=08

2020

Poisson Process Def35-1 An approximate Poisson process with parameter λgt0

The numbers of changes occurring in non-overlapping intervals are independent

The probability of exactly one change in a sufficiently short interval of length h is approximately λh

The probability of two or more changes in a sufficiently short interval is essentially zero

Determine pmf During the unit interval of length 1 there are x changes For nraquox we partition the unit interval into n subintervals of length 1n The probability of x changes in the unit interval equivThe probability of one

change in each of exactly x of these n subintervals The probability of one change in each subinterval is roughly λ(1n) The probability of two or more changes in each subinterval is essentially 0 The change occurrence or not in each subinterval becomes a Bernoulli trial

Thus for a sequence of n Bernoulli trials with probability p = λnP(X=x) can be approximated by (binomial)

2121

Poisson DistributionPoisson Distribution

2222

ExamplesExamples Ex35-1 X has a Poisson distribution with a mean of λ=5Ex35-1 X has a Poisson distribution with a mean of λ=5

Table III on p 652 lists selected values of the distributionTable III on p 652 lists selected values of the distribution P(Xle6) = 0762P(Xle6) = 0762 P(Xgt5) = 1-P(Xle5) = 1-0616 =0384P(Xgt5) = 1-P(Xle5) = 1-0616 =0384 P(X=6) = P(Xle6)-P(Xle5) = 0762-0616 = 0146P(X=6) = P(Xle6)-P(Xle5) = 0762-0616 = 0146

Ex35-2 The Poisson probability histogramsEx35-2 The Poisson probability histograms

232310

)()(

)(

xx

etxf

tx

9900559

9936192

1)1(

1

1

21

12100559 2

)(

n

iinn

n

iin xxsx

More ExamplesMore Examples Empirical data vs Theoretical formula (Ex35-3)Empirical data vs Theoretical formula (Ex35-3)

X is the number of αparticles emittedX is the number of αparticles emitted by barium-133 in 1 sec and counted by barium-133 in 1 sec and counted by a Geiger counterby a Geiger counter

100 observations are made100 observations are made

Generally with unit interval of Generally with unit interval of length t the Poisson pmf islength t the Poisson pmf is

10486101048610 Ex35-4 Assume a tape flaw is a Poisson distribution with a Ex35-4 Assume a tape flaw is a Poisson distribution with a

mean of λ= 11200 flaw per feetmean of λ= 11200 flaw per feet What is the distribution of X the number of flaws in a 4800-foot rollWhat is the distribution of X the number of flaws in a 4800-foot roll E(X)=48001200=4E(X)=48001200=4 P(X=0) = eP(X=0) = e-4-4= 0018= 0018 By Table III on p 652 P(Xle4) By Table III on p 652 P(Xle4) =0 629=0 629

10

)4()(

)4(

xx

exf

x

2424

8570

)2()3( 210020

3

0

)2(

x

x

x

eXP

Poisson with λ=np can simulate Binomial for large n aPoisson with λ=np can simulate Binomial for large n and small pnd small p (μ σ(μ σ22)=(λ λ) asymp(np npq))=(λ λ) asymp(np npq)

Ex35-6 Bulbs with 2 defective rate Ex35-6 Bulbs with 2 defective rate The probability that a box of 100 bulbs contains at most 3 The probability that a box of 100 bulbs contains at most 3 defdef

ective bulbs isective bulbs is

By Binomial the tedious computation will result inBy Binomial the tedious computation will result in

Ex35-7 p160 the comparisons of Binomial and PoisEx35-7 p160 the comparisons of Binomial and Poisson distributionsson distributions

When Poisson asymp BinomialWhen Poisson asymp Binomial

3

0

100100 8590)980()020()3(x

xxxCXP

2525

xnxqpx

nxXPxf

)()(

Ex35-8 Among a lot of 1000 parts n=100 parts are taken at randoEx35-8 Among a lot of 1000 parts n=100 parts are taken at random wo replacement The lot is accepted if no more than 2 of100 parts m wo replacement The lot is accepted if no more than 2 of100 parts taken are defective Assume p is the defective ratetaken are defective Assume p is the defective rate Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=

2)2)HypergeometricHypergeometric

NN11=1000p=1000pN=1000 N=1000 NN22=N-N=N-N11

Since N is large it makes little difference if sampling is done with or withSince N is large it makes little difference if sampling is done with or without replacement simulated by BinomrArrout replacement simulated by BinomrArr ialial

n Bernoulli trialsn Bernoulli trials For small p p=0~1 simulated by PoissorArrFor small p p=0~1 simulated by PoissorArr nn

OC(001)=092OC(001)=092OC(002)=0677OC(002)=0677OC(003)=0423OC(003)=0423OC(005)=0125OC(005)=0125OC(010)=0003OC(010)=0003

When Poisson asymp Binomial Binomial When Poisson asymp Binomial Binomial asymp Hypergeometric Hypergeometricasymp Hypergeometric Hypergeometric

21 )()(21

21

NxnNxnxxXPxfn

NN

xn

N

x

N

2

0

)100(

)100()2( 100

x

px

x

epXPpnp

33

Random Variables of the Discrete Type When S contains a countable number of points

X can be defined to correspond each point in S to a positive integer S is a set of discrete points or a discrete outcome space X is a random variable of discrete type The probability mass function (pmf) f(x) denotes P(X=x)

f(x) is also called probability function probability density function frequency function

Def31-2 f(x) of a discrete random variable X is a function f(x)gt0 x Sisin sumx Sisin f(x) = 1 P(X A) = sumisin x Aisin f(x) where A Ssub

f(x)=0 if x S S is referred to as the notin support of X and the space of X A distribution is uniform if its pmf is constant over the spa

ce For instance f(x)=16 in rolling a fair 6-sided dice Ex31-3 Roll a 4-sided die twice and let X be the largest of two outc

omes S=(i j) i j=14 P(X=1)=P((11))=116 P(X=2)=P((12) (21) (22))=316 hellip f(x)=P(X=x)=(2x-1)16 for x=14 f(x)=0 otherwise

44

Graphing the Distribution (a) For a discrete probability distribution we

simply plot the points (x f(x)) for all x R (b) To get a better picture of the distribution we

use bar graphs and histograms (c) A bar graph is simply a set of lines connecting

(x 0) and (x f(x)) (d) If X takes on only integer values then a

histogram can be used by using rectangles centered at each x R of height f(x) and width one

55

Graphic Representation of f(x) The graph consists of the set of points (x f(x) x Sisin

Better visual appreciation a bar graph a probability histogram P(X=x)=f(x)=(2x-1)16 x=14

Hyper-geometric distribution a collection has N1 objects of type 1 and N2 objects of type 2 X is the number of type-1 objects in the n objects taken from the collection wo replacement

66

Examples Ex31-5 In a pond of 50 fish with 10 tagged 7 fish are caught at random wo

replacement The probability of 2 tagged fish caught is

Ex31-7 In a lot of 25 items with unknown defective 5 items are selected at random wo replacement for exam If no defective found the lot is accepted otherwise rejected Given N1 defective the acceptance probability operating characteristic curve is

Ex31-8 Roll a 4-sided die twice X is the sum 28 f(x)=(4-|x-5|)16 1000 experiments simulated on a computer

77

Mathematical Expectation Def32-1 For a random variable X with pmf f(x) if sumx Sisin u(x)f(x) exists it

is called the mathematical expectation or the expected value of the function u(X) denoted by E[u(X)] It is the weighted mean of u(x) x Sisin The function u(X) is also a random variable say Y with pmf g(y)

Ex32-2 For a random variable X with f(x)=13 x S=-1 0 1isin Let u(X)=X2 Then E[u(X)]= E[X2]=23 The support of the random variable Y=X2 is S1=0 1

and P(Y=0)=P(X=0) P(Y=1)=P(X=-1)+P(X=1) so its pmf Hence sumy S1isin yg(y) = 23 too

Thm32-1 mathematical expectation E if exists satisfies For a constant c E(c)=c For a constant c amp a function u E[c u(X)]=c E[u(X)] For constants a amp b and functions u amp v E[a u(X) + b v(X)]=a E[u(X)] + b

E[v(X)] It can be applied for 2+terms since E is a linear or distributive operator

88

Examples Ex32-3 f(x)=x10 x=1234

Ex32-4 u(x)=(x-b)2 where b is a constant Suppose E[(X-b)2] exists What is the value of b to minimize it

Ex32-5 X has a hypergeometric distribution

99

More Examples Ex32-6 f(x)=x6 x=123

E(X)=μ=1(16)+2(16)+3(16)=73 Var(X)=σ2=E[(X-μ)2]=E(X2)-μ2=hellip=59 σ=0745rArr

Ex32-7 f(x)=13 x=-101 E(X)=μ=-1(13)+0(13)+1(13)=0 Var(X)=σ2=E(X2)-μ2=hellip=23 The standard deviation σ=0816rArr Comparatively g(y)=13 y=-202

Its mean is also zero but Var(Y)=83 and σY=2σ more spread outrArr

Ex32-8 uniform f(x)=1m x=1m E(X)=μ=1(1m)+hellip+m(1m)=(m+1)2 Var(X)=σ2=E(X2)-μ2=hellip=(m2-1)12

For instance m=6 when rolling a 6-sided die

1010

Derived Random Variables Linear Combination X has a mean μX and variance σX

2 Y=aX+b μrArr Y = aμX+b Var(Y)=E[(Y-μY)2] =hellip=a2σX

2 σX=|a|σX a=2 b=0 mean2 variance4 standard deviation2rArr a=1 b=-1 mean-1 variance1 standard deviation1 rArr Var(X-1) = Var(X)

The rth moment of the distribution about b E[(X-b)r] The rth factorial moment E[(X)r]=E[X(X-1)(X-2)hellip(X-r+1)]

E[(X)2] = E[X(X-1)] = E(X2)-E(X) = E(X2)-μE[(X)2]+μ-μ2= E(X2)-μ+μ-μ2= E(X2)-μ2= Var(X)=σ2

Ex32-9 X has a hypergeometric distribution (ref Ex32-5)

1111

Bernoulli Trials A Bernoulli experiment is a random experiment whose outcome can

be classified in one of two mutually exclusive and exhaustive ways success or failure

A series of Bernoulli trials occurs after independent experiments Probabilities of success p and failure q remain the same (p+q=1)

Random variable X follows a Bernoulli distribution X(success)=1 and X(failure)=0 The pmf of X is f(x)=pxq(1-x) x=01 (μ σ2)=(p pq)

A series of n Bernoulli trials a random sample will be an n-tuple of 01rsquos Ex33-4 Plant 5 seeds and observe the outcome (10101) 1st 3rd 5th seed

s germinated If the germination probability is 8 the probability of this outcome is (8)(2)(8)(2)(8) assuming independence

Let X be the number of successes in n trials X follows a binomial distribution denoted as b(n p) The pmf of X is

1212

Example Ex33-5 For lottery with 2 winning if X equals the number of winning

tickets among n=8 purchases The probability of having 2 winning tickets is

Ex33-6 The effect of n and p is illustrated as follows

1313

Cumulative Distribution Function The cumulative probability F(x) defined as P(Xle x) is called the cumulat

ive distribution function or the distribution function

Ex33-7 Assume the distribution of X is b(10 08) F(8) = P(Xle8) = 1-P(X=9)-P(X=10) = 1-10(8)9(2)-(8)10=6242 F(6) = P(Xle6) = sumx=06Cx

10 (8)x(2)10-x

Ex33-9 Y follows b(8 065) If X=8-Y X has b(8 35) whose distribution function is in Table II (p647)

Eg P(Yge6) = P(8-Yle8-6) = P(Xle2)=04278 from table lookup Likewise P(Yle5) = P(8-Yge8-5) = P(Xge3) = 1-P(Xle2) = 1-04278 =0 5722 P(Y=5) = P(X=3) = P(Xle3)-P(Xle2) = 07064-04278 = 02786

The mean and variance of the binomial distribution is (μ σ2)=(np npq) Ex34-2 details the computations

1414

Comparisons Empirical Data vs Analytical Formula

Ex33-11 b(5 5) has μ= np=

25 σ2= npq= 125 Simulate the model for 100 times

2 3 2 hellip = 247 srArr 2=15243

Suppose an urn has N1success balls and N2 failure balls N = N1+ N2

Let p = N1N and X be the number of success balls in a random sample of size n taken from this urn

If the sampling is done one at a time with replacement X follows b(n p) If the sampling is done without replacement X has a hypergeometric distrib

ution with pmf

If N is large and n is relative small it makes little difference if the sampling is done with or without replacement (See Fig33-4)

x

1515

Moment-Generating Function (mgf )

Def34-1 X is a random variable of the discrete type with pmf f(x) and space S If there is a positive integer h st

E(etX) = sumx Sisin etxf(x) exists and is finite for ndashhlttlth thenM(t) = E(etX) is called the moment-generating function of X ndashE(etX) exists and is finite for ndashhlttlth hArrM(r)(t) exist at t=0 r=123hellip ndashUnique association pmf hArrmgf

Sharing the same mgf two random variables have the same distribution of probability

Ex34-1 X has mgf M(t) = et(36)+ e2t(26)+ e3t(16) From eldquoxrdquot Its pmf has f(0)=0 f(1)=36 f(2)=26 f(3)=16 f(4)=0 hellip Therefore f(x)=(4-x)6 x=123

Ex34-2 X has mgf M(t) = et2(1-et2) tltln2 (1-z)-1= 1 + z + z2+ z3+ hellip |z|lt1

1616

Application of Application of mgf

M(t) = E(etX) = sumx Sisin etxf(x) exists and is finite for ndashhlttlthM(t) = sumx Sisin xetxf(x) M(0) = sumx Sisin xf(x) = E(X) M(t) = sumx Sisin x2etxf(x) hellip M(0) = sumx Sisin x2f(x) = E(X2) hellip M(r)(t) = sumx Sisin xretxf(x) M(r)(0) = sumx Sisin xrf(x) = E(Xr) as t=0

M(t) must be formulated (in closed form) to get its derivatives of higher order Ex34-3 X has a binomial distribution b(n p)

Thus its mgf is

When n=1 X has a Bernoulli distribution

1717

Negative Binomial Distribution Let X be the number of Bernoulli trials to observe the rth succ

ess X has a negative binomial distribution

Its pmf g(x) is

If r=1 X has a geometric distribution with

1818

Geometric Distribution X has a geometric distribution with the pmf

P(X gt k) P(X lek)

Memory-less (EX3412)

Ex34-4 Fruit fliesrsquo eyes with frac14white and frac34red The probability of checking at least 4 flies to observe a white eye is

P(Xge4) = P(Xgt3) = (frac34)3= 04219 The probability of checking at most 4 flies to observe a white eye

isP(Xle4) = 1-(frac34)3=06836 The probability of finding the first white eye on the 4th fly checked is

P(X=4) = pq4-1= 01055 lt= P(Xle4) -P(Xle3)

Ex34-4 For a basketball player with 80 free throw X is the minimum number of throws for a total of 10 free throws Its pmf is μ= rp= 1008 = 125 σ2= rqp2= 10(02)(08)2= 3125

1919

mgf rArr pdf By Maclaurinrsquosseries expansion (ref p632)

If the moments of X E(Xr) = M(r)(0) are known M(t) is thus determined

rArr pdf can be obtained by rewriting M(t) as the weighted sum of eldquoxrdquot

Ex34-7 If the moments of X are E(Xr) = 08 r=123hellip Then M(t) can be determined as

Therefore P(X=0)=02 and P(X=1)=08

2020

Poisson Process Def35-1 An approximate Poisson process with parameter λgt0

The numbers of changes occurring in non-overlapping intervals are independent

The probability of exactly one change in a sufficiently short interval of length h is approximately λh

The probability of two or more changes in a sufficiently short interval is essentially zero

Determine pmf During the unit interval of length 1 there are x changes For nraquox we partition the unit interval into n subintervals of length 1n The probability of x changes in the unit interval equivThe probability of one

change in each of exactly x of these n subintervals The probability of one change in each subinterval is roughly λ(1n) The probability of two or more changes in each subinterval is essentially 0 The change occurrence or not in each subinterval becomes a Bernoulli trial

Thus for a sequence of n Bernoulli trials with probability p = λnP(X=x) can be approximated by (binomial)

2121

Poisson DistributionPoisson Distribution

2222

ExamplesExamples Ex35-1 X has a Poisson distribution with a mean of λ=5Ex35-1 X has a Poisson distribution with a mean of λ=5

Table III on p 652 lists selected values of the distributionTable III on p 652 lists selected values of the distribution P(Xle6) = 0762P(Xle6) = 0762 P(Xgt5) = 1-P(Xle5) = 1-0616 =0384P(Xgt5) = 1-P(Xle5) = 1-0616 =0384 P(X=6) = P(Xle6)-P(Xle5) = 0762-0616 = 0146P(X=6) = P(Xle6)-P(Xle5) = 0762-0616 = 0146

Ex35-2 The Poisson probability histogramsEx35-2 The Poisson probability histograms

232310

)()(

)(

xx

etxf

tx

9900559

9936192

1)1(

1

1

21

12100559 2

)(

n

iinn

n

iin xxsx

More ExamplesMore Examples Empirical data vs Theoretical formula (Ex35-3)Empirical data vs Theoretical formula (Ex35-3)

X is the number of αparticles emittedX is the number of αparticles emitted by barium-133 in 1 sec and counted by barium-133 in 1 sec and counted by a Geiger counterby a Geiger counter

100 observations are made100 observations are made

Generally with unit interval of Generally with unit interval of length t the Poisson pmf islength t the Poisson pmf is

10486101048610 Ex35-4 Assume a tape flaw is a Poisson distribution with a Ex35-4 Assume a tape flaw is a Poisson distribution with a

mean of λ= 11200 flaw per feetmean of λ= 11200 flaw per feet What is the distribution of X the number of flaws in a 4800-foot rollWhat is the distribution of X the number of flaws in a 4800-foot roll E(X)=48001200=4E(X)=48001200=4 P(X=0) = eP(X=0) = e-4-4= 0018= 0018 By Table III on p 652 P(Xle4) By Table III on p 652 P(Xle4) =0 629=0 629

10

)4()(

)4(

xx

exf

x

2424

8570

)2()3( 210020

3

0

)2(

x

x

x

eXP

Poisson with λ=np can simulate Binomial for large n aPoisson with λ=np can simulate Binomial for large n and small pnd small p (μ σ(μ σ22)=(λ λ) asymp(np npq))=(λ λ) asymp(np npq)

Ex35-6 Bulbs with 2 defective rate Ex35-6 Bulbs with 2 defective rate The probability that a box of 100 bulbs contains at most 3 The probability that a box of 100 bulbs contains at most 3 defdef

ective bulbs isective bulbs is

By Binomial the tedious computation will result inBy Binomial the tedious computation will result in

Ex35-7 p160 the comparisons of Binomial and PoisEx35-7 p160 the comparisons of Binomial and Poisson distributionsson distributions

When Poisson asymp BinomialWhen Poisson asymp Binomial

3

0

100100 8590)980()020()3(x

xxxCXP

2525

xnxqpx

nxXPxf

)()(

Ex35-8 Among a lot of 1000 parts n=100 parts are taken at randoEx35-8 Among a lot of 1000 parts n=100 parts are taken at random wo replacement The lot is accepted if no more than 2 of100 parts m wo replacement The lot is accepted if no more than 2 of100 parts taken are defective Assume p is the defective ratetaken are defective Assume p is the defective rate Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=

2)2)HypergeometricHypergeometric

NN11=1000p=1000pN=1000 N=1000 NN22=N-N=N-N11

Since N is large it makes little difference if sampling is done with or withSince N is large it makes little difference if sampling is done with or without replacement simulated by BinomrArrout replacement simulated by BinomrArr ialial

n Bernoulli trialsn Bernoulli trials For small p p=0~1 simulated by PoissorArrFor small p p=0~1 simulated by PoissorArr nn

OC(001)=092OC(001)=092OC(002)=0677OC(002)=0677OC(003)=0423OC(003)=0423OC(005)=0125OC(005)=0125OC(010)=0003OC(010)=0003

When Poisson asymp Binomial Binomial When Poisson asymp Binomial Binomial asymp Hypergeometric Hypergeometricasymp Hypergeometric Hypergeometric

21 )()(21

21

NxnNxnxxXPxfn

NN

xn

N

x

N

2

0

)100(

)100()2( 100

x

px

x

epXPpnp

44

Graphing the Distribution (a) For a discrete probability distribution we

simply plot the points (x f(x)) for all x R (b) To get a better picture of the distribution we

use bar graphs and histograms (c) A bar graph is simply a set of lines connecting

(x 0) and (x f(x)) (d) If X takes on only integer values then a

histogram can be used by using rectangles centered at each x R of height f(x) and width one

55

Graphic Representation of f(x) The graph consists of the set of points (x f(x) x Sisin

Better visual appreciation a bar graph a probability histogram P(X=x)=f(x)=(2x-1)16 x=14

Hyper-geometric distribution a collection has N1 objects of type 1 and N2 objects of type 2 X is the number of type-1 objects in the n objects taken from the collection wo replacement

66

Examples Ex31-5 In a pond of 50 fish with 10 tagged 7 fish are caught at random wo

replacement The probability of 2 tagged fish caught is

Ex31-7 In a lot of 25 items with unknown defective 5 items are selected at random wo replacement for exam If no defective found the lot is accepted otherwise rejected Given N1 defective the acceptance probability operating characteristic curve is

Ex31-8 Roll a 4-sided die twice X is the sum 28 f(x)=(4-|x-5|)16 1000 experiments simulated on a computer

77

Mathematical Expectation Def32-1 For a random variable X with pmf f(x) if sumx Sisin u(x)f(x) exists it

is called the mathematical expectation or the expected value of the function u(X) denoted by E[u(X)] It is the weighted mean of u(x) x Sisin The function u(X) is also a random variable say Y with pmf g(y)

Ex32-2 For a random variable X with f(x)=13 x S=-1 0 1isin Let u(X)=X2 Then E[u(X)]= E[X2]=23 The support of the random variable Y=X2 is S1=0 1

and P(Y=0)=P(X=0) P(Y=1)=P(X=-1)+P(X=1) so its pmf Hence sumy S1isin yg(y) = 23 too

Thm32-1 mathematical expectation E if exists satisfies For a constant c E(c)=c For a constant c amp a function u E[c u(X)]=c E[u(X)] For constants a amp b and functions u amp v E[a u(X) + b v(X)]=a E[u(X)] + b

E[v(X)] It can be applied for 2+terms since E is a linear or distributive operator

88

Examples Ex32-3 f(x)=x10 x=1234

Ex32-4 u(x)=(x-b)2 where b is a constant Suppose E[(X-b)2] exists What is the value of b to minimize it

Ex32-5 X has a hypergeometric distribution

99

More Examples Ex32-6 f(x)=x6 x=123

E(X)=μ=1(16)+2(16)+3(16)=73 Var(X)=σ2=E[(X-μ)2]=E(X2)-μ2=hellip=59 σ=0745rArr

Ex32-7 f(x)=13 x=-101 E(X)=μ=-1(13)+0(13)+1(13)=0 Var(X)=σ2=E(X2)-μ2=hellip=23 The standard deviation σ=0816rArr Comparatively g(y)=13 y=-202

Its mean is also zero but Var(Y)=83 and σY=2σ more spread outrArr

Ex32-8 uniform f(x)=1m x=1m E(X)=μ=1(1m)+hellip+m(1m)=(m+1)2 Var(X)=σ2=E(X2)-μ2=hellip=(m2-1)12

For instance m=6 when rolling a 6-sided die

1010

Derived Random Variables Linear Combination X has a mean μX and variance σX

2 Y=aX+b μrArr Y = aμX+b Var(Y)=E[(Y-μY)2] =hellip=a2σX

2 σX=|a|σX a=2 b=0 mean2 variance4 standard deviation2rArr a=1 b=-1 mean-1 variance1 standard deviation1 rArr Var(X-1) = Var(X)

The rth moment of the distribution about b E[(X-b)r] The rth factorial moment E[(X)r]=E[X(X-1)(X-2)hellip(X-r+1)]

E[(X)2] = E[X(X-1)] = E(X2)-E(X) = E(X2)-μE[(X)2]+μ-μ2= E(X2)-μ+μ-μ2= E(X2)-μ2= Var(X)=σ2

Ex32-9 X has a hypergeometric distribution (ref Ex32-5)

1111

Bernoulli Trials A Bernoulli experiment is a random experiment whose outcome can

be classified in one of two mutually exclusive and exhaustive ways success or failure

A series of Bernoulli trials occurs after independent experiments Probabilities of success p and failure q remain the same (p+q=1)

Random variable X follows a Bernoulli distribution X(success)=1 and X(failure)=0 The pmf of X is f(x)=pxq(1-x) x=01 (μ σ2)=(p pq)

A series of n Bernoulli trials a random sample will be an n-tuple of 01rsquos Ex33-4 Plant 5 seeds and observe the outcome (10101) 1st 3rd 5th seed

s germinated If the germination probability is 8 the probability of this outcome is (8)(2)(8)(2)(8) assuming independence

Let X be the number of successes in n trials X follows a binomial distribution denoted as b(n p) The pmf of X is

1212

Example Ex33-5 For lottery with 2 winning if X equals the number of winning

tickets among n=8 purchases The probability of having 2 winning tickets is

Ex33-6 The effect of n and p is illustrated as follows

1313

Cumulative Distribution Function The cumulative probability F(x) defined as P(Xle x) is called the cumulat

ive distribution function or the distribution function

Ex33-7 Assume the distribution of X is b(10 08) F(8) = P(Xle8) = 1-P(X=9)-P(X=10) = 1-10(8)9(2)-(8)10=6242 F(6) = P(Xle6) = sumx=06Cx

10 (8)x(2)10-x

Ex33-9 Y follows b(8 065) If X=8-Y X has b(8 35) whose distribution function is in Table II (p647)

Eg P(Yge6) = P(8-Yle8-6) = P(Xle2)=04278 from table lookup Likewise P(Yle5) = P(8-Yge8-5) = P(Xge3) = 1-P(Xle2) = 1-04278 =0 5722 P(Y=5) = P(X=3) = P(Xle3)-P(Xle2) = 07064-04278 = 02786

The mean and variance of the binomial distribution is (μ σ2)=(np npq) Ex34-2 details the computations

1414

Comparisons Empirical Data vs Analytical Formula

Ex33-11 b(5 5) has μ= np=

25 σ2= npq= 125 Simulate the model for 100 times

2 3 2 hellip = 247 srArr 2=15243

Suppose an urn has N1success balls and N2 failure balls N = N1+ N2

Let p = N1N and X be the number of success balls in a random sample of size n taken from this urn

If the sampling is done one at a time with replacement X follows b(n p) If the sampling is done without replacement X has a hypergeometric distrib

ution with pmf

If N is large and n is relative small it makes little difference if the sampling is done with or without replacement (See Fig33-4)

x

1515

Moment-Generating Function (mgf )

Def34-1 X is a random variable of the discrete type with pmf f(x) and space S If there is a positive integer h st

E(etX) = sumx Sisin etxf(x) exists and is finite for ndashhlttlth thenM(t) = E(etX) is called the moment-generating function of X ndashE(etX) exists and is finite for ndashhlttlth hArrM(r)(t) exist at t=0 r=123hellip ndashUnique association pmf hArrmgf

Sharing the same mgf two random variables have the same distribution of probability

Ex34-1 X has mgf M(t) = et(36)+ e2t(26)+ e3t(16) From eldquoxrdquot Its pmf has f(0)=0 f(1)=36 f(2)=26 f(3)=16 f(4)=0 hellip Therefore f(x)=(4-x)6 x=123

Ex34-2 X has mgf M(t) = et2(1-et2) tltln2 (1-z)-1= 1 + z + z2+ z3+ hellip |z|lt1

1616

Application of Application of mgf

M(t) = E(etX) = sumx Sisin etxf(x) exists and is finite for ndashhlttlthM(t) = sumx Sisin xetxf(x) M(0) = sumx Sisin xf(x) = E(X) M(t) = sumx Sisin x2etxf(x) hellip M(0) = sumx Sisin x2f(x) = E(X2) hellip M(r)(t) = sumx Sisin xretxf(x) M(r)(0) = sumx Sisin xrf(x) = E(Xr) as t=0

M(t) must be formulated (in closed form) to get its derivatives of higher order Ex34-3 X has a binomial distribution b(n p)

Thus its mgf is

When n=1 X has a Bernoulli distribution

1717

Negative Binomial Distribution Let X be the number of Bernoulli trials to observe the rth succ

ess X has a negative binomial distribution

Its pmf g(x) is

If r=1 X has a geometric distribution with

1818

Geometric Distribution X has a geometric distribution with the pmf

P(X gt k) P(X lek)

Memory-less (EX3412)

Ex34-4 Fruit fliesrsquo eyes with frac14white and frac34red The probability of checking at least 4 flies to observe a white eye is

P(Xge4) = P(Xgt3) = (frac34)3= 04219 The probability of checking at most 4 flies to observe a white eye

isP(Xle4) = 1-(frac34)3=06836 The probability of finding the first white eye on the 4th fly checked is

P(X=4) = pq4-1= 01055 lt= P(Xle4) -P(Xle3)

Ex34-4 For a basketball player with 80 free throw X is the minimum number of throws for a total of 10 free throws Its pmf is μ= rp= 1008 = 125 σ2= rqp2= 10(02)(08)2= 3125

1919

mgf rArr pdf By Maclaurinrsquosseries expansion (ref p632)

If the moments of X E(Xr) = M(r)(0) are known M(t) is thus determined

rArr pdf can be obtained by rewriting M(t) as the weighted sum of eldquoxrdquot

Ex34-7 If the moments of X are E(Xr) = 08 r=123hellip Then M(t) can be determined as

Therefore P(X=0)=02 and P(X=1)=08

2020

Poisson Process Def35-1 An approximate Poisson process with parameter λgt0

The numbers of changes occurring in non-overlapping intervals are independent

The probability of exactly one change in a sufficiently short interval of length h is approximately λh

The probability of two or more changes in a sufficiently short interval is essentially zero

Determine pmf During the unit interval of length 1 there are x changes For nraquox we partition the unit interval into n subintervals of length 1n The probability of x changes in the unit interval equivThe probability of one

change in each of exactly x of these n subintervals The probability of one change in each subinterval is roughly λ(1n) The probability of two or more changes in each subinterval is essentially 0 The change occurrence or not in each subinterval becomes a Bernoulli trial

Thus for a sequence of n Bernoulli trials with probability p = λnP(X=x) can be approximated by (binomial)

2121

Poisson DistributionPoisson Distribution

2222

ExamplesExamples Ex35-1 X has a Poisson distribution with a mean of λ=5Ex35-1 X has a Poisson distribution with a mean of λ=5

Table III on p 652 lists selected values of the distributionTable III on p 652 lists selected values of the distribution P(Xle6) = 0762P(Xle6) = 0762 P(Xgt5) = 1-P(Xle5) = 1-0616 =0384P(Xgt5) = 1-P(Xle5) = 1-0616 =0384 P(X=6) = P(Xle6)-P(Xle5) = 0762-0616 = 0146P(X=6) = P(Xle6)-P(Xle5) = 0762-0616 = 0146

Ex35-2 The Poisson probability histogramsEx35-2 The Poisson probability histograms

232310

)()(

)(

xx

etxf

tx

9900559

9936192

1)1(

1

1

21

12100559 2

)(

n

iinn

n

iin xxsx

More ExamplesMore Examples Empirical data vs Theoretical formula (Ex35-3)Empirical data vs Theoretical formula (Ex35-3)

X is the number of αparticles emittedX is the number of αparticles emitted by barium-133 in 1 sec and counted by barium-133 in 1 sec and counted by a Geiger counterby a Geiger counter

100 observations are made100 observations are made

Generally with unit interval of Generally with unit interval of length t the Poisson pmf islength t the Poisson pmf is

10486101048610 Ex35-4 Assume a tape flaw is a Poisson distribution with a Ex35-4 Assume a tape flaw is a Poisson distribution with a

mean of λ= 11200 flaw per feetmean of λ= 11200 flaw per feet What is the distribution of X the number of flaws in a 4800-foot rollWhat is the distribution of X the number of flaws in a 4800-foot roll E(X)=48001200=4E(X)=48001200=4 P(X=0) = eP(X=0) = e-4-4= 0018= 0018 By Table III on p 652 P(Xle4) By Table III on p 652 P(Xle4) =0 629=0 629

10

)4()(

)4(

xx

exf

x

2424

8570

)2()3( 210020

3

0

)2(

x

x

x

eXP

Poisson with λ=np can simulate Binomial for large n aPoisson with λ=np can simulate Binomial for large n and small pnd small p (μ σ(μ σ22)=(λ λ) asymp(np npq))=(λ λ) asymp(np npq)

Ex35-6 Bulbs with 2 defective rate Ex35-6 Bulbs with 2 defective rate The probability that a box of 100 bulbs contains at most 3 The probability that a box of 100 bulbs contains at most 3 defdef

ective bulbs isective bulbs is

By Binomial the tedious computation will result inBy Binomial the tedious computation will result in

Ex35-7 p160 the comparisons of Binomial and PoisEx35-7 p160 the comparisons of Binomial and Poisson distributionsson distributions

When Poisson asymp BinomialWhen Poisson asymp Binomial

3

0

100100 8590)980()020()3(x

xxxCXP

2525

xnxqpx

nxXPxf

)()(

Ex35-8 Among a lot of 1000 parts n=100 parts are taken at randoEx35-8 Among a lot of 1000 parts n=100 parts are taken at random wo replacement The lot is accepted if no more than 2 of100 parts m wo replacement The lot is accepted if no more than 2 of100 parts taken are defective Assume p is the defective ratetaken are defective Assume p is the defective rate Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=

2)2)HypergeometricHypergeometric

NN11=1000p=1000pN=1000 N=1000 NN22=N-N=N-N11

Since N is large it makes little difference if sampling is done with or withSince N is large it makes little difference if sampling is done with or without replacement simulated by BinomrArrout replacement simulated by BinomrArr ialial

n Bernoulli trialsn Bernoulli trials For small p p=0~1 simulated by PoissorArrFor small p p=0~1 simulated by PoissorArr nn

OC(001)=092OC(001)=092OC(002)=0677OC(002)=0677OC(003)=0423OC(003)=0423OC(005)=0125OC(005)=0125OC(010)=0003OC(010)=0003

When Poisson asymp Binomial Binomial When Poisson asymp Binomial Binomial asymp Hypergeometric Hypergeometricasymp Hypergeometric Hypergeometric

21 )()(21

21

NxnNxnxxXPxfn

NN

xn

N

x

N

2

0

)100(

)100()2( 100

x

px

x

epXPpnp

55

Graphic Representation of f(x) The graph consists of the set of points (x f(x) x Sisin

Better visual appreciation a bar graph a probability histogram P(X=x)=f(x)=(2x-1)16 x=14

Hyper-geometric distribution a collection has N1 objects of type 1 and N2 objects of type 2 X is the number of type-1 objects in the n objects taken from the collection wo replacement

66

Examples Ex31-5 In a pond of 50 fish with 10 tagged 7 fish are caught at random wo

replacement The probability of 2 tagged fish caught is

Ex31-7 In a lot of 25 items with unknown defective 5 items are selected at random wo replacement for exam If no defective found the lot is accepted otherwise rejected Given N1 defective the acceptance probability operating characteristic curve is

Ex31-8 Roll a 4-sided die twice X is the sum 28 f(x)=(4-|x-5|)16 1000 experiments simulated on a computer

77

Mathematical Expectation Def32-1 For a random variable X with pmf f(x) if sumx Sisin u(x)f(x) exists it

is called the mathematical expectation or the expected value of the function u(X) denoted by E[u(X)] It is the weighted mean of u(x) x Sisin The function u(X) is also a random variable say Y with pmf g(y)

Ex32-2 For a random variable X with f(x)=13 x S=-1 0 1isin Let u(X)=X2 Then E[u(X)]= E[X2]=23 The support of the random variable Y=X2 is S1=0 1

and P(Y=0)=P(X=0) P(Y=1)=P(X=-1)+P(X=1) so its pmf Hence sumy S1isin yg(y) = 23 too

Thm32-1 mathematical expectation E if exists satisfies For a constant c E(c)=c For a constant c amp a function u E[c u(X)]=c E[u(X)] For constants a amp b and functions u amp v E[a u(X) + b v(X)]=a E[u(X)] + b

E[v(X)] It can be applied for 2+terms since E is a linear or distributive operator

88

Examples Ex32-3 f(x)=x10 x=1234

Ex32-4 u(x)=(x-b)2 where b is a constant Suppose E[(X-b)2] exists What is the value of b to minimize it

Ex32-5 X has a hypergeometric distribution

99

More Examples Ex32-6 f(x)=x6 x=123

E(X)=μ=1(16)+2(16)+3(16)=73 Var(X)=σ2=E[(X-μ)2]=E(X2)-μ2=hellip=59 σ=0745rArr

Ex32-7 f(x)=13 x=-101 E(X)=μ=-1(13)+0(13)+1(13)=0 Var(X)=σ2=E(X2)-μ2=hellip=23 The standard deviation σ=0816rArr Comparatively g(y)=13 y=-202

Its mean is also zero but Var(Y)=83 and σY=2σ more spread outrArr

Ex32-8 uniform f(x)=1m x=1m E(X)=μ=1(1m)+hellip+m(1m)=(m+1)2 Var(X)=σ2=E(X2)-μ2=hellip=(m2-1)12

For instance m=6 when rolling a 6-sided die

1010

Derived Random Variables Linear Combination X has a mean μX and variance σX

2 Y=aX+b μrArr Y = aμX+b Var(Y)=E[(Y-μY)2] =hellip=a2σX

2 σX=|a|σX a=2 b=0 mean2 variance4 standard deviation2rArr a=1 b=-1 mean-1 variance1 standard deviation1 rArr Var(X-1) = Var(X)

The rth moment of the distribution about b E[(X-b)r] The rth factorial moment E[(X)r]=E[X(X-1)(X-2)hellip(X-r+1)]

E[(X)2] = E[X(X-1)] = E(X2)-E(X) = E(X2)-μE[(X)2]+μ-μ2= E(X2)-μ+μ-μ2= E(X2)-μ2= Var(X)=σ2

Ex32-9 X has a hypergeometric distribution (ref Ex32-5)

1111

Bernoulli Trials A Bernoulli experiment is a random experiment whose outcome can

be classified in one of two mutually exclusive and exhaustive ways success or failure

A series of Bernoulli trials occurs after independent experiments Probabilities of success p and failure q remain the same (p+q=1)

Random variable X follows a Bernoulli distribution X(success)=1 and X(failure)=0 The pmf of X is f(x)=pxq(1-x) x=01 (μ σ2)=(p pq)

A series of n Bernoulli trials a random sample will be an n-tuple of 01rsquos Ex33-4 Plant 5 seeds and observe the outcome (10101) 1st 3rd 5th seed

s germinated If the germination probability is 8 the probability of this outcome is (8)(2)(8)(2)(8) assuming independence

Let X be the number of successes in n trials X follows a binomial distribution denoted as b(n p) The pmf of X is

1212

Example Ex33-5 For lottery with 2 winning if X equals the number of winning

tickets among n=8 purchases The probability of having 2 winning tickets is

Ex33-6 The effect of n and p is illustrated as follows

1313

Cumulative Distribution Function The cumulative probability F(x) defined as P(Xle x) is called the cumulat

ive distribution function or the distribution function

Ex33-7 Assume the distribution of X is b(10 08) F(8) = P(Xle8) = 1-P(X=9)-P(X=10) = 1-10(8)9(2)-(8)10=6242 F(6) = P(Xle6) = sumx=06Cx

10 (8)x(2)10-x

Ex33-9 Y follows b(8 065) If X=8-Y X has b(8 35) whose distribution function is in Table II (p647)

Eg P(Yge6) = P(8-Yle8-6) = P(Xle2)=04278 from table lookup Likewise P(Yle5) = P(8-Yge8-5) = P(Xge3) = 1-P(Xle2) = 1-04278 =0 5722 P(Y=5) = P(X=3) = P(Xle3)-P(Xle2) = 07064-04278 = 02786

The mean and variance of the binomial distribution is (μ σ2)=(np npq) Ex34-2 details the computations

1414

Comparisons Empirical Data vs Analytical Formula

Ex33-11 b(5 5) has μ= np=

25 σ2= npq= 125 Simulate the model for 100 times

2 3 2 hellip = 247 srArr 2=15243

Suppose an urn has N1success balls and N2 failure balls N = N1+ N2

Let p = N1N and X be the number of success balls in a random sample of size n taken from this urn

If the sampling is done one at a time with replacement X follows b(n p) If the sampling is done without replacement X has a hypergeometric distrib

ution with pmf

If N is large and n is relative small it makes little difference if the sampling is done with or without replacement (See Fig33-4)

x

1515

Moment-Generating Function (mgf )

Def34-1 X is a random variable of the discrete type with pmf f(x) and space S If there is a positive integer h st

E(etX) = sumx Sisin etxf(x) exists and is finite for ndashhlttlth thenM(t) = E(etX) is called the moment-generating function of X ndashE(etX) exists and is finite for ndashhlttlth hArrM(r)(t) exist at t=0 r=123hellip ndashUnique association pmf hArrmgf

Sharing the same mgf two random variables have the same distribution of probability

Ex34-1 X has mgf M(t) = et(36)+ e2t(26)+ e3t(16) From eldquoxrdquot Its pmf has f(0)=0 f(1)=36 f(2)=26 f(3)=16 f(4)=0 hellip Therefore f(x)=(4-x)6 x=123

Ex34-2 X has mgf M(t) = et2(1-et2) tltln2 (1-z)-1= 1 + z + z2+ z3+ hellip |z|lt1

1616

Application of Application of mgf

M(t) = E(etX) = sumx Sisin etxf(x) exists and is finite for ndashhlttlthM(t) = sumx Sisin xetxf(x) M(0) = sumx Sisin xf(x) = E(X) M(t) = sumx Sisin x2etxf(x) hellip M(0) = sumx Sisin x2f(x) = E(X2) hellip M(r)(t) = sumx Sisin xretxf(x) M(r)(0) = sumx Sisin xrf(x) = E(Xr) as t=0

M(t) must be formulated (in closed form) to get its derivatives of higher order Ex34-3 X has a binomial distribution b(n p)

Thus its mgf is

When n=1 X has a Bernoulli distribution

1717

Negative Binomial Distribution Let X be the number of Bernoulli trials to observe the rth succ

ess X has a negative binomial distribution

Its pmf g(x) is

If r=1 X has a geometric distribution with

1818

Geometric Distribution X has a geometric distribution with the pmf

P(X gt k) P(X lek)

Memory-less (EX3412)

Ex34-4 Fruit fliesrsquo eyes with frac14white and frac34red The probability of checking at least 4 flies to observe a white eye is

P(Xge4) = P(Xgt3) = (frac34)3= 04219 The probability of checking at most 4 flies to observe a white eye

isP(Xle4) = 1-(frac34)3=06836 The probability of finding the first white eye on the 4th fly checked is

P(X=4) = pq4-1= 01055 lt= P(Xle4) -P(Xle3)

Ex34-4 For a basketball player with 80 free throw X is the minimum number of throws for a total of 10 free throws Its pmf is μ= rp= 1008 = 125 σ2= rqp2= 10(02)(08)2= 3125

1919

mgf rArr pdf By Maclaurinrsquosseries expansion (ref p632)

If the moments of X E(Xr) = M(r)(0) are known M(t) is thus determined

rArr pdf can be obtained by rewriting M(t) as the weighted sum of eldquoxrdquot

Ex34-7 If the moments of X are E(Xr) = 08 r=123hellip Then M(t) can be determined as

Therefore P(X=0)=02 and P(X=1)=08

2020

Poisson Process Def35-1 An approximate Poisson process with parameter λgt0

The numbers of changes occurring in non-overlapping intervals are independent

The probability of exactly one change in a sufficiently short interval of length h is approximately λh

The probability of two or more changes in a sufficiently short interval is essentially zero

Determine pmf During the unit interval of length 1 there are x changes For nraquox we partition the unit interval into n subintervals of length 1n The probability of x changes in the unit interval equivThe probability of one

change in each of exactly x of these n subintervals The probability of one change in each subinterval is roughly λ(1n) The probability of two or more changes in each subinterval is essentially 0 The change occurrence or not in each subinterval becomes a Bernoulli trial

Thus for a sequence of n Bernoulli trials with probability p = λnP(X=x) can be approximated by (binomial)

2121

Poisson DistributionPoisson Distribution

2222

ExamplesExamples Ex35-1 X has a Poisson distribution with a mean of λ=5Ex35-1 X has a Poisson distribution with a mean of λ=5

Table III on p 652 lists selected values of the distributionTable III on p 652 lists selected values of the distribution P(Xle6) = 0762P(Xle6) = 0762 P(Xgt5) = 1-P(Xle5) = 1-0616 =0384P(Xgt5) = 1-P(Xle5) = 1-0616 =0384 P(X=6) = P(Xle6)-P(Xle5) = 0762-0616 = 0146P(X=6) = P(Xle6)-P(Xle5) = 0762-0616 = 0146

Ex35-2 The Poisson probability histogramsEx35-2 The Poisson probability histograms

232310

)()(

)(

xx

etxf

tx

9900559

9936192

1)1(

1

1

21

12100559 2

)(

n

iinn

n

iin xxsx

More ExamplesMore Examples Empirical data vs Theoretical formula (Ex35-3)Empirical data vs Theoretical formula (Ex35-3)

X is the number of αparticles emittedX is the number of αparticles emitted by barium-133 in 1 sec and counted by barium-133 in 1 sec and counted by a Geiger counterby a Geiger counter

100 observations are made100 observations are made

Generally with unit interval of Generally with unit interval of length t the Poisson pmf islength t the Poisson pmf is

10486101048610 Ex35-4 Assume a tape flaw is a Poisson distribution with a Ex35-4 Assume a tape flaw is a Poisson distribution with a

mean of λ= 11200 flaw per feetmean of λ= 11200 flaw per feet What is the distribution of X the number of flaws in a 4800-foot rollWhat is the distribution of X the number of flaws in a 4800-foot roll E(X)=48001200=4E(X)=48001200=4 P(X=0) = eP(X=0) = e-4-4= 0018= 0018 By Table III on p 652 P(Xle4) By Table III on p 652 P(Xle4) =0 629=0 629

10

)4()(

)4(

xx

exf

x

2424

8570

)2()3( 210020

3

0

)2(

x

x

x

eXP

Poisson with λ=np can simulate Binomial for large n aPoisson with λ=np can simulate Binomial for large n and small pnd small p (μ σ(μ σ22)=(λ λ) asymp(np npq))=(λ λ) asymp(np npq)

Ex35-6 Bulbs with 2 defective rate Ex35-6 Bulbs with 2 defective rate The probability that a box of 100 bulbs contains at most 3 The probability that a box of 100 bulbs contains at most 3 defdef

ective bulbs isective bulbs is

By Binomial the tedious computation will result inBy Binomial the tedious computation will result in

Ex35-7 p160 the comparisons of Binomial and PoisEx35-7 p160 the comparisons of Binomial and Poisson distributionsson distributions

When Poisson asymp BinomialWhen Poisson asymp Binomial

3

0

100100 8590)980()020()3(x

xxxCXP

2525

xnxqpx

nxXPxf

)()(

Ex35-8 Among a lot of 1000 parts n=100 parts are taken at randoEx35-8 Among a lot of 1000 parts n=100 parts are taken at random wo replacement The lot is accepted if no more than 2 of100 parts m wo replacement The lot is accepted if no more than 2 of100 parts taken are defective Assume p is the defective ratetaken are defective Assume p is the defective rate Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=

2)2)HypergeometricHypergeometric

NN11=1000p=1000pN=1000 N=1000 NN22=N-N=N-N11

Since N is large it makes little difference if sampling is done with or withSince N is large it makes little difference if sampling is done with or without replacement simulated by BinomrArrout replacement simulated by BinomrArr ialial

n Bernoulli trialsn Bernoulli trials For small p p=0~1 simulated by PoissorArrFor small p p=0~1 simulated by PoissorArr nn

OC(001)=092OC(001)=092OC(002)=0677OC(002)=0677OC(003)=0423OC(003)=0423OC(005)=0125OC(005)=0125OC(010)=0003OC(010)=0003

When Poisson asymp Binomial Binomial When Poisson asymp Binomial Binomial asymp Hypergeometric Hypergeometricasymp Hypergeometric Hypergeometric

21 )()(21

21

NxnNxnxxXPxfn

NN

xn

N

x

N

2

0

)100(

)100()2( 100

x

px

x

epXPpnp

66

Examples Ex31-5 In a pond of 50 fish with 10 tagged 7 fish are caught at random wo

replacement The probability of 2 tagged fish caught is

Ex31-7 In a lot of 25 items with unknown defective 5 items are selected at random wo replacement for exam If no defective found the lot is accepted otherwise rejected Given N1 defective the acceptance probability operating characteristic curve is

Ex31-8 Roll a 4-sided die twice X is the sum 28 f(x)=(4-|x-5|)16 1000 experiments simulated on a computer

77

Mathematical Expectation Def32-1 For a random variable X with pmf f(x) if sumx Sisin u(x)f(x) exists it

is called the mathematical expectation or the expected value of the function u(X) denoted by E[u(X)] It is the weighted mean of u(x) x Sisin The function u(X) is also a random variable say Y with pmf g(y)

Ex32-2 For a random variable X with f(x)=13 x S=-1 0 1isin Let u(X)=X2 Then E[u(X)]= E[X2]=23 The support of the random variable Y=X2 is S1=0 1

and P(Y=0)=P(X=0) P(Y=1)=P(X=-1)+P(X=1) so its pmf Hence sumy S1isin yg(y) = 23 too

Thm32-1 mathematical expectation E if exists satisfies For a constant c E(c)=c For a constant c amp a function u E[c u(X)]=c E[u(X)] For constants a amp b and functions u amp v E[a u(X) + b v(X)]=a E[u(X)] + b

E[v(X)] It can be applied for 2+terms since E is a linear or distributive operator

88

Examples Ex32-3 f(x)=x10 x=1234

Ex32-4 u(x)=(x-b)2 where b is a constant Suppose E[(X-b)2] exists What is the value of b to minimize it

Ex32-5 X has a hypergeometric distribution

99

More Examples Ex32-6 f(x)=x6 x=123

E(X)=μ=1(16)+2(16)+3(16)=73 Var(X)=σ2=E[(X-μ)2]=E(X2)-μ2=hellip=59 σ=0745rArr

Ex32-7 f(x)=13 x=-101 E(X)=μ=-1(13)+0(13)+1(13)=0 Var(X)=σ2=E(X2)-μ2=hellip=23 The standard deviation σ=0816rArr Comparatively g(y)=13 y=-202

Its mean is also zero but Var(Y)=83 and σY=2σ more spread outrArr

Ex32-8 uniform f(x)=1m x=1m E(X)=μ=1(1m)+hellip+m(1m)=(m+1)2 Var(X)=σ2=E(X2)-μ2=hellip=(m2-1)12

For instance m=6 when rolling a 6-sided die

1010

Derived Random Variables Linear Combination X has a mean μX and variance σX

2 Y=aX+b μrArr Y = aμX+b Var(Y)=E[(Y-μY)2] =hellip=a2σX

2 σX=|a|σX a=2 b=0 mean2 variance4 standard deviation2rArr a=1 b=-1 mean-1 variance1 standard deviation1 rArr Var(X-1) = Var(X)

The rth moment of the distribution about b E[(X-b)r] The rth factorial moment E[(X)r]=E[X(X-1)(X-2)hellip(X-r+1)]

E[(X)2] = E[X(X-1)] = E(X2)-E(X) = E(X2)-μE[(X)2]+μ-μ2= E(X2)-μ+μ-μ2= E(X2)-μ2= Var(X)=σ2

Ex32-9 X has a hypergeometric distribution (ref Ex32-5)

1111

Bernoulli Trials A Bernoulli experiment is a random experiment whose outcome can

be classified in one of two mutually exclusive and exhaustive ways success or failure

A series of Bernoulli trials occurs after independent experiments Probabilities of success p and failure q remain the same (p+q=1)

Random variable X follows a Bernoulli distribution X(success)=1 and X(failure)=0 The pmf of X is f(x)=pxq(1-x) x=01 (μ σ2)=(p pq)

A series of n Bernoulli trials a random sample will be an n-tuple of 01rsquos Ex33-4 Plant 5 seeds and observe the outcome (10101) 1st 3rd 5th seed

s germinated If the germination probability is 8 the probability of this outcome is (8)(2)(8)(2)(8) assuming independence

Let X be the number of successes in n trials X follows a binomial distribution denoted as b(n p) The pmf of X is

1212

Example Ex33-5 For lottery with 2 winning if X equals the number of winning

tickets among n=8 purchases The probability of having 2 winning tickets is

Ex33-6 The effect of n and p is illustrated as follows

1313

Cumulative Distribution Function The cumulative probability F(x) defined as P(Xle x) is called the cumulat

ive distribution function or the distribution function

Ex33-7 Assume the distribution of X is b(10 08) F(8) = P(Xle8) = 1-P(X=9)-P(X=10) = 1-10(8)9(2)-(8)10=6242 F(6) = P(Xle6) = sumx=06Cx

10 (8)x(2)10-x

Ex33-9 Y follows b(8 065) If X=8-Y X has b(8 35) whose distribution function is in Table II (p647)

Eg P(Yge6) = P(8-Yle8-6) = P(Xle2)=04278 from table lookup Likewise P(Yle5) = P(8-Yge8-5) = P(Xge3) = 1-P(Xle2) = 1-04278 =0 5722 P(Y=5) = P(X=3) = P(Xle3)-P(Xle2) = 07064-04278 = 02786

The mean and variance of the binomial distribution is (μ σ2)=(np npq) Ex34-2 details the computations

1414

Comparisons Empirical Data vs Analytical Formula

Ex33-11 b(5 5) has μ= np=

25 σ2= npq= 125 Simulate the model for 100 times

2 3 2 hellip = 247 srArr 2=15243

Suppose an urn has N1success balls and N2 failure balls N = N1+ N2

Let p = N1N and X be the number of success balls in a random sample of size n taken from this urn

If the sampling is done one at a time with replacement X follows b(n p) If the sampling is done without replacement X has a hypergeometric distrib

ution with pmf

If N is large and n is relative small it makes little difference if the sampling is done with or without replacement (See Fig33-4)

x

1515

Moment-Generating Function (mgf )

Def34-1 X is a random variable of the discrete type with pmf f(x) and space S If there is a positive integer h st

E(etX) = sumx Sisin etxf(x) exists and is finite for ndashhlttlth thenM(t) = E(etX) is called the moment-generating function of X ndashE(etX) exists and is finite for ndashhlttlth hArrM(r)(t) exist at t=0 r=123hellip ndashUnique association pmf hArrmgf

Sharing the same mgf two random variables have the same distribution of probability

Ex34-1 X has mgf M(t) = et(36)+ e2t(26)+ e3t(16) From eldquoxrdquot Its pmf has f(0)=0 f(1)=36 f(2)=26 f(3)=16 f(4)=0 hellip Therefore f(x)=(4-x)6 x=123

Ex34-2 X has mgf M(t) = et2(1-et2) tltln2 (1-z)-1= 1 + z + z2+ z3+ hellip |z|lt1

1616

Application of Application of mgf

M(t) = E(etX) = sumx Sisin etxf(x) exists and is finite for ndashhlttlthM(t) = sumx Sisin xetxf(x) M(0) = sumx Sisin xf(x) = E(X) M(t) = sumx Sisin x2etxf(x) hellip M(0) = sumx Sisin x2f(x) = E(X2) hellip M(r)(t) = sumx Sisin xretxf(x) M(r)(0) = sumx Sisin xrf(x) = E(Xr) as t=0

M(t) must be formulated (in closed form) to get its derivatives of higher order Ex34-3 X has a binomial distribution b(n p)

Thus its mgf is

When n=1 X has a Bernoulli distribution

1717

Negative Binomial Distribution Let X be the number of Bernoulli trials to observe the rth succ

ess X has a negative binomial distribution

Its pmf g(x) is

If r=1 X has a geometric distribution with

1818

Geometric Distribution X has a geometric distribution with the pmf

P(X gt k) P(X lek)

Memory-less (EX3412)

Ex34-4 Fruit fliesrsquo eyes with frac14white and frac34red The probability of checking at least 4 flies to observe a white eye is

P(Xge4) = P(Xgt3) = (frac34)3= 04219 The probability of checking at most 4 flies to observe a white eye

isP(Xle4) = 1-(frac34)3=06836 The probability of finding the first white eye on the 4th fly checked is

P(X=4) = pq4-1= 01055 lt= P(Xle4) -P(Xle3)

Ex34-4 For a basketball player with 80 free throw X is the minimum number of throws for a total of 10 free throws Its pmf is μ= rp= 1008 = 125 σ2= rqp2= 10(02)(08)2= 3125

1919

mgf rArr pdf By Maclaurinrsquosseries expansion (ref p632)

If the moments of X E(Xr) = M(r)(0) are known M(t) is thus determined

rArr pdf can be obtained by rewriting M(t) as the weighted sum of eldquoxrdquot

Ex34-7 If the moments of X are E(Xr) = 08 r=123hellip Then M(t) can be determined as

Therefore P(X=0)=02 and P(X=1)=08

2020

Poisson Process Def35-1 An approximate Poisson process with parameter λgt0

The numbers of changes occurring in non-overlapping intervals are independent

The probability of exactly one change in a sufficiently short interval of length h is approximately λh

The probability of two or more changes in a sufficiently short interval is essentially zero

Determine pmf During the unit interval of length 1 there are x changes For nraquox we partition the unit interval into n subintervals of length 1n The probability of x changes in the unit interval equivThe probability of one

change in each of exactly x of these n subintervals The probability of one change in each subinterval is roughly λ(1n) The probability of two or more changes in each subinterval is essentially 0 The change occurrence or not in each subinterval becomes a Bernoulli trial

Thus for a sequence of n Bernoulli trials with probability p = λnP(X=x) can be approximated by (binomial)

2121

Poisson DistributionPoisson Distribution

2222

ExamplesExamples Ex35-1 X has a Poisson distribution with a mean of λ=5Ex35-1 X has a Poisson distribution with a mean of λ=5

Table III on p 652 lists selected values of the distributionTable III on p 652 lists selected values of the distribution P(Xle6) = 0762P(Xle6) = 0762 P(Xgt5) = 1-P(Xle5) = 1-0616 =0384P(Xgt5) = 1-P(Xle5) = 1-0616 =0384 P(X=6) = P(Xle6)-P(Xle5) = 0762-0616 = 0146P(X=6) = P(Xle6)-P(Xle5) = 0762-0616 = 0146

Ex35-2 The Poisson probability histogramsEx35-2 The Poisson probability histograms

232310

)()(

)(

xx

etxf

tx

9900559

9936192

1)1(

1

1

21

12100559 2

)(

n

iinn

n

iin xxsx

More ExamplesMore Examples Empirical data vs Theoretical formula (Ex35-3)Empirical data vs Theoretical formula (Ex35-3)

X is the number of αparticles emittedX is the number of αparticles emitted by barium-133 in 1 sec and counted by barium-133 in 1 sec and counted by a Geiger counterby a Geiger counter

100 observations are made100 observations are made

Generally with unit interval of Generally with unit interval of length t the Poisson pmf islength t the Poisson pmf is

10486101048610 Ex35-4 Assume a tape flaw is a Poisson distribution with a Ex35-4 Assume a tape flaw is a Poisson distribution with a

mean of λ= 11200 flaw per feetmean of λ= 11200 flaw per feet What is the distribution of X the number of flaws in a 4800-foot rollWhat is the distribution of X the number of flaws in a 4800-foot roll E(X)=48001200=4E(X)=48001200=4 P(X=0) = eP(X=0) = e-4-4= 0018= 0018 By Table III on p 652 P(Xle4) By Table III on p 652 P(Xle4) =0 629=0 629

10

)4()(

)4(

xx

exf

x

2424

8570

)2()3( 210020

3

0

)2(

x

x

x

eXP

Poisson with λ=np can simulate Binomial for large n aPoisson with λ=np can simulate Binomial for large n and small pnd small p (μ σ(μ σ22)=(λ λ) asymp(np npq))=(λ λ) asymp(np npq)

Ex35-6 Bulbs with 2 defective rate Ex35-6 Bulbs with 2 defective rate The probability that a box of 100 bulbs contains at most 3 The probability that a box of 100 bulbs contains at most 3 defdef

ective bulbs isective bulbs is

By Binomial the tedious computation will result inBy Binomial the tedious computation will result in

Ex35-7 p160 the comparisons of Binomial and PoisEx35-7 p160 the comparisons of Binomial and Poisson distributionsson distributions

When Poisson asymp BinomialWhen Poisson asymp Binomial

3

0

100100 8590)980()020()3(x

xxxCXP

2525

xnxqpx

nxXPxf

)()(

Ex35-8 Among a lot of 1000 parts n=100 parts are taken at randoEx35-8 Among a lot of 1000 parts n=100 parts are taken at random wo replacement The lot is accepted if no more than 2 of100 parts m wo replacement The lot is accepted if no more than 2 of100 parts taken are defective Assume p is the defective ratetaken are defective Assume p is the defective rate Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=

2)2)HypergeometricHypergeometric

NN11=1000p=1000pN=1000 N=1000 NN22=N-N=N-N11

Since N is large it makes little difference if sampling is done with or withSince N is large it makes little difference if sampling is done with or without replacement simulated by BinomrArrout replacement simulated by BinomrArr ialial

n Bernoulli trialsn Bernoulli trials For small p p=0~1 simulated by PoissorArrFor small p p=0~1 simulated by PoissorArr nn

OC(001)=092OC(001)=092OC(002)=0677OC(002)=0677OC(003)=0423OC(003)=0423OC(005)=0125OC(005)=0125OC(010)=0003OC(010)=0003

When Poisson asymp Binomial Binomial When Poisson asymp Binomial Binomial asymp Hypergeometric Hypergeometricasymp Hypergeometric Hypergeometric

21 )()(21

21

NxnNxnxxXPxfn

NN

xn

N

x

N

2

0

)100(

)100()2( 100

x

px

x

epXPpnp

77

Mathematical Expectation Def32-1 For a random variable X with pmf f(x) if sumx Sisin u(x)f(x) exists it

is called the mathematical expectation or the expected value of the function u(X) denoted by E[u(X)] It is the weighted mean of u(x) x Sisin The function u(X) is also a random variable say Y with pmf g(y)

Ex32-2 For a random variable X with f(x)=13 x S=-1 0 1isin Let u(X)=X2 Then E[u(X)]= E[X2]=23 The support of the random variable Y=X2 is S1=0 1

and P(Y=0)=P(X=0) P(Y=1)=P(X=-1)+P(X=1) so its pmf Hence sumy S1isin yg(y) = 23 too

Thm32-1 mathematical expectation E if exists satisfies For a constant c E(c)=c For a constant c amp a function u E[c u(X)]=c E[u(X)] For constants a amp b and functions u amp v E[a u(X) + b v(X)]=a E[u(X)] + b

E[v(X)] It can be applied for 2+terms since E is a linear or distributive operator

88

Examples Ex32-3 f(x)=x10 x=1234

Ex32-4 u(x)=(x-b)2 where b is a constant Suppose E[(X-b)2] exists What is the value of b to minimize it

Ex32-5 X has a hypergeometric distribution

99

More Examples Ex32-6 f(x)=x6 x=123

E(X)=μ=1(16)+2(16)+3(16)=73 Var(X)=σ2=E[(X-μ)2]=E(X2)-μ2=hellip=59 σ=0745rArr

Ex32-7 f(x)=13 x=-101 E(X)=μ=-1(13)+0(13)+1(13)=0 Var(X)=σ2=E(X2)-μ2=hellip=23 The standard deviation σ=0816rArr Comparatively g(y)=13 y=-202

Its mean is also zero but Var(Y)=83 and σY=2σ more spread outrArr

Ex32-8 uniform f(x)=1m x=1m E(X)=μ=1(1m)+hellip+m(1m)=(m+1)2 Var(X)=σ2=E(X2)-μ2=hellip=(m2-1)12

For instance m=6 when rolling a 6-sided die

1010

Derived Random Variables Linear Combination X has a mean μX and variance σX

2 Y=aX+b μrArr Y = aμX+b Var(Y)=E[(Y-μY)2] =hellip=a2σX

2 σX=|a|σX a=2 b=0 mean2 variance4 standard deviation2rArr a=1 b=-1 mean-1 variance1 standard deviation1 rArr Var(X-1) = Var(X)

The rth moment of the distribution about b E[(X-b)r] The rth factorial moment E[(X)r]=E[X(X-1)(X-2)hellip(X-r+1)]

E[(X)2] = E[X(X-1)] = E(X2)-E(X) = E(X2)-μE[(X)2]+μ-μ2= E(X2)-μ+μ-μ2= E(X2)-μ2= Var(X)=σ2

Ex32-9 X has a hypergeometric distribution (ref Ex32-5)

1111

Bernoulli Trials A Bernoulli experiment is a random experiment whose outcome can

be classified in one of two mutually exclusive and exhaustive ways success or failure

A series of Bernoulli trials occurs after independent experiments Probabilities of success p and failure q remain the same (p+q=1)

Random variable X follows a Bernoulli distribution X(success)=1 and X(failure)=0 The pmf of X is f(x)=pxq(1-x) x=01 (μ σ2)=(p pq)

A series of n Bernoulli trials a random sample will be an n-tuple of 01rsquos Ex33-4 Plant 5 seeds and observe the outcome (10101) 1st 3rd 5th seed

s germinated If the germination probability is 8 the probability of this outcome is (8)(2)(8)(2)(8) assuming independence

Let X be the number of successes in n trials X follows a binomial distribution denoted as b(n p) The pmf of X is

1212

Example Ex33-5 For lottery with 2 winning if X equals the number of winning

tickets among n=8 purchases The probability of having 2 winning tickets is

Ex33-6 The effect of n and p is illustrated as follows

1313

Cumulative Distribution Function The cumulative probability F(x) defined as P(Xle x) is called the cumulat

ive distribution function or the distribution function

Ex33-7 Assume the distribution of X is b(10 08) F(8) = P(Xle8) = 1-P(X=9)-P(X=10) = 1-10(8)9(2)-(8)10=6242 F(6) = P(Xle6) = sumx=06Cx

10 (8)x(2)10-x

Ex33-9 Y follows b(8 065) If X=8-Y X has b(8 35) whose distribution function is in Table II (p647)

Eg P(Yge6) = P(8-Yle8-6) = P(Xle2)=04278 from table lookup Likewise P(Yle5) = P(8-Yge8-5) = P(Xge3) = 1-P(Xle2) = 1-04278 =0 5722 P(Y=5) = P(X=3) = P(Xle3)-P(Xle2) = 07064-04278 = 02786

The mean and variance of the binomial distribution is (μ σ2)=(np npq) Ex34-2 details the computations

1414

Comparisons Empirical Data vs Analytical Formula

Ex33-11 b(5 5) has μ= np=

25 σ2= npq= 125 Simulate the model for 100 times

2 3 2 hellip = 247 srArr 2=15243

Suppose an urn has N1success balls and N2 failure balls N = N1+ N2

Let p = N1N and X be the number of success balls in a random sample of size n taken from this urn

If the sampling is done one at a time with replacement X follows b(n p) If the sampling is done without replacement X has a hypergeometric distrib

ution with pmf

If N is large and n is relative small it makes little difference if the sampling is done with or without replacement (See Fig33-4)

x

1515

Moment-Generating Function (mgf )

Def34-1 X is a random variable of the discrete type with pmf f(x) and space S If there is a positive integer h st

E(etX) = sumx Sisin etxf(x) exists and is finite for ndashhlttlth thenM(t) = E(etX) is called the moment-generating function of X ndashE(etX) exists and is finite for ndashhlttlth hArrM(r)(t) exist at t=0 r=123hellip ndashUnique association pmf hArrmgf

Sharing the same mgf two random variables have the same distribution of probability

Ex34-1 X has mgf M(t) = et(36)+ e2t(26)+ e3t(16) From eldquoxrdquot Its pmf has f(0)=0 f(1)=36 f(2)=26 f(3)=16 f(4)=0 hellip Therefore f(x)=(4-x)6 x=123

Ex34-2 X has mgf M(t) = et2(1-et2) tltln2 (1-z)-1= 1 + z + z2+ z3+ hellip |z|lt1

1616

Application of Application of mgf

M(t) = E(etX) = sumx Sisin etxf(x) exists and is finite for ndashhlttlthM(t) = sumx Sisin xetxf(x) M(0) = sumx Sisin xf(x) = E(X) M(t) = sumx Sisin x2etxf(x) hellip M(0) = sumx Sisin x2f(x) = E(X2) hellip M(r)(t) = sumx Sisin xretxf(x) M(r)(0) = sumx Sisin xrf(x) = E(Xr) as t=0

M(t) must be formulated (in closed form) to get its derivatives of higher order Ex34-3 X has a binomial distribution b(n p)

Thus its mgf is

When n=1 X has a Bernoulli distribution

1717

Negative Binomial Distribution Let X be the number of Bernoulli trials to observe the rth succ

ess X has a negative binomial distribution

Its pmf g(x) is

If r=1 X has a geometric distribution with

1818

Geometric Distribution X has a geometric distribution with the pmf

P(X gt k) P(X lek)

Memory-less (EX3412)

Ex34-4 Fruit fliesrsquo eyes with frac14white and frac34red The probability of checking at least 4 flies to observe a white eye is

P(Xge4) = P(Xgt3) = (frac34)3= 04219 The probability of checking at most 4 flies to observe a white eye

isP(Xle4) = 1-(frac34)3=06836 The probability of finding the first white eye on the 4th fly checked is

P(X=4) = pq4-1= 01055 lt= P(Xle4) -P(Xle3)

Ex34-4 For a basketball player with 80 free throw X is the minimum number of throws for a total of 10 free throws Its pmf is μ= rp= 1008 = 125 σ2= rqp2= 10(02)(08)2= 3125

1919

mgf rArr pdf By Maclaurinrsquosseries expansion (ref p632)

If the moments of X E(Xr) = M(r)(0) are known M(t) is thus determined

rArr pdf can be obtained by rewriting M(t) as the weighted sum of eldquoxrdquot

Ex34-7 If the moments of X are E(Xr) = 08 r=123hellip Then M(t) can be determined as

Therefore P(X=0)=02 and P(X=1)=08

2020

Poisson Process Def35-1 An approximate Poisson process with parameter λgt0

The numbers of changes occurring in non-overlapping intervals are independent

The probability of exactly one change in a sufficiently short interval of length h is approximately λh

The probability of two or more changes in a sufficiently short interval is essentially zero

Determine pmf During the unit interval of length 1 there are x changes For nraquox we partition the unit interval into n subintervals of length 1n The probability of x changes in the unit interval equivThe probability of one

change in each of exactly x of these n subintervals The probability of one change in each subinterval is roughly λ(1n) The probability of two or more changes in each subinterval is essentially 0 The change occurrence or not in each subinterval becomes a Bernoulli trial

Thus for a sequence of n Bernoulli trials with probability p = λnP(X=x) can be approximated by (binomial)

2121

Poisson DistributionPoisson Distribution

2222

ExamplesExamples Ex35-1 X has a Poisson distribution with a mean of λ=5Ex35-1 X has a Poisson distribution with a mean of λ=5

Table III on p 652 lists selected values of the distributionTable III on p 652 lists selected values of the distribution P(Xle6) = 0762P(Xle6) = 0762 P(Xgt5) = 1-P(Xle5) = 1-0616 =0384P(Xgt5) = 1-P(Xle5) = 1-0616 =0384 P(X=6) = P(Xle6)-P(Xle5) = 0762-0616 = 0146P(X=6) = P(Xle6)-P(Xle5) = 0762-0616 = 0146

Ex35-2 The Poisson probability histogramsEx35-2 The Poisson probability histograms

232310

)()(

)(

xx

etxf

tx

9900559

9936192

1)1(

1

1

21

12100559 2

)(

n

iinn

n

iin xxsx

More ExamplesMore Examples Empirical data vs Theoretical formula (Ex35-3)Empirical data vs Theoretical formula (Ex35-3)

X is the number of αparticles emittedX is the number of αparticles emitted by barium-133 in 1 sec and counted by barium-133 in 1 sec and counted by a Geiger counterby a Geiger counter

100 observations are made100 observations are made

Generally with unit interval of Generally with unit interval of length t the Poisson pmf islength t the Poisson pmf is

10486101048610 Ex35-4 Assume a tape flaw is a Poisson distribution with a Ex35-4 Assume a tape flaw is a Poisson distribution with a

mean of λ= 11200 flaw per feetmean of λ= 11200 flaw per feet What is the distribution of X the number of flaws in a 4800-foot rollWhat is the distribution of X the number of flaws in a 4800-foot roll E(X)=48001200=4E(X)=48001200=4 P(X=0) = eP(X=0) = e-4-4= 0018= 0018 By Table III on p 652 P(Xle4) By Table III on p 652 P(Xle4) =0 629=0 629

10

)4()(

)4(

xx

exf

x

2424

8570

)2()3( 210020

3

0

)2(

x

x

x

eXP

Poisson with λ=np can simulate Binomial for large n aPoisson with λ=np can simulate Binomial for large n and small pnd small p (μ σ(μ σ22)=(λ λ) asymp(np npq))=(λ λ) asymp(np npq)

Ex35-6 Bulbs with 2 defective rate Ex35-6 Bulbs with 2 defective rate The probability that a box of 100 bulbs contains at most 3 The probability that a box of 100 bulbs contains at most 3 defdef

ective bulbs isective bulbs is

By Binomial the tedious computation will result inBy Binomial the tedious computation will result in

Ex35-7 p160 the comparisons of Binomial and PoisEx35-7 p160 the comparisons of Binomial and Poisson distributionsson distributions

When Poisson asymp BinomialWhen Poisson asymp Binomial

3

0

100100 8590)980()020()3(x

xxxCXP

2525

xnxqpx

nxXPxf

)()(

Ex35-8 Among a lot of 1000 parts n=100 parts are taken at randoEx35-8 Among a lot of 1000 parts n=100 parts are taken at random wo replacement The lot is accepted if no more than 2 of100 parts m wo replacement The lot is accepted if no more than 2 of100 parts taken are defective Assume p is the defective ratetaken are defective Assume p is the defective rate Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=

2)2)HypergeometricHypergeometric

NN11=1000p=1000pN=1000 N=1000 NN22=N-N=N-N11

Since N is large it makes little difference if sampling is done with or withSince N is large it makes little difference if sampling is done with or without replacement simulated by BinomrArrout replacement simulated by BinomrArr ialial

n Bernoulli trialsn Bernoulli trials For small p p=0~1 simulated by PoissorArrFor small p p=0~1 simulated by PoissorArr nn

OC(001)=092OC(001)=092OC(002)=0677OC(002)=0677OC(003)=0423OC(003)=0423OC(005)=0125OC(005)=0125OC(010)=0003OC(010)=0003

When Poisson asymp Binomial Binomial When Poisson asymp Binomial Binomial asymp Hypergeometric Hypergeometricasymp Hypergeometric Hypergeometric

21 )()(21

21

NxnNxnxxXPxfn

NN

xn

N

x

N

2

0

)100(

)100()2( 100

x

px

x

epXPpnp

88

Examples Ex32-3 f(x)=x10 x=1234

Ex32-4 u(x)=(x-b)2 where b is a constant Suppose E[(X-b)2] exists What is the value of b to minimize it

Ex32-5 X has a hypergeometric distribution

99

More Examples Ex32-6 f(x)=x6 x=123

E(X)=μ=1(16)+2(16)+3(16)=73 Var(X)=σ2=E[(X-μ)2]=E(X2)-μ2=hellip=59 σ=0745rArr

Ex32-7 f(x)=13 x=-101 E(X)=μ=-1(13)+0(13)+1(13)=0 Var(X)=σ2=E(X2)-μ2=hellip=23 The standard deviation σ=0816rArr Comparatively g(y)=13 y=-202

Its mean is also zero but Var(Y)=83 and σY=2σ more spread outrArr

Ex32-8 uniform f(x)=1m x=1m E(X)=μ=1(1m)+hellip+m(1m)=(m+1)2 Var(X)=σ2=E(X2)-μ2=hellip=(m2-1)12

For instance m=6 when rolling a 6-sided die

1010

Derived Random Variables Linear Combination X has a mean μX and variance σX

2 Y=aX+b μrArr Y = aμX+b Var(Y)=E[(Y-μY)2] =hellip=a2σX

2 σX=|a|σX a=2 b=0 mean2 variance4 standard deviation2rArr a=1 b=-1 mean-1 variance1 standard deviation1 rArr Var(X-1) = Var(X)

The rth moment of the distribution about b E[(X-b)r] The rth factorial moment E[(X)r]=E[X(X-1)(X-2)hellip(X-r+1)]

E[(X)2] = E[X(X-1)] = E(X2)-E(X) = E(X2)-μE[(X)2]+μ-μ2= E(X2)-μ+μ-μ2= E(X2)-μ2= Var(X)=σ2

Ex32-9 X has a hypergeometric distribution (ref Ex32-5)

1111

Bernoulli Trials A Bernoulli experiment is a random experiment whose outcome can

be classified in one of two mutually exclusive and exhaustive ways success or failure

A series of Bernoulli trials occurs after independent experiments Probabilities of success p and failure q remain the same (p+q=1)

Random variable X follows a Bernoulli distribution X(success)=1 and X(failure)=0 The pmf of X is f(x)=pxq(1-x) x=01 (μ σ2)=(p pq)

A series of n Bernoulli trials a random sample will be an n-tuple of 01rsquos Ex33-4 Plant 5 seeds and observe the outcome (10101) 1st 3rd 5th seed

s germinated If the germination probability is 8 the probability of this outcome is (8)(2)(8)(2)(8) assuming independence

Let X be the number of successes in n trials X follows a binomial distribution denoted as b(n p) The pmf of X is

1212

Example Ex33-5 For lottery with 2 winning if X equals the number of winning

tickets among n=8 purchases The probability of having 2 winning tickets is

Ex33-6 The effect of n and p is illustrated as follows

1313

Cumulative Distribution Function The cumulative probability F(x) defined as P(Xle x) is called the cumulat

ive distribution function or the distribution function

Ex33-7 Assume the distribution of X is b(10 08) F(8) = P(Xle8) = 1-P(X=9)-P(X=10) = 1-10(8)9(2)-(8)10=6242 F(6) = P(Xle6) = sumx=06Cx

10 (8)x(2)10-x

Ex33-9 Y follows b(8 065) If X=8-Y X has b(8 35) whose distribution function is in Table II (p647)

Eg P(Yge6) = P(8-Yle8-6) = P(Xle2)=04278 from table lookup Likewise P(Yle5) = P(8-Yge8-5) = P(Xge3) = 1-P(Xle2) = 1-04278 =0 5722 P(Y=5) = P(X=3) = P(Xle3)-P(Xle2) = 07064-04278 = 02786

The mean and variance of the binomial distribution is (μ σ2)=(np npq) Ex34-2 details the computations

1414

Comparisons Empirical Data vs Analytical Formula

Ex33-11 b(5 5) has μ= np=

25 σ2= npq= 125 Simulate the model for 100 times

2 3 2 hellip = 247 srArr 2=15243

Suppose an urn has N1success balls and N2 failure balls N = N1+ N2

Let p = N1N and X be the number of success balls in a random sample of size n taken from this urn

If the sampling is done one at a time with replacement X follows b(n p) If the sampling is done without replacement X has a hypergeometric distrib

ution with pmf

If N is large and n is relative small it makes little difference if the sampling is done with or without replacement (See Fig33-4)

x

1515

Moment-Generating Function (mgf )

Def34-1 X is a random variable of the discrete type with pmf f(x) and space S If there is a positive integer h st

E(etX) = sumx Sisin etxf(x) exists and is finite for ndashhlttlth thenM(t) = E(etX) is called the moment-generating function of X ndashE(etX) exists and is finite for ndashhlttlth hArrM(r)(t) exist at t=0 r=123hellip ndashUnique association pmf hArrmgf

Sharing the same mgf two random variables have the same distribution of probability

Ex34-1 X has mgf M(t) = et(36)+ e2t(26)+ e3t(16) From eldquoxrdquot Its pmf has f(0)=0 f(1)=36 f(2)=26 f(3)=16 f(4)=0 hellip Therefore f(x)=(4-x)6 x=123

Ex34-2 X has mgf M(t) = et2(1-et2) tltln2 (1-z)-1= 1 + z + z2+ z3+ hellip |z|lt1

1616

Application of Application of mgf

M(t) = E(etX) = sumx Sisin etxf(x) exists and is finite for ndashhlttlthM(t) = sumx Sisin xetxf(x) M(0) = sumx Sisin xf(x) = E(X) M(t) = sumx Sisin x2etxf(x) hellip M(0) = sumx Sisin x2f(x) = E(X2) hellip M(r)(t) = sumx Sisin xretxf(x) M(r)(0) = sumx Sisin xrf(x) = E(Xr) as t=0

M(t) must be formulated (in closed form) to get its derivatives of higher order Ex34-3 X has a binomial distribution b(n p)

Thus its mgf is

When n=1 X has a Bernoulli distribution

1717

Negative Binomial Distribution Let X be the number of Bernoulli trials to observe the rth succ

ess X has a negative binomial distribution

Its pmf g(x) is

If r=1 X has a geometric distribution with

1818

Geometric Distribution X has a geometric distribution with the pmf

P(X gt k) P(X lek)

Memory-less (EX3412)

Ex34-4 Fruit fliesrsquo eyes with frac14white and frac34red The probability of checking at least 4 flies to observe a white eye is

P(Xge4) = P(Xgt3) = (frac34)3= 04219 The probability of checking at most 4 flies to observe a white eye

isP(Xle4) = 1-(frac34)3=06836 The probability of finding the first white eye on the 4th fly checked is

P(X=4) = pq4-1= 01055 lt= P(Xle4) -P(Xle3)

Ex34-4 For a basketball player with 80 free throw X is the minimum number of throws for a total of 10 free throws Its pmf is μ= rp= 1008 = 125 σ2= rqp2= 10(02)(08)2= 3125

1919

mgf rArr pdf By Maclaurinrsquosseries expansion (ref p632)

If the moments of X E(Xr) = M(r)(0) are known M(t) is thus determined

rArr pdf can be obtained by rewriting M(t) as the weighted sum of eldquoxrdquot

Ex34-7 If the moments of X are E(Xr) = 08 r=123hellip Then M(t) can be determined as

Therefore P(X=0)=02 and P(X=1)=08

2020

Poisson Process Def35-1 An approximate Poisson process with parameter λgt0

The numbers of changes occurring in non-overlapping intervals are independent

The probability of exactly one change in a sufficiently short interval of length h is approximately λh

The probability of two or more changes in a sufficiently short interval is essentially zero

Determine pmf During the unit interval of length 1 there are x changes For nraquox we partition the unit interval into n subintervals of length 1n The probability of x changes in the unit interval equivThe probability of one

change in each of exactly x of these n subintervals The probability of one change in each subinterval is roughly λ(1n) The probability of two or more changes in each subinterval is essentially 0 The change occurrence or not in each subinterval becomes a Bernoulli trial

Thus for a sequence of n Bernoulli trials with probability p = λnP(X=x) can be approximated by (binomial)

2121

Poisson DistributionPoisson Distribution

2222

ExamplesExamples Ex35-1 X has a Poisson distribution with a mean of λ=5Ex35-1 X has a Poisson distribution with a mean of λ=5

Table III on p 652 lists selected values of the distributionTable III on p 652 lists selected values of the distribution P(Xle6) = 0762P(Xle6) = 0762 P(Xgt5) = 1-P(Xle5) = 1-0616 =0384P(Xgt5) = 1-P(Xle5) = 1-0616 =0384 P(X=6) = P(Xle6)-P(Xle5) = 0762-0616 = 0146P(X=6) = P(Xle6)-P(Xle5) = 0762-0616 = 0146

Ex35-2 The Poisson probability histogramsEx35-2 The Poisson probability histograms

232310

)()(

)(

xx

etxf

tx

9900559

9936192

1)1(

1

1

21

12100559 2

)(

n

iinn

n

iin xxsx

More ExamplesMore Examples Empirical data vs Theoretical formula (Ex35-3)Empirical data vs Theoretical formula (Ex35-3)

X is the number of αparticles emittedX is the number of αparticles emitted by barium-133 in 1 sec and counted by barium-133 in 1 sec and counted by a Geiger counterby a Geiger counter

100 observations are made100 observations are made

Generally with unit interval of Generally with unit interval of length t the Poisson pmf islength t the Poisson pmf is

10486101048610 Ex35-4 Assume a tape flaw is a Poisson distribution with a Ex35-4 Assume a tape flaw is a Poisson distribution with a

mean of λ= 11200 flaw per feetmean of λ= 11200 flaw per feet What is the distribution of X the number of flaws in a 4800-foot rollWhat is the distribution of X the number of flaws in a 4800-foot roll E(X)=48001200=4E(X)=48001200=4 P(X=0) = eP(X=0) = e-4-4= 0018= 0018 By Table III on p 652 P(Xle4) By Table III on p 652 P(Xle4) =0 629=0 629

10

)4()(

)4(

xx

exf

x

2424

8570

)2()3( 210020

3

0

)2(

x

x

x

eXP

Poisson with λ=np can simulate Binomial for large n aPoisson with λ=np can simulate Binomial for large n and small pnd small p (μ σ(μ σ22)=(λ λ) asymp(np npq))=(λ λ) asymp(np npq)

Ex35-6 Bulbs with 2 defective rate Ex35-6 Bulbs with 2 defective rate The probability that a box of 100 bulbs contains at most 3 The probability that a box of 100 bulbs contains at most 3 defdef

ective bulbs isective bulbs is

By Binomial the tedious computation will result inBy Binomial the tedious computation will result in

Ex35-7 p160 the comparisons of Binomial and PoisEx35-7 p160 the comparisons of Binomial and Poisson distributionsson distributions

When Poisson asymp BinomialWhen Poisson asymp Binomial

3

0

100100 8590)980()020()3(x

xxxCXP

2525

xnxqpx

nxXPxf

)()(

Ex35-8 Among a lot of 1000 parts n=100 parts are taken at randoEx35-8 Among a lot of 1000 parts n=100 parts are taken at random wo replacement The lot is accepted if no more than 2 of100 parts m wo replacement The lot is accepted if no more than 2 of100 parts taken are defective Assume p is the defective ratetaken are defective Assume p is the defective rate Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=

2)2)HypergeometricHypergeometric

NN11=1000p=1000pN=1000 N=1000 NN22=N-N=N-N11

Since N is large it makes little difference if sampling is done with or withSince N is large it makes little difference if sampling is done with or without replacement simulated by BinomrArrout replacement simulated by BinomrArr ialial

n Bernoulli trialsn Bernoulli trials For small p p=0~1 simulated by PoissorArrFor small p p=0~1 simulated by PoissorArr nn

OC(001)=092OC(001)=092OC(002)=0677OC(002)=0677OC(003)=0423OC(003)=0423OC(005)=0125OC(005)=0125OC(010)=0003OC(010)=0003

When Poisson asymp Binomial Binomial When Poisson asymp Binomial Binomial asymp Hypergeometric Hypergeometricasymp Hypergeometric Hypergeometric

21 )()(21

21

NxnNxnxxXPxfn

NN

xn

N

x

N

2

0

)100(

)100()2( 100

x

px

x

epXPpnp

99

More Examples Ex32-6 f(x)=x6 x=123

E(X)=μ=1(16)+2(16)+3(16)=73 Var(X)=σ2=E[(X-μ)2]=E(X2)-μ2=hellip=59 σ=0745rArr

Ex32-7 f(x)=13 x=-101 E(X)=μ=-1(13)+0(13)+1(13)=0 Var(X)=σ2=E(X2)-μ2=hellip=23 The standard deviation σ=0816rArr Comparatively g(y)=13 y=-202

Its mean is also zero but Var(Y)=83 and σY=2σ more spread outrArr

Ex32-8 uniform f(x)=1m x=1m E(X)=μ=1(1m)+hellip+m(1m)=(m+1)2 Var(X)=σ2=E(X2)-μ2=hellip=(m2-1)12

For instance m=6 when rolling a 6-sided die

1010

Derived Random Variables Linear Combination X has a mean μX and variance σX

2 Y=aX+b μrArr Y = aμX+b Var(Y)=E[(Y-μY)2] =hellip=a2σX

2 σX=|a|σX a=2 b=0 mean2 variance4 standard deviation2rArr a=1 b=-1 mean-1 variance1 standard deviation1 rArr Var(X-1) = Var(X)

The rth moment of the distribution about b E[(X-b)r] The rth factorial moment E[(X)r]=E[X(X-1)(X-2)hellip(X-r+1)]

E[(X)2] = E[X(X-1)] = E(X2)-E(X) = E(X2)-μE[(X)2]+μ-μ2= E(X2)-μ+μ-μ2= E(X2)-μ2= Var(X)=σ2

Ex32-9 X has a hypergeometric distribution (ref Ex32-5)

1111

Bernoulli Trials A Bernoulli experiment is a random experiment whose outcome can

be classified in one of two mutually exclusive and exhaustive ways success or failure

A series of Bernoulli trials occurs after independent experiments Probabilities of success p and failure q remain the same (p+q=1)

Random variable X follows a Bernoulli distribution X(success)=1 and X(failure)=0 The pmf of X is f(x)=pxq(1-x) x=01 (μ σ2)=(p pq)

A series of n Bernoulli trials a random sample will be an n-tuple of 01rsquos Ex33-4 Plant 5 seeds and observe the outcome (10101) 1st 3rd 5th seed

s germinated If the germination probability is 8 the probability of this outcome is (8)(2)(8)(2)(8) assuming independence

Let X be the number of successes in n trials X follows a binomial distribution denoted as b(n p) The pmf of X is

1212

Example Ex33-5 For lottery with 2 winning if X equals the number of winning

tickets among n=8 purchases The probability of having 2 winning tickets is

Ex33-6 The effect of n and p is illustrated as follows

1313

Cumulative Distribution Function The cumulative probability F(x) defined as P(Xle x) is called the cumulat

ive distribution function or the distribution function

Ex33-7 Assume the distribution of X is b(10 08) F(8) = P(Xle8) = 1-P(X=9)-P(X=10) = 1-10(8)9(2)-(8)10=6242 F(6) = P(Xle6) = sumx=06Cx

10 (8)x(2)10-x

Ex33-9 Y follows b(8 065) If X=8-Y X has b(8 35) whose distribution function is in Table II (p647)

Eg P(Yge6) = P(8-Yle8-6) = P(Xle2)=04278 from table lookup Likewise P(Yle5) = P(8-Yge8-5) = P(Xge3) = 1-P(Xle2) = 1-04278 =0 5722 P(Y=5) = P(X=3) = P(Xle3)-P(Xle2) = 07064-04278 = 02786

The mean and variance of the binomial distribution is (μ σ2)=(np npq) Ex34-2 details the computations

1414

Comparisons Empirical Data vs Analytical Formula

Ex33-11 b(5 5) has μ= np=

25 σ2= npq= 125 Simulate the model for 100 times

2 3 2 hellip = 247 srArr 2=15243

Suppose an urn has N1success balls and N2 failure balls N = N1+ N2

Let p = N1N and X be the number of success balls in a random sample of size n taken from this urn

If the sampling is done one at a time with replacement X follows b(n p) If the sampling is done without replacement X has a hypergeometric distrib

ution with pmf

If N is large and n is relative small it makes little difference if the sampling is done with or without replacement (See Fig33-4)

x

1515

Moment-Generating Function (mgf )

Def34-1 X is a random variable of the discrete type with pmf f(x) and space S If there is a positive integer h st

E(etX) = sumx Sisin etxf(x) exists and is finite for ndashhlttlth thenM(t) = E(etX) is called the moment-generating function of X ndashE(etX) exists and is finite for ndashhlttlth hArrM(r)(t) exist at t=0 r=123hellip ndashUnique association pmf hArrmgf

Sharing the same mgf two random variables have the same distribution of probability

Ex34-1 X has mgf M(t) = et(36)+ e2t(26)+ e3t(16) From eldquoxrdquot Its pmf has f(0)=0 f(1)=36 f(2)=26 f(3)=16 f(4)=0 hellip Therefore f(x)=(4-x)6 x=123

Ex34-2 X has mgf M(t) = et2(1-et2) tltln2 (1-z)-1= 1 + z + z2+ z3+ hellip |z|lt1

1616

Application of Application of mgf

M(t) = E(etX) = sumx Sisin etxf(x) exists and is finite for ndashhlttlthM(t) = sumx Sisin xetxf(x) M(0) = sumx Sisin xf(x) = E(X) M(t) = sumx Sisin x2etxf(x) hellip M(0) = sumx Sisin x2f(x) = E(X2) hellip M(r)(t) = sumx Sisin xretxf(x) M(r)(0) = sumx Sisin xrf(x) = E(Xr) as t=0

M(t) must be formulated (in closed form) to get its derivatives of higher order Ex34-3 X has a binomial distribution b(n p)

Thus its mgf is

When n=1 X has a Bernoulli distribution

1717

Negative Binomial Distribution Let X be the number of Bernoulli trials to observe the rth succ

ess X has a negative binomial distribution

Its pmf g(x) is

If r=1 X has a geometric distribution with

1818

Geometric Distribution X has a geometric distribution with the pmf

P(X gt k) P(X lek)

Memory-less (EX3412)

Ex34-4 Fruit fliesrsquo eyes with frac14white and frac34red The probability of checking at least 4 flies to observe a white eye is

P(Xge4) = P(Xgt3) = (frac34)3= 04219 The probability of checking at most 4 flies to observe a white eye

isP(Xle4) = 1-(frac34)3=06836 The probability of finding the first white eye on the 4th fly checked is

P(X=4) = pq4-1= 01055 lt= P(Xle4) -P(Xle3)

Ex34-4 For a basketball player with 80 free throw X is the minimum number of throws for a total of 10 free throws Its pmf is μ= rp= 1008 = 125 σ2= rqp2= 10(02)(08)2= 3125

1919

mgf rArr pdf By Maclaurinrsquosseries expansion (ref p632)

If the moments of X E(Xr) = M(r)(0) are known M(t) is thus determined

rArr pdf can be obtained by rewriting M(t) as the weighted sum of eldquoxrdquot

Ex34-7 If the moments of X are E(Xr) = 08 r=123hellip Then M(t) can be determined as

Therefore P(X=0)=02 and P(X=1)=08

2020

Poisson Process Def35-1 An approximate Poisson process with parameter λgt0

The numbers of changes occurring in non-overlapping intervals are independent

The probability of exactly one change in a sufficiently short interval of length h is approximately λh

The probability of two or more changes in a sufficiently short interval is essentially zero

Determine pmf During the unit interval of length 1 there are x changes For nraquox we partition the unit interval into n subintervals of length 1n The probability of x changes in the unit interval equivThe probability of one

change in each of exactly x of these n subintervals The probability of one change in each subinterval is roughly λ(1n) The probability of two or more changes in each subinterval is essentially 0 The change occurrence or not in each subinterval becomes a Bernoulli trial

Thus for a sequence of n Bernoulli trials with probability p = λnP(X=x) can be approximated by (binomial)

2121

Poisson DistributionPoisson Distribution

2222

ExamplesExamples Ex35-1 X has a Poisson distribution with a mean of λ=5Ex35-1 X has a Poisson distribution with a mean of λ=5

Table III on p 652 lists selected values of the distributionTable III on p 652 lists selected values of the distribution P(Xle6) = 0762P(Xle6) = 0762 P(Xgt5) = 1-P(Xle5) = 1-0616 =0384P(Xgt5) = 1-P(Xle5) = 1-0616 =0384 P(X=6) = P(Xle6)-P(Xle5) = 0762-0616 = 0146P(X=6) = P(Xle6)-P(Xle5) = 0762-0616 = 0146

Ex35-2 The Poisson probability histogramsEx35-2 The Poisson probability histograms

232310

)()(

)(

xx

etxf

tx

9900559

9936192

1)1(

1

1

21

12100559 2

)(

n

iinn

n

iin xxsx

More ExamplesMore Examples Empirical data vs Theoretical formula (Ex35-3)Empirical data vs Theoretical formula (Ex35-3)

X is the number of αparticles emittedX is the number of αparticles emitted by barium-133 in 1 sec and counted by barium-133 in 1 sec and counted by a Geiger counterby a Geiger counter

100 observations are made100 observations are made

Generally with unit interval of Generally with unit interval of length t the Poisson pmf islength t the Poisson pmf is

10486101048610 Ex35-4 Assume a tape flaw is a Poisson distribution with a Ex35-4 Assume a tape flaw is a Poisson distribution with a

mean of λ= 11200 flaw per feetmean of λ= 11200 flaw per feet What is the distribution of X the number of flaws in a 4800-foot rollWhat is the distribution of X the number of flaws in a 4800-foot roll E(X)=48001200=4E(X)=48001200=4 P(X=0) = eP(X=0) = e-4-4= 0018= 0018 By Table III on p 652 P(Xle4) By Table III on p 652 P(Xle4) =0 629=0 629

10

)4()(

)4(

xx

exf

x

2424

8570

)2()3( 210020

3

0

)2(

x

x

x

eXP

Poisson with λ=np can simulate Binomial for large n aPoisson with λ=np can simulate Binomial for large n and small pnd small p (μ σ(μ σ22)=(λ λ) asymp(np npq))=(λ λ) asymp(np npq)

Ex35-6 Bulbs with 2 defective rate Ex35-6 Bulbs with 2 defective rate The probability that a box of 100 bulbs contains at most 3 The probability that a box of 100 bulbs contains at most 3 defdef

ective bulbs isective bulbs is

By Binomial the tedious computation will result inBy Binomial the tedious computation will result in

Ex35-7 p160 the comparisons of Binomial and PoisEx35-7 p160 the comparisons of Binomial and Poisson distributionsson distributions

When Poisson asymp BinomialWhen Poisson asymp Binomial

3

0

100100 8590)980()020()3(x

xxxCXP

2525

xnxqpx

nxXPxf

)()(

Ex35-8 Among a lot of 1000 parts n=100 parts are taken at randoEx35-8 Among a lot of 1000 parts n=100 parts are taken at random wo replacement The lot is accepted if no more than 2 of100 parts m wo replacement The lot is accepted if no more than 2 of100 parts taken are defective Assume p is the defective ratetaken are defective Assume p is the defective rate Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=

2)2)HypergeometricHypergeometric

NN11=1000p=1000pN=1000 N=1000 NN22=N-N=N-N11

Since N is large it makes little difference if sampling is done with or withSince N is large it makes little difference if sampling is done with or without replacement simulated by BinomrArrout replacement simulated by BinomrArr ialial

n Bernoulli trialsn Bernoulli trials For small p p=0~1 simulated by PoissorArrFor small p p=0~1 simulated by PoissorArr nn

OC(001)=092OC(001)=092OC(002)=0677OC(002)=0677OC(003)=0423OC(003)=0423OC(005)=0125OC(005)=0125OC(010)=0003OC(010)=0003

When Poisson asymp Binomial Binomial When Poisson asymp Binomial Binomial asymp Hypergeometric Hypergeometricasymp Hypergeometric Hypergeometric

21 )()(21

21

NxnNxnxxXPxfn

NN

xn

N

x

N

2

0

)100(

)100()2( 100

x

px

x

epXPpnp

1010

Derived Random Variables Linear Combination X has a mean μX and variance σX

2 Y=aX+b μrArr Y = aμX+b Var(Y)=E[(Y-μY)2] =hellip=a2σX

2 σX=|a|σX a=2 b=0 mean2 variance4 standard deviation2rArr a=1 b=-1 mean-1 variance1 standard deviation1 rArr Var(X-1) = Var(X)

The rth moment of the distribution about b E[(X-b)r] The rth factorial moment E[(X)r]=E[X(X-1)(X-2)hellip(X-r+1)]

E[(X)2] = E[X(X-1)] = E(X2)-E(X) = E(X2)-μE[(X)2]+μ-μ2= E(X2)-μ+μ-μ2= E(X2)-μ2= Var(X)=σ2

Ex32-9 X has a hypergeometric distribution (ref Ex32-5)

1111

Bernoulli Trials A Bernoulli experiment is a random experiment whose outcome can

be classified in one of two mutually exclusive and exhaustive ways success or failure

A series of Bernoulli trials occurs after independent experiments Probabilities of success p and failure q remain the same (p+q=1)

Random variable X follows a Bernoulli distribution X(success)=1 and X(failure)=0 The pmf of X is f(x)=pxq(1-x) x=01 (μ σ2)=(p pq)

A series of n Bernoulli trials a random sample will be an n-tuple of 01rsquos Ex33-4 Plant 5 seeds and observe the outcome (10101) 1st 3rd 5th seed

s germinated If the germination probability is 8 the probability of this outcome is (8)(2)(8)(2)(8) assuming independence

Let X be the number of successes in n trials X follows a binomial distribution denoted as b(n p) The pmf of X is

1212

Example Ex33-5 For lottery with 2 winning if X equals the number of winning

tickets among n=8 purchases The probability of having 2 winning tickets is

Ex33-6 The effect of n and p is illustrated as follows

1313

Cumulative Distribution Function The cumulative probability F(x) defined as P(Xle x) is called the cumulat

ive distribution function or the distribution function

Ex33-7 Assume the distribution of X is b(10 08) F(8) = P(Xle8) = 1-P(X=9)-P(X=10) = 1-10(8)9(2)-(8)10=6242 F(6) = P(Xle6) = sumx=06Cx

10 (8)x(2)10-x

Ex33-9 Y follows b(8 065) If X=8-Y X has b(8 35) whose distribution function is in Table II (p647)

Eg P(Yge6) = P(8-Yle8-6) = P(Xle2)=04278 from table lookup Likewise P(Yle5) = P(8-Yge8-5) = P(Xge3) = 1-P(Xle2) = 1-04278 =0 5722 P(Y=5) = P(X=3) = P(Xle3)-P(Xle2) = 07064-04278 = 02786

The mean and variance of the binomial distribution is (μ σ2)=(np npq) Ex34-2 details the computations

1414

Comparisons Empirical Data vs Analytical Formula

Ex33-11 b(5 5) has μ= np=

25 σ2= npq= 125 Simulate the model for 100 times

2 3 2 hellip = 247 srArr 2=15243

Suppose an urn has N1success balls and N2 failure balls N = N1+ N2

Let p = N1N and X be the number of success balls in a random sample of size n taken from this urn

If the sampling is done one at a time with replacement X follows b(n p) If the sampling is done without replacement X has a hypergeometric distrib

ution with pmf

If N is large and n is relative small it makes little difference if the sampling is done with or without replacement (See Fig33-4)

x

1515

Moment-Generating Function (mgf )

Def34-1 X is a random variable of the discrete type with pmf f(x) and space S If there is a positive integer h st

E(etX) = sumx Sisin etxf(x) exists and is finite for ndashhlttlth thenM(t) = E(etX) is called the moment-generating function of X ndashE(etX) exists and is finite for ndashhlttlth hArrM(r)(t) exist at t=0 r=123hellip ndashUnique association pmf hArrmgf

Sharing the same mgf two random variables have the same distribution of probability

Ex34-1 X has mgf M(t) = et(36)+ e2t(26)+ e3t(16) From eldquoxrdquot Its pmf has f(0)=0 f(1)=36 f(2)=26 f(3)=16 f(4)=0 hellip Therefore f(x)=(4-x)6 x=123

Ex34-2 X has mgf M(t) = et2(1-et2) tltln2 (1-z)-1= 1 + z + z2+ z3+ hellip |z|lt1

1616

Application of Application of mgf

M(t) = E(etX) = sumx Sisin etxf(x) exists and is finite for ndashhlttlthM(t) = sumx Sisin xetxf(x) M(0) = sumx Sisin xf(x) = E(X) M(t) = sumx Sisin x2etxf(x) hellip M(0) = sumx Sisin x2f(x) = E(X2) hellip M(r)(t) = sumx Sisin xretxf(x) M(r)(0) = sumx Sisin xrf(x) = E(Xr) as t=0

M(t) must be formulated (in closed form) to get its derivatives of higher order Ex34-3 X has a binomial distribution b(n p)

Thus its mgf is

When n=1 X has a Bernoulli distribution

1717

Negative Binomial Distribution Let X be the number of Bernoulli trials to observe the rth succ

ess X has a negative binomial distribution

Its pmf g(x) is

If r=1 X has a geometric distribution with

1818

Geometric Distribution X has a geometric distribution with the pmf

P(X gt k) P(X lek)

Memory-less (EX3412)

Ex34-4 Fruit fliesrsquo eyes with frac14white and frac34red The probability of checking at least 4 flies to observe a white eye is

P(Xge4) = P(Xgt3) = (frac34)3= 04219 The probability of checking at most 4 flies to observe a white eye

isP(Xle4) = 1-(frac34)3=06836 The probability of finding the first white eye on the 4th fly checked is

P(X=4) = pq4-1= 01055 lt= P(Xle4) -P(Xle3)

Ex34-4 For a basketball player with 80 free throw X is the minimum number of throws for a total of 10 free throws Its pmf is μ= rp= 1008 = 125 σ2= rqp2= 10(02)(08)2= 3125

1919

mgf rArr pdf By Maclaurinrsquosseries expansion (ref p632)

If the moments of X E(Xr) = M(r)(0) are known M(t) is thus determined

rArr pdf can be obtained by rewriting M(t) as the weighted sum of eldquoxrdquot

Ex34-7 If the moments of X are E(Xr) = 08 r=123hellip Then M(t) can be determined as

Therefore P(X=0)=02 and P(X=1)=08

2020

Poisson Process Def35-1 An approximate Poisson process with parameter λgt0

The numbers of changes occurring in non-overlapping intervals are independent

The probability of exactly one change in a sufficiently short interval of length h is approximately λh

The probability of two or more changes in a sufficiently short interval is essentially zero

Determine pmf During the unit interval of length 1 there are x changes For nraquox we partition the unit interval into n subintervals of length 1n The probability of x changes in the unit interval equivThe probability of one

change in each of exactly x of these n subintervals The probability of one change in each subinterval is roughly λ(1n) The probability of two or more changes in each subinterval is essentially 0 The change occurrence or not in each subinterval becomes a Bernoulli trial

Thus for a sequence of n Bernoulli trials with probability p = λnP(X=x) can be approximated by (binomial)

2121

Poisson DistributionPoisson Distribution

2222

ExamplesExamples Ex35-1 X has a Poisson distribution with a mean of λ=5Ex35-1 X has a Poisson distribution with a mean of λ=5

Table III on p 652 lists selected values of the distributionTable III on p 652 lists selected values of the distribution P(Xle6) = 0762P(Xle6) = 0762 P(Xgt5) = 1-P(Xle5) = 1-0616 =0384P(Xgt5) = 1-P(Xle5) = 1-0616 =0384 P(X=6) = P(Xle6)-P(Xle5) = 0762-0616 = 0146P(X=6) = P(Xle6)-P(Xle5) = 0762-0616 = 0146

Ex35-2 The Poisson probability histogramsEx35-2 The Poisson probability histograms

232310

)()(

)(

xx

etxf

tx

9900559

9936192

1)1(

1

1

21

12100559 2

)(

n

iinn

n

iin xxsx

More ExamplesMore Examples Empirical data vs Theoretical formula (Ex35-3)Empirical data vs Theoretical formula (Ex35-3)

X is the number of αparticles emittedX is the number of αparticles emitted by barium-133 in 1 sec and counted by barium-133 in 1 sec and counted by a Geiger counterby a Geiger counter

100 observations are made100 observations are made

Generally with unit interval of Generally with unit interval of length t the Poisson pmf islength t the Poisson pmf is

10486101048610 Ex35-4 Assume a tape flaw is a Poisson distribution with a Ex35-4 Assume a tape flaw is a Poisson distribution with a

mean of λ= 11200 flaw per feetmean of λ= 11200 flaw per feet What is the distribution of X the number of flaws in a 4800-foot rollWhat is the distribution of X the number of flaws in a 4800-foot roll E(X)=48001200=4E(X)=48001200=4 P(X=0) = eP(X=0) = e-4-4= 0018= 0018 By Table III on p 652 P(Xle4) By Table III on p 652 P(Xle4) =0 629=0 629

10

)4()(

)4(

xx

exf

x

2424

8570

)2()3( 210020

3

0

)2(

x

x

x

eXP

Poisson with λ=np can simulate Binomial for large n aPoisson with λ=np can simulate Binomial for large n and small pnd small p (μ σ(μ σ22)=(λ λ) asymp(np npq))=(λ λ) asymp(np npq)

Ex35-6 Bulbs with 2 defective rate Ex35-6 Bulbs with 2 defective rate The probability that a box of 100 bulbs contains at most 3 The probability that a box of 100 bulbs contains at most 3 defdef

ective bulbs isective bulbs is

By Binomial the tedious computation will result inBy Binomial the tedious computation will result in

Ex35-7 p160 the comparisons of Binomial and PoisEx35-7 p160 the comparisons of Binomial and Poisson distributionsson distributions

When Poisson asymp BinomialWhen Poisson asymp Binomial

3

0

100100 8590)980()020()3(x

xxxCXP

2525

xnxqpx

nxXPxf

)()(

Ex35-8 Among a lot of 1000 parts n=100 parts are taken at randoEx35-8 Among a lot of 1000 parts n=100 parts are taken at random wo replacement The lot is accepted if no more than 2 of100 parts m wo replacement The lot is accepted if no more than 2 of100 parts taken are defective Assume p is the defective ratetaken are defective Assume p is the defective rate Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=

2)2)HypergeometricHypergeometric

NN11=1000p=1000pN=1000 N=1000 NN22=N-N=N-N11

Since N is large it makes little difference if sampling is done with or withSince N is large it makes little difference if sampling is done with or without replacement simulated by BinomrArrout replacement simulated by BinomrArr ialial

n Bernoulli trialsn Bernoulli trials For small p p=0~1 simulated by PoissorArrFor small p p=0~1 simulated by PoissorArr nn

OC(001)=092OC(001)=092OC(002)=0677OC(002)=0677OC(003)=0423OC(003)=0423OC(005)=0125OC(005)=0125OC(010)=0003OC(010)=0003

When Poisson asymp Binomial Binomial When Poisson asymp Binomial Binomial asymp Hypergeometric Hypergeometricasymp Hypergeometric Hypergeometric

21 )()(21

21

NxnNxnxxXPxfn

NN

xn

N

x

N

2

0

)100(

)100()2( 100

x

px

x

epXPpnp

1111

Bernoulli Trials A Bernoulli experiment is a random experiment whose outcome can

be classified in one of two mutually exclusive and exhaustive ways success or failure

A series of Bernoulli trials occurs after independent experiments Probabilities of success p and failure q remain the same (p+q=1)

Random variable X follows a Bernoulli distribution X(success)=1 and X(failure)=0 The pmf of X is f(x)=pxq(1-x) x=01 (μ σ2)=(p pq)

A series of n Bernoulli trials a random sample will be an n-tuple of 01rsquos Ex33-4 Plant 5 seeds and observe the outcome (10101) 1st 3rd 5th seed

s germinated If the germination probability is 8 the probability of this outcome is (8)(2)(8)(2)(8) assuming independence

Let X be the number of successes in n trials X follows a binomial distribution denoted as b(n p) The pmf of X is

1212

Example Ex33-5 For lottery with 2 winning if X equals the number of winning

tickets among n=8 purchases The probability of having 2 winning tickets is

Ex33-6 The effect of n and p is illustrated as follows

1313

Cumulative Distribution Function The cumulative probability F(x) defined as P(Xle x) is called the cumulat

ive distribution function or the distribution function

Ex33-7 Assume the distribution of X is b(10 08) F(8) = P(Xle8) = 1-P(X=9)-P(X=10) = 1-10(8)9(2)-(8)10=6242 F(6) = P(Xle6) = sumx=06Cx

10 (8)x(2)10-x

Ex33-9 Y follows b(8 065) If X=8-Y X has b(8 35) whose distribution function is in Table II (p647)

Eg P(Yge6) = P(8-Yle8-6) = P(Xle2)=04278 from table lookup Likewise P(Yle5) = P(8-Yge8-5) = P(Xge3) = 1-P(Xle2) = 1-04278 =0 5722 P(Y=5) = P(X=3) = P(Xle3)-P(Xle2) = 07064-04278 = 02786

The mean and variance of the binomial distribution is (μ σ2)=(np npq) Ex34-2 details the computations

1414

Comparisons Empirical Data vs Analytical Formula

Ex33-11 b(5 5) has μ= np=

25 σ2= npq= 125 Simulate the model for 100 times

2 3 2 hellip = 247 srArr 2=15243

Suppose an urn has N1success balls and N2 failure balls N = N1+ N2

Let p = N1N and X be the number of success balls in a random sample of size n taken from this urn

If the sampling is done one at a time with replacement X follows b(n p) If the sampling is done without replacement X has a hypergeometric distrib

ution with pmf

If N is large and n is relative small it makes little difference if the sampling is done with or without replacement (See Fig33-4)

x

1515

Moment-Generating Function (mgf )

Def34-1 X is a random variable of the discrete type with pmf f(x) and space S If there is a positive integer h st

E(etX) = sumx Sisin etxf(x) exists and is finite for ndashhlttlth thenM(t) = E(etX) is called the moment-generating function of X ndashE(etX) exists and is finite for ndashhlttlth hArrM(r)(t) exist at t=0 r=123hellip ndashUnique association pmf hArrmgf

Sharing the same mgf two random variables have the same distribution of probability

Ex34-1 X has mgf M(t) = et(36)+ e2t(26)+ e3t(16) From eldquoxrdquot Its pmf has f(0)=0 f(1)=36 f(2)=26 f(3)=16 f(4)=0 hellip Therefore f(x)=(4-x)6 x=123

Ex34-2 X has mgf M(t) = et2(1-et2) tltln2 (1-z)-1= 1 + z + z2+ z3+ hellip |z|lt1

1616

Application of Application of mgf

M(t) = E(etX) = sumx Sisin etxf(x) exists and is finite for ndashhlttlthM(t) = sumx Sisin xetxf(x) M(0) = sumx Sisin xf(x) = E(X) M(t) = sumx Sisin x2etxf(x) hellip M(0) = sumx Sisin x2f(x) = E(X2) hellip M(r)(t) = sumx Sisin xretxf(x) M(r)(0) = sumx Sisin xrf(x) = E(Xr) as t=0

M(t) must be formulated (in closed form) to get its derivatives of higher order Ex34-3 X has a binomial distribution b(n p)

Thus its mgf is

When n=1 X has a Bernoulli distribution

1717

Negative Binomial Distribution Let X be the number of Bernoulli trials to observe the rth succ

ess X has a negative binomial distribution

Its pmf g(x) is

If r=1 X has a geometric distribution with

1818

Geometric Distribution X has a geometric distribution with the pmf

P(X gt k) P(X lek)

Memory-less (EX3412)

Ex34-4 Fruit fliesrsquo eyes with frac14white and frac34red The probability of checking at least 4 flies to observe a white eye is

P(Xge4) = P(Xgt3) = (frac34)3= 04219 The probability of checking at most 4 flies to observe a white eye

isP(Xle4) = 1-(frac34)3=06836 The probability of finding the first white eye on the 4th fly checked is

P(X=4) = pq4-1= 01055 lt= P(Xle4) -P(Xle3)

Ex34-4 For a basketball player with 80 free throw X is the minimum number of throws for a total of 10 free throws Its pmf is μ= rp= 1008 = 125 σ2= rqp2= 10(02)(08)2= 3125

1919

mgf rArr pdf By Maclaurinrsquosseries expansion (ref p632)

If the moments of X E(Xr) = M(r)(0) are known M(t) is thus determined

rArr pdf can be obtained by rewriting M(t) as the weighted sum of eldquoxrdquot

Ex34-7 If the moments of X are E(Xr) = 08 r=123hellip Then M(t) can be determined as

Therefore P(X=0)=02 and P(X=1)=08

2020

Poisson Process Def35-1 An approximate Poisson process with parameter λgt0

The numbers of changes occurring in non-overlapping intervals are independent

The probability of exactly one change in a sufficiently short interval of length h is approximately λh

The probability of two or more changes in a sufficiently short interval is essentially zero

Determine pmf During the unit interval of length 1 there are x changes For nraquox we partition the unit interval into n subintervals of length 1n The probability of x changes in the unit interval equivThe probability of one

change in each of exactly x of these n subintervals The probability of one change in each subinterval is roughly λ(1n) The probability of two or more changes in each subinterval is essentially 0 The change occurrence or not in each subinterval becomes a Bernoulli trial

Thus for a sequence of n Bernoulli trials with probability p = λnP(X=x) can be approximated by (binomial)

2121

Poisson DistributionPoisson Distribution

2222

ExamplesExamples Ex35-1 X has a Poisson distribution with a mean of λ=5Ex35-1 X has a Poisson distribution with a mean of λ=5

Table III on p 652 lists selected values of the distributionTable III on p 652 lists selected values of the distribution P(Xle6) = 0762P(Xle6) = 0762 P(Xgt5) = 1-P(Xle5) = 1-0616 =0384P(Xgt5) = 1-P(Xle5) = 1-0616 =0384 P(X=6) = P(Xle6)-P(Xle5) = 0762-0616 = 0146P(X=6) = P(Xle6)-P(Xle5) = 0762-0616 = 0146

Ex35-2 The Poisson probability histogramsEx35-2 The Poisson probability histograms

232310

)()(

)(

xx

etxf

tx

9900559

9936192

1)1(

1

1

21

12100559 2

)(

n

iinn

n

iin xxsx

More ExamplesMore Examples Empirical data vs Theoretical formula (Ex35-3)Empirical data vs Theoretical formula (Ex35-3)

X is the number of αparticles emittedX is the number of αparticles emitted by barium-133 in 1 sec and counted by barium-133 in 1 sec and counted by a Geiger counterby a Geiger counter

100 observations are made100 observations are made

Generally with unit interval of Generally with unit interval of length t the Poisson pmf islength t the Poisson pmf is

10486101048610 Ex35-4 Assume a tape flaw is a Poisson distribution with a Ex35-4 Assume a tape flaw is a Poisson distribution with a

mean of λ= 11200 flaw per feetmean of λ= 11200 flaw per feet What is the distribution of X the number of flaws in a 4800-foot rollWhat is the distribution of X the number of flaws in a 4800-foot roll E(X)=48001200=4E(X)=48001200=4 P(X=0) = eP(X=0) = e-4-4= 0018= 0018 By Table III on p 652 P(Xle4) By Table III on p 652 P(Xle4) =0 629=0 629

10

)4()(

)4(

xx

exf

x

2424

8570

)2()3( 210020

3

0

)2(

x

x

x

eXP

Poisson with λ=np can simulate Binomial for large n aPoisson with λ=np can simulate Binomial for large n and small pnd small p (μ σ(μ σ22)=(λ λ) asymp(np npq))=(λ λ) asymp(np npq)

Ex35-6 Bulbs with 2 defective rate Ex35-6 Bulbs with 2 defective rate The probability that a box of 100 bulbs contains at most 3 The probability that a box of 100 bulbs contains at most 3 defdef

ective bulbs isective bulbs is

By Binomial the tedious computation will result inBy Binomial the tedious computation will result in

Ex35-7 p160 the comparisons of Binomial and PoisEx35-7 p160 the comparisons of Binomial and Poisson distributionsson distributions

When Poisson asymp BinomialWhen Poisson asymp Binomial

3

0

100100 8590)980()020()3(x

xxxCXP

2525

xnxqpx

nxXPxf

)()(

Ex35-8 Among a lot of 1000 parts n=100 parts are taken at randoEx35-8 Among a lot of 1000 parts n=100 parts are taken at random wo replacement The lot is accepted if no more than 2 of100 parts m wo replacement The lot is accepted if no more than 2 of100 parts taken are defective Assume p is the defective ratetaken are defective Assume p is the defective rate Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=

2)2)HypergeometricHypergeometric

NN11=1000p=1000pN=1000 N=1000 NN22=N-N=N-N11

Since N is large it makes little difference if sampling is done with or withSince N is large it makes little difference if sampling is done with or without replacement simulated by BinomrArrout replacement simulated by BinomrArr ialial

n Bernoulli trialsn Bernoulli trials For small p p=0~1 simulated by PoissorArrFor small p p=0~1 simulated by PoissorArr nn

OC(001)=092OC(001)=092OC(002)=0677OC(002)=0677OC(003)=0423OC(003)=0423OC(005)=0125OC(005)=0125OC(010)=0003OC(010)=0003

When Poisson asymp Binomial Binomial When Poisson asymp Binomial Binomial asymp Hypergeometric Hypergeometricasymp Hypergeometric Hypergeometric

21 )()(21

21

NxnNxnxxXPxfn

NN

xn

N

x

N

2

0

)100(

)100()2( 100

x

px

x

epXPpnp

1212

Example Ex33-5 For lottery with 2 winning if X equals the number of winning

tickets among n=8 purchases The probability of having 2 winning tickets is

Ex33-6 The effect of n and p is illustrated as follows

1313

Cumulative Distribution Function The cumulative probability F(x) defined as P(Xle x) is called the cumulat

ive distribution function or the distribution function

Ex33-7 Assume the distribution of X is b(10 08) F(8) = P(Xle8) = 1-P(X=9)-P(X=10) = 1-10(8)9(2)-(8)10=6242 F(6) = P(Xle6) = sumx=06Cx

10 (8)x(2)10-x

Ex33-9 Y follows b(8 065) If X=8-Y X has b(8 35) whose distribution function is in Table II (p647)

Eg P(Yge6) = P(8-Yle8-6) = P(Xle2)=04278 from table lookup Likewise P(Yle5) = P(8-Yge8-5) = P(Xge3) = 1-P(Xle2) = 1-04278 =0 5722 P(Y=5) = P(X=3) = P(Xle3)-P(Xle2) = 07064-04278 = 02786

The mean and variance of the binomial distribution is (μ σ2)=(np npq) Ex34-2 details the computations

1414

Comparisons Empirical Data vs Analytical Formula

Ex33-11 b(5 5) has μ= np=

25 σ2= npq= 125 Simulate the model for 100 times

2 3 2 hellip = 247 srArr 2=15243

Suppose an urn has N1success balls and N2 failure balls N = N1+ N2

Let p = N1N and X be the number of success balls in a random sample of size n taken from this urn

If the sampling is done one at a time with replacement X follows b(n p) If the sampling is done without replacement X has a hypergeometric distrib

ution with pmf

If N is large and n is relative small it makes little difference if the sampling is done with or without replacement (See Fig33-4)

x

1515

Moment-Generating Function (mgf )

Def34-1 X is a random variable of the discrete type with pmf f(x) and space S If there is a positive integer h st

E(etX) = sumx Sisin etxf(x) exists and is finite for ndashhlttlth thenM(t) = E(etX) is called the moment-generating function of X ndashE(etX) exists and is finite for ndashhlttlth hArrM(r)(t) exist at t=0 r=123hellip ndashUnique association pmf hArrmgf

Sharing the same mgf two random variables have the same distribution of probability

Ex34-1 X has mgf M(t) = et(36)+ e2t(26)+ e3t(16) From eldquoxrdquot Its pmf has f(0)=0 f(1)=36 f(2)=26 f(3)=16 f(4)=0 hellip Therefore f(x)=(4-x)6 x=123

Ex34-2 X has mgf M(t) = et2(1-et2) tltln2 (1-z)-1= 1 + z + z2+ z3+ hellip |z|lt1

1616

Application of Application of mgf

M(t) = E(etX) = sumx Sisin etxf(x) exists and is finite for ndashhlttlthM(t) = sumx Sisin xetxf(x) M(0) = sumx Sisin xf(x) = E(X) M(t) = sumx Sisin x2etxf(x) hellip M(0) = sumx Sisin x2f(x) = E(X2) hellip M(r)(t) = sumx Sisin xretxf(x) M(r)(0) = sumx Sisin xrf(x) = E(Xr) as t=0

M(t) must be formulated (in closed form) to get its derivatives of higher order Ex34-3 X has a binomial distribution b(n p)

Thus its mgf is

When n=1 X has a Bernoulli distribution

1717

Negative Binomial Distribution Let X be the number of Bernoulli trials to observe the rth succ

ess X has a negative binomial distribution

Its pmf g(x) is

If r=1 X has a geometric distribution with

1818

Geometric Distribution X has a geometric distribution with the pmf

P(X gt k) P(X lek)

Memory-less (EX3412)

Ex34-4 Fruit fliesrsquo eyes with frac14white and frac34red The probability of checking at least 4 flies to observe a white eye is

P(Xge4) = P(Xgt3) = (frac34)3= 04219 The probability of checking at most 4 flies to observe a white eye

isP(Xle4) = 1-(frac34)3=06836 The probability of finding the first white eye on the 4th fly checked is

P(X=4) = pq4-1= 01055 lt= P(Xle4) -P(Xle3)

Ex34-4 For a basketball player with 80 free throw X is the minimum number of throws for a total of 10 free throws Its pmf is μ= rp= 1008 = 125 σ2= rqp2= 10(02)(08)2= 3125

1919

mgf rArr pdf By Maclaurinrsquosseries expansion (ref p632)

If the moments of X E(Xr) = M(r)(0) are known M(t) is thus determined

rArr pdf can be obtained by rewriting M(t) as the weighted sum of eldquoxrdquot

Ex34-7 If the moments of X are E(Xr) = 08 r=123hellip Then M(t) can be determined as

Therefore P(X=0)=02 and P(X=1)=08

2020

Poisson Process Def35-1 An approximate Poisson process with parameter λgt0

The numbers of changes occurring in non-overlapping intervals are independent

The probability of exactly one change in a sufficiently short interval of length h is approximately λh

The probability of two or more changes in a sufficiently short interval is essentially zero

Determine pmf During the unit interval of length 1 there are x changes For nraquox we partition the unit interval into n subintervals of length 1n The probability of x changes in the unit interval equivThe probability of one

change in each of exactly x of these n subintervals The probability of one change in each subinterval is roughly λ(1n) The probability of two or more changes in each subinterval is essentially 0 The change occurrence or not in each subinterval becomes a Bernoulli trial

Thus for a sequence of n Bernoulli trials with probability p = λnP(X=x) can be approximated by (binomial)

2121

Poisson DistributionPoisson Distribution

2222

ExamplesExamples Ex35-1 X has a Poisson distribution with a mean of λ=5Ex35-1 X has a Poisson distribution with a mean of λ=5

Table III on p 652 lists selected values of the distributionTable III on p 652 lists selected values of the distribution P(Xle6) = 0762P(Xle6) = 0762 P(Xgt5) = 1-P(Xle5) = 1-0616 =0384P(Xgt5) = 1-P(Xle5) = 1-0616 =0384 P(X=6) = P(Xle6)-P(Xle5) = 0762-0616 = 0146P(X=6) = P(Xle6)-P(Xle5) = 0762-0616 = 0146

Ex35-2 The Poisson probability histogramsEx35-2 The Poisson probability histograms

232310

)()(

)(

xx

etxf

tx

9900559

9936192

1)1(

1

1

21

12100559 2

)(

n

iinn

n

iin xxsx

More ExamplesMore Examples Empirical data vs Theoretical formula (Ex35-3)Empirical data vs Theoretical formula (Ex35-3)

X is the number of αparticles emittedX is the number of αparticles emitted by barium-133 in 1 sec and counted by barium-133 in 1 sec and counted by a Geiger counterby a Geiger counter

100 observations are made100 observations are made

Generally with unit interval of Generally with unit interval of length t the Poisson pmf islength t the Poisson pmf is

10486101048610 Ex35-4 Assume a tape flaw is a Poisson distribution with a Ex35-4 Assume a tape flaw is a Poisson distribution with a

mean of λ= 11200 flaw per feetmean of λ= 11200 flaw per feet What is the distribution of X the number of flaws in a 4800-foot rollWhat is the distribution of X the number of flaws in a 4800-foot roll E(X)=48001200=4E(X)=48001200=4 P(X=0) = eP(X=0) = e-4-4= 0018= 0018 By Table III on p 652 P(Xle4) By Table III on p 652 P(Xle4) =0 629=0 629

10

)4()(

)4(

xx

exf

x

2424

8570

)2()3( 210020

3

0

)2(

x

x

x

eXP

Poisson with λ=np can simulate Binomial for large n aPoisson with λ=np can simulate Binomial for large n and small pnd small p (μ σ(μ σ22)=(λ λ) asymp(np npq))=(λ λ) asymp(np npq)

Ex35-6 Bulbs with 2 defective rate Ex35-6 Bulbs with 2 defective rate The probability that a box of 100 bulbs contains at most 3 The probability that a box of 100 bulbs contains at most 3 defdef

ective bulbs isective bulbs is

By Binomial the tedious computation will result inBy Binomial the tedious computation will result in

Ex35-7 p160 the comparisons of Binomial and PoisEx35-7 p160 the comparisons of Binomial and Poisson distributionsson distributions

When Poisson asymp BinomialWhen Poisson asymp Binomial

3

0

100100 8590)980()020()3(x

xxxCXP

2525

xnxqpx

nxXPxf

)()(

Ex35-8 Among a lot of 1000 parts n=100 parts are taken at randoEx35-8 Among a lot of 1000 parts n=100 parts are taken at random wo replacement The lot is accepted if no more than 2 of100 parts m wo replacement The lot is accepted if no more than 2 of100 parts taken are defective Assume p is the defective ratetaken are defective Assume p is the defective rate Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=

2)2)HypergeometricHypergeometric

NN11=1000p=1000pN=1000 N=1000 NN22=N-N=N-N11

Since N is large it makes little difference if sampling is done with or withSince N is large it makes little difference if sampling is done with or without replacement simulated by BinomrArrout replacement simulated by BinomrArr ialial

n Bernoulli trialsn Bernoulli trials For small p p=0~1 simulated by PoissorArrFor small p p=0~1 simulated by PoissorArr nn

OC(001)=092OC(001)=092OC(002)=0677OC(002)=0677OC(003)=0423OC(003)=0423OC(005)=0125OC(005)=0125OC(010)=0003OC(010)=0003

When Poisson asymp Binomial Binomial When Poisson asymp Binomial Binomial asymp Hypergeometric Hypergeometricasymp Hypergeometric Hypergeometric

21 )()(21

21

NxnNxnxxXPxfn

NN

xn

N

x

N

2

0

)100(

)100()2( 100

x

px

x

epXPpnp

1313

Cumulative Distribution Function The cumulative probability F(x) defined as P(Xle x) is called the cumulat

ive distribution function or the distribution function

Ex33-7 Assume the distribution of X is b(10 08) F(8) = P(Xle8) = 1-P(X=9)-P(X=10) = 1-10(8)9(2)-(8)10=6242 F(6) = P(Xle6) = sumx=06Cx

10 (8)x(2)10-x

Ex33-9 Y follows b(8 065) If X=8-Y X has b(8 35) whose distribution function is in Table II (p647)

Eg P(Yge6) = P(8-Yle8-6) = P(Xle2)=04278 from table lookup Likewise P(Yle5) = P(8-Yge8-5) = P(Xge3) = 1-P(Xle2) = 1-04278 =0 5722 P(Y=5) = P(X=3) = P(Xle3)-P(Xle2) = 07064-04278 = 02786

The mean and variance of the binomial distribution is (μ σ2)=(np npq) Ex34-2 details the computations

1414

Comparisons Empirical Data vs Analytical Formula

Ex33-11 b(5 5) has μ= np=

25 σ2= npq= 125 Simulate the model for 100 times

2 3 2 hellip = 247 srArr 2=15243

Suppose an urn has N1success balls and N2 failure balls N = N1+ N2

Let p = N1N and X be the number of success balls in a random sample of size n taken from this urn

If the sampling is done one at a time with replacement X follows b(n p) If the sampling is done without replacement X has a hypergeometric distrib

ution with pmf

If N is large and n is relative small it makes little difference if the sampling is done with or without replacement (See Fig33-4)

x

1515

Moment-Generating Function (mgf )

Def34-1 X is a random variable of the discrete type with pmf f(x) and space S If there is a positive integer h st

E(etX) = sumx Sisin etxf(x) exists and is finite for ndashhlttlth thenM(t) = E(etX) is called the moment-generating function of X ndashE(etX) exists and is finite for ndashhlttlth hArrM(r)(t) exist at t=0 r=123hellip ndashUnique association pmf hArrmgf

Sharing the same mgf two random variables have the same distribution of probability

Ex34-1 X has mgf M(t) = et(36)+ e2t(26)+ e3t(16) From eldquoxrdquot Its pmf has f(0)=0 f(1)=36 f(2)=26 f(3)=16 f(4)=0 hellip Therefore f(x)=(4-x)6 x=123

Ex34-2 X has mgf M(t) = et2(1-et2) tltln2 (1-z)-1= 1 + z + z2+ z3+ hellip |z|lt1

1616

Application of Application of mgf

M(t) = E(etX) = sumx Sisin etxf(x) exists and is finite for ndashhlttlthM(t) = sumx Sisin xetxf(x) M(0) = sumx Sisin xf(x) = E(X) M(t) = sumx Sisin x2etxf(x) hellip M(0) = sumx Sisin x2f(x) = E(X2) hellip M(r)(t) = sumx Sisin xretxf(x) M(r)(0) = sumx Sisin xrf(x) = E(Xr) as t=0

M(t) must be formulated (in closed form) to get its derivatives of higher order Ex34-3 X has a binomial distribution b(n p)

Thus its mgf is

When n=1 X has a Bernoulli distribution

1717

Negative Binomial Distribution Let X be the number of Bernoulli trials to observe the rth succ

ess X has a negative binomial distribution

Its pmf g(x) is

If r=1 X has a geometric distribution with

1818

Geometric Distribution X has a geometric distribution with the pmf

P(X gt k) P(X lek)

Memory-less (EX3412)

Ex34-4 Fruit fliesrsquo eyes with frac14white and frac34red The probability of checking at least 4 flies to observe a white eye is

P(Xge4) = P(Xgt3) = (frac34)3= 04219 The probability of checking at most 4 flies to observe a white eye

isP(Xle4) = 1-(frac34)3=06836 The probability of finding the first white eye on the 4th fly checked is

P(X=4) = pq4-1= 01055 lt= P(Xle4) -P(Xle3)

Ex34-4 For a basketball player with 80 free throw X is the minimum number of throws for a total of 10 free throws Its pmf is μ= rp= 1008 = 125 σ2= rqp2= 10(02)(08)2= 3125

1919

mgf rArr pdf By Maclaurinrsquosseries expansion (ref p632)

If the moments of X E(Xr) = M(r)(0) are known M(t) is thus determined

rArr pdf can be obtained by rewriting M(t) as the weighted sum of eldquoxrdquot

Ex34-7 If the moments of X are E(Xr) = 08 r=123hellip Then M(t) can be determined as

Therefore P(X=0)=02 and P(X=1)=08

2020

Poisson Process Def35-1 An approximate Poisson process with parameter λgt0

The numbers of changes occurring in non-overlapping intervals are independent

The probability of exactly one change in a sufficiently short interval of length h is approximately λh

The probability of two or more changes in a sufficiently short interval is essentially zero

Determine pmf During the unit interval of length 1 there are x changes For nraquox we partition the unit interval into n subintervals of length 1n The probability of x changes in the unit interval equivThe probability of one

change in each of exactly x of these n subintervals The probability of one change in each subinterval is roughly λ(1n) The probability of two or more changes in each subinterval is essentially 0 The change occurrence or not in each subinterval becomes a Bernoulli trial

Thus for a sequence of n Bernoulli trials with probability p = λnP(X=x) can be approximated by (binomial)

2121

Poisson DistributionPoisson Distribution

2222

ExamplesExamples Ex35-1 X has a Poisson distribution with a mean of λ=5Ex35-1 X has a Poisson distribution with a mean of λ=5

Table III on p 652 lists selected values of the distributionTable III on p 652 lists selected values of the distribution P(Xle6) = 0762P(Xle6) = 0762 P(Xgt5) = 1-P(Xle5) = 1-0616 =0384P(Xgt5) = 1-P(Xle5) = 1-0616 =0384 P(X=6) = P(Xle6)-P(Xle5) = 0762-0616 = 0146P(X=6) = P(Xle6)-P(Xle5) = 0762-0616 = 0146

Ex35-2 The Poisson probability histogramsEx35-2 The Poisson probability histograms

232310

)()(

)(

xx

etxf

tx

9900559

9936192

1)1(

1

1

21

12100559 2

)(

n

iinn

n

iin xxsx

More ExamplesMore Examples Empirical data vs Theoretical formula (Ex35-3)Empirical data vs Theoretical formula (Ex35-3)

X is the number of αparticles emittedX is the number of αparticles emitted by barium-133 in 1 sec and counted by barium-133 in 1 sec and counted by a Geiger counterby a Geiger counter

100 observations are made100 observations are made

Generally with unit interval of Generally with unit interval of length t the Poisson pmf islength t the Poisson pmf is

10486101048610 Ex35-4 Assume a tape flaw is a Poisson distribution with a Ex35-4 Assume a tape flaw is a Poisson distribution with a

mean of λ= 11200 flaw per feetmean of λ= 11200 flaw per feet What is the distribution of X the number of flaws in a 4800-foot rollWhat is the distribution of X the number of flaws in a 4800-foot roll E(X)=48001200=4E(X)=48001200=4 P(X=0) = eP(X=0) = e-4-4= 0018= 0018 By Table III on p 652 P(Xle4) By Table III on p 652 P(Xle4) =0 629=0 629

10

)4()(

)4(

xx

exf

x

2424

8570

)2()3( 210020

3

0

)2(

x

x

x

eXP

Poisson with λ=np can simulate Binomial for large n aPoisson with λ=np can simulate Binomial for large n and small pnd small p (μ σ(μ σ22)=(λ λ) asymp(np npq))=(λ λ) asymp(np npq)

Ex35-6 Bulbs with 2 defective rate Ex35-6 Bulbs with 2 defective rate The probability that a box of 100 bulbs contains at most 3 The probability that a box of 100 bulbs contains at most 3 defdef

ective bulbs isective bulbs is

By Binomial the tedious computation will result inBy Binomial the tedious computation will result in

Ex35-7 p160 the comparisons of Binomial and PoisEx35-7 p160 the comparisons of Binomial and Poisson distributionsson distributions

When Poisson asymp BinomialWhen Poisson asymp Binomial

3

0

100100 8590)980()020()3(x

xxxCXP

2525

xnxqpx

nxXPxf

)()(

Ex35-8 Among a lot of 1000 parts n=100 parts are taken at randoEx35-8 Among a lot of 1000 parts n=100 parts are taken at random wo replacement The lot is accepted if no more than 2 of100 parts m wo replacement The lot is accepted if no more than 2 of100 parts taken are defective Assume p is the defective ratetaken are defective Assume p is the defective rate Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=

2)2)HypergeometricHypergeometric

NN11=1000p=1000pN=1000 N=1000 NN22=N-N=N-N11

Since N is large it makes little difference if sampling is done with or withSince N is large it makes little difference if sampling is done with or without replacement simulated by BinomrArrout replacement simulated by BinomrArr ialial

n Bernoulli trialsn Bernoulli trials For small p p=0~1 simulated by PoissorArrFor small p p=0~1 simulated by PoissorArr nn

OC(001)=092OC(001)=092OC(002)=0677OC(002)=0677OC(003)=0423OC(003)=0423OC(005)=0125OC(005)=0125OC(010)=0003OC(010)=0003

When Poisson asymp Binomial Binomial When Poisson asymp Binomial Binomial asymp Hypergeometric Hypergeometricasymp Hypergeometric Hypergeometric

21 )()(21

21

NxnNxnxxXPxfn

NN

xn

N

x

N

2

0

)100(

)100()2( 100

x

px

x

epXPpnp

1414

Comparisons Empirical Data vs Analytical Formula

Ex33-11 b(5 5) has μ= np=

25 σ2= npq= 125 Simulate the model for 100 times

2 3 2 hellip = 247 srArr 2=15243

Suppose an urn has N1success balls and N2 failure balls N = N1+ N2

Let p = N1N and X be the number of success balls in a random sample of size n taken from this urn

If the sampling is done one at a time with replacement X follows b(n p) If the sampling is done without replacement X has a hypergeometric distrib

ution with pmf

If N is large and n is relative small it makes little difference if the sampling is done with or without replacement (See Fig33-4)

x

1515

Moment-Generating Function (mgf )

Def34-1 X is a random variable of the discrete type with pmf f(x) and space S If there is a positive integer h st

E(etX) = sumx Sisin etxf(x) exists and is finite for ndashhlttlth thenM(t) = E(etX) is called the moment-generating function of X ndashE(etX) exists and is finite for ndashhlttlth hArrM(r)(t) exist at t=0 r=123hellip ndashUnique association pmf hArrmgf

Sharing the same mgf two random variables have the same distribution of probability

Ex34-1 X has mgf M(t) = et(36)+ e2t(26)+ e3t(16) From eldquoxrdquot Its pmf has f(0)=0 f(1)=36 f(2)=26 f(3)=16 f(4)=0 hellip Therefore f(x)=(4-x)6 x=123

Ex34-2 X has mgf M(t) = et2(1-et2) tltln2 (1-z)-1= 1 + z + z2+ z3+ hellip |z|lt1

1616

Application of Application of mgf

M(t) = E(etX) = sumx Sisin etxf(x) exists and is finite for ndashhlttlthM(t) = sumx Sisin xetxf(x) M(0) = sumx Sisin xf(x) = E(X) M(t) = sumx Sisin x2etxf(x) hellip M(0) = sumx Sisin x2f(x) = E(X2) hellip M(r)(t) = sumx Sisin xretxf(x) M(r)(0) = sumx Sisin xrf(x) = E(Xr) as t=0

M(t) must be formulated (in closed form) to get its derivatives of higher order Ex34-3 X has a binomial distribution b(n p)

Thus its mgf is

When n=1 X has a Bernoulli distribution

1717

Negative Binomial Distribution Let X be the number of Bernoulli trials to observe the rth succ

ess X has a negative binomial distribution

Its pmf g(x) is

If r=1 X has a geometric distribution with

1818

Geometric Distribution X has a geometric distribution with the pmf

P(X gt k) P(X lek)

Memory-less (EX3412)

Ex34-4 Fruit fliesrsquo eyes with frac14white and frac34red The probability of checking at least 4 flies to observe a white eye is

P(Xge4) = P(Xgt3) = (frac34)3= 04219 The probability of checking at most 4 flies to observe a white eye

isP(Xle4) = 1-(frac34)3=06836 The probability of finding the first white eye on the 4th fly checked is

P(X=4) = pq4-1= 01055 lt= P(Xle4) -P(Xle3)

Ex34-4 For a basketball player with 80 free throw X is the minimum number of throws for a total of 10 free throws Its pmf is μ= rp= 1008 = 125 σ2= rqp2= 10(02)(08)2= 3125

1919

mgf rArr pdf By Maclaurinrsquosseries expansion (ref p632)

If the moments of X E(Xr) = M(r)(0) are known M(t) is thus determined

rArr pdf can be obtained by rewriting M(t) as the weighted sum of eldquoxrdquot

Ex34-7 If the moments of X are E(Xr) = 08 r=123hellip Then M(t) can be determined as

Therefore P(X=0)=02 and P(X=1)=08

2020

Poisson Process Def35-1 An approximate Poisson process with parameter λgt0

The numbers of changes occurring in non-overlapping intervals are independent

The probability of exactly one change in a sufficiently short interval of length h is approximately λh

The probability of two or more changes in a sufficiently short interval is essentially zero

Determine pmf During the unit interval of length 1 there are x changes For nraquox we partition the unit interval into n subintervals of length 1n The probability of x changes in the unit interval equivThe probability of one

change in each of exactly x of these n subintervals The probability of one change in each subinterval is roughly λ(1n) The probability of two or more changes in each subinterval is essentially 0 The change occurrence or not in each subinterval becomes a Bernoulli trial

Thus for a sequence of n Bernoulli trials with probability p = λnP(X=x) can be approximated by (binomial)

2121

Poisson DistributionPoisson Distribution

2222

ExamplesExamples Ex35-1 X has a Poisson distribution with a mean of λ=5Ex35-1 X has a Poisson distribution with a mean of λ=5

Table III on p 652 lists selected values of the distributionTable III on p 652 lists selected values of the distribution P(Xle6) = 0762P(Xle6) = 0762 P(Xgt5) = 1-P(Xle5) = 1-0616 =0384P(Xgt5) = 1-P(Xle5) = 1-0616 =0384 P(X=6) = P(Xle6)-P(Xle5) = 0762-0616 = 0146P(X=6) = P(Xle6)-P(Xle5) = 0762-0616 = 0146

Ex35-2 The Poisson probability histogramsEx35-2 The Poisson probability histograms

232310

)()(

)(

xx

etxf

tx

9900559

9936192

1)1(

1

1

21

12100559 2

)(

n

iinn

n

iin xxsx

More ExamplesMore Examples Empirical data vs Theoretical formula (Ex35-3)Empirical data vs Theoretical formula (Ex35-3)

X is the number of αparticles emittedX is the number of αparticles emitted by barium-133 in 1 sec and counted by barium-133 in 1 sec and counted by a Geiger counterby a Geiger counter

100 observations are made100 observations are made

Generally with unit interval of Generally with unit interval of length t the Poisson pmf islength t the Poisson pmf is

10486101048610 Ex35-4 Assume a tape flaw is a Poisson distribution with a Ex35-4 Assume a tape flaw is a Poisson distribution with a

mean of λ= 11200 flaw per feetmean of λ= 11200 flaw per feet What is the distribution of X the number of flaws in a 4800-foot rollWhat is the distribution of X the number of flaws in a 4800-foot roll E(X)=48001200=4E(X)=48001200=4 P(X=0) = eP(X=0) = e-4-4= 0018= 0018 By Table III on p 652 P(Xle4) By Table III on p 652 P(Xle4) =0 629=0 629

10

)4()(

)4(

xx

exf

x

2424

8570

)2()3( 210020

3

0

)2(

x

x

x

eXP

Poisson with λ=np can simulate Binomial for large n aPoisson with λ=np can simulate Binomial for large n and small pnd small p (μ σ(μ σ22)=(λ λ) asymp(np npq))=(λ λ) asymp(np npq)

Ex35-6 Bulbs with 2 defective rate Ex35-6 Bulbs with 2 defective rate The probability that a box of 100 bulbs contains at most 3 The probability that a box of 100 bulbs contains at most 3 defdef

ective bulbs isective bulbs is

By Binomial the tedious computation will result inBy Binomial the tedious computation will result in

Ex35-7 p160 the comparisons of Binomial and PoisEx35-7 p160 the comparisons of Binomial and Poisson distributionsson distributions

When Poisson asymp BinomialWhen Poisson asymp Binomial

3

0

100100 8590)980()020()3(x

xxxCXP

2525

xnxqpx

nxXPxf

)()(

Ex35-8 Among a lot of 1000 parts n=100 parts are taken at randoEx35-8 Among a lot of 1000 parts n=100 parts are taken at random wo replacement The lot is accepted if no more than 2 of100 parts m wo replacement The lot is accepted if no more than 2 of100 parts taken are defective Assume p is the defective ratetaken are defective Assume p is the defective rate Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=

2)2)HypergeometricHypergeometric

NN11=1000p=1000pN=1000 N=1000 NN22=N-N=N-N11

Since N is large it makes little difference if sampling is done with or withSince N is large it makes little difference if sampling is done with or without replacement simulated by BinomrArrout replacement simulated by BinomrArr ialial

n Bernoulli trialsn Bernoulli trials For small p p=0~1 simulated by PoissorArrFor small p p=0~1 simulated by PoissorArr nn

OC(001)=092OC(001)=092OC(002)=0677OC(002)=0677OC(003)=0423OC(003)=0423OC(005)=0125OC(005)=0125OC(010)=0003OC(010)=0003

When Poisson asymp Binomial Binomial When Poisson asymp Binomial Binomial asymp Hypergeometric Hypergeometricasymp Hypergeometric Hypergeometric

21 )()(21

21

NxnNxnxxXPxfn

NN

xn

N

x

N

2

0

)100(

)100()2( 100

x

px

x

epXPpnp

1515

Moment-Generating Function (mgf )

Def34-1 X is a random variable of the discrete type with pmf f(x) and space S If there is a positive integer h st

E(etX) = sumx Sisin etxf(x) exists and is finite for ndashhlttlth thenM(t) = E(etX) is called the moment-generating function of X ndashE(etX) exists and is finite for ndashhlttlth hArrM(r)(t) exist at t=0 r=123hellip ndashUnique association pmf hArrmgf

Sharing the same mgf two random variables have the same distribution of probability

Ex34-1 X has mgf M(t) = et(36)+ e2t(26)+ e3t(16) From eldquoxrdquot Its pmf has f(0)=0 f(1)=36 f(2)=26 f(3)=16 f(4)=0 hellip Therefore f(x)=(4-x)6 x=123

Ex34-2 X has mgf M(t) = et2(1-et2) tltln2 (1-z)-1= 1 + z + z2+ z3+ hellip |z|lt1

1616

Application of Application of mgf

M(t) = E(etX) = sumx Sisin etxf(x) exists and is finite for ndashhlttlthM(t) = sumx Sisin xetxf(x) M(0) = sumx Sisin xf(x) = E(X) M(t) = sumx Sisin x2etxf(x) hellip M(0) = sumx Sisin x2f(x) = E(X2) hellip M(r)(t) = sumx Sisin xretxf(x) M(r)(0) = sumx Sisin xrf(x) = E(Xr) as t=0

M(t) must be formulated (in closed form) to get its derivatives of higher order Ex34-3 X has a binomial distribution b(n p)

Thus its mgf is

When n=1 X has a Bernoulli distribution

1717

Negative Binomial Distribution Let X be the number of Bernoulli trials to observe the rth succ

ess X has a negative binomial distribution

Its pmf g(x) is

If r=1 X has a geometric distribution with

1818

Geometric Distribution X has a geometric distribution with the pmf

P(X gt k) P(X lek)

Memory-less (EX3412)

Ex34-4 Fruit fliesrsquo eyes with frac14white and frac34red The probability of checking at least 4 flies to observe a white eye is

P(Xge4) = P(Xgt3) = (frac34)3= 04219 The probability of checking at most 4 flies to observe a white eye

isP(Xle4) = 1-(frac34)3=06836 The probability of finding the first white eye on the 4th fly checked is

P(X=4) = pq4-1= 01055 lt= P(Xle4) -P(Xle3)

Ex34-4 For a basketball player with 80 free throw X is the minimum number of throws for a total of 10 free throws Its pmf is μ= rp= 1008 = 125 σ2= rqp2= 10(02)(08)2= 3125

1919

mgf rArr pdf By Maclaurinrsquosseries expansion (ref p632)

If the moments of X E(Xr) = M(r)(0) are known M(t) is thus determined

rArr pdf can be obtained by rewriting M(t) as the weighted sum of eldquoxrdquot

Ex34-7 If the moments of X are E(Xr) = 08 r=123hellip Then M(t) can be determined as

Therefore P(X=0)=02 and P(X=1)=08

2020

Poisson Process Def35-1 An approximate Poisson process with parameter λgt0

The numbers of changes occurring in non-overlapping intervals are independent

The probability of exactly one change in a sufficiently short interval of length h is approximately λh

The probability of two or more changes in a sufficiently short interval is essentially zero

Determine pmf During the unit interval of length 1 there are x changes For nraquox we partition the unit interval into n subintervals of length 1n The probability of x changes in the unit interval equivThe probability of one

change in each of exactly x of these n subintervals The probability of one change in each subinterval is roughly λ(1n) The probability of two or more changes in each subinterval is essentially 0 The change occurrence or not in each subinterval becomes a Bernoulli trial

Thus for a sequence of n Bernoulli trials with probability p = λnP(X=x) can be approximated by (binomial)

2121

Poisson DistributionPoisson Distribution

2222

ExamplesExamples Ex35-1 X has a Poisson distribution with a mean of λ=5Ex35-1 X has a Poisson distribution with a mean of λ=5

Table III on p 652 lists selected values of the distributionTable III on p 652 lists selected values of the distribution P(Xle6) = 0762P(Xle6) = 0762 P(Xgt5) = 1-P(Xle5) = 1-0616 =0384P(Xgt5) = 1-P(Xle5) = 1-0616 =0384 P(X=6) = P(Xle6)-P(Xle5) = 0762-0616 = 0146P(X=6) = P(Xle6)-P(Xle5) = 0762-0616 = 0146

Ex35-2 The Poisson probability histogramsEx35-2 The Poisson probability histograms

232310

)()(

)(

xx

etxf

tx

9900559

9936192

1)1(

1

1

21

12100559 2

)(

n

iinn

n

iin xxsx

More ExamplesMore Examples Empirical data vs Theoretical formula (Ex35-3)Empirical data vs Theoretical formula (Ex35-3)

X is the number of αparticles emittedX is the number of αparticles emitted by barium-133 in 1 sec and counted by barium-133 in 1 sec and counted by a Geiger counterby a Geiger counter

100 observations are made100 observations are made

Generally with unit interval of Generally with unit interval of length t the Poisson pmf islength t the Poisson pmf is

10486101048610 Ex35-4 Assume a tape flaw is a Poisson distribution with a Ex35-4 Assume a tape flaw is a Poisson distribution with a

mean of λ= 11200 flaw per feetmean of λ= 11200 flaw per feet What is the distribution of X the number of flaws in a 4800-foot rollWhat is the distribution of X the number of flaws in a 4800-foot roll E(X)=48001200=4E(X)=48001200=4 P(X=0) = eP(X=0) = e-4-4= 0018= 0018 By Table III on p 652 P(Xle4) By Table III on p 652 P(Xle4) =0 629=0 629

10

)4()(

)4(

xx

exf

x

2424

8570

)2()3( 210020

3

0

)2(

x

x

x

eXP

Poisson with λ=np can simulate Binomial for large n aPoisson with λ=np can simulate Binomial for large n and small pnd small p (μ σ(μ σ22)=(λ λ) asymp(np npq))=(λ λ) asymp(np npq)

Ex35-6 Bulbs with 2 defective rate Ex35-6 Bulbs with 2 defective rate The probability that a box of 100 bulbs contains at most 3 The probability that a box of 100 bulbs contains at most 3 defdef

ective bulbs isective bulbs is

By Binomial the tedious computation will result inBy Binomial the tedious computation will result in

Ex35-7 p160 the comparisons of Binomial and PoisEx35-7 p160 the comparisons of Binomial and Poisson distributionsson distributions

When Poisson asymp BinomialWhen Poisson asymp Binomial

3

0

100100 8590)980()020()3(x

xxxCXP

2525

xnxqpx

nxXPxf

)()(

Ex35-8 Among a lot of 1000 parts n=100 parts are taken at randoEx35-8 Among a lot of 1000 parts n=100 parts are taken at random wo replacement The lot is accepted if no more than 2 of100 parts m wo replacement The lot is accepted if no more than 2 of100 parts taken are defective Assume p is the defective ratetaken are defective Assume p is the defective rate Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=

2)2)HypergeometricHypergeometric

NN11=1000p=1000pN=1000 N=1000 NN22=N-N=N-N11

Since N is large it makes little difference if sampling is done with or withSince N is large it makes little difference if sampling is done with or without replacement simulated by BinomrArrout replacement simulated by BinomrArr ialial

n Bernoulli trialsn Bernoulli trials For small p p=0~1 simulated by PoissorArrFor small p p=0~1 simulated by PoissorArr nn

OC(001)=092OC(001)=092OC(002)=0677OC(002)=0677OC(003)=0423OC(003)=0423OC(005)=0125OC(005)=0125OC(010)=0003OC(010)=0003

When Poisson asymp Binomial Binomial When Poisson asymp Binomial Binomial asymp Hypergeometric Hypergeometricasymp Hypergeometric Hypergeometric

21 )()(21

21

NxnNxnxxXPxfn

NN

xn

N

x

N

2

0

)100(

)100()2( 100

x

px

x

epXPpnp

1616

Application of Application of mgf

M(t) = E(etX) = sumx Sisin etxf(x) exists and is finite for ndashhlttlthM(t) = sumx Sisin xetxf(x) M(0) = sumx Sisin xf(x) = E(X) M(t) = sumx Sisin x2etxf(x) hellip M(0) = sumx Sisin x2f(x) = E(X2) hellip M(r)(t) = sumx Sisin xretxf(x) M(r)(0) = sumx Sisin xrf(x) = E(Xr) as t=0

M(t) must be formulated (in closed form) to get its derivatives of higher order Ex34-3 X has a binomial distribution b(n p)

Thus its mgf is

When n=1 X has a Bernoulli distribution

1717

Negative Binomial Distribution Let X be the number of Bernoulli trials to observe the rth succ

ess X has a negative binomial distribution

Its pmf g(x) is

If r=1 X has a geometric distribution with

1818

Geometric Distribution X has a geometric distribution with the pmf

P(X gt k) P(X lek)

Memory-less (EX3412)

Ex34-4 Fruit fliesrsquo eyes with frac14white and frac34red The probability of checking at least 4 flies to observe a white eye is

P(Xge4) = P(Xgt3) = (frac34)3= 04219 The probability of checking at most 4 flies to observe a white eye

isP(Xle4) = 1-(frac34)3=06836 The probability of finding the first white eye on the 4th fly checked is

P(X=4) = pq4-1= 01055 lt= P(Xle4) -P(Xle3)

Ex34-4 For a basketball player with 80 free throw X is the minimum number of throws for a total of 10 free throws Its pmf is μ= rp= 1008 = 125 σ2= rqp2= 10(02)(08)2= 3125

1919

mgf rArr pdf By Maclaurinrsquosseries expansion (ref p632)

If the moments of X E(Xr) = M(r)(0) are known M(t) is thus determined

rArr pdf can be obtained by rewriting M(t) as the weighted sum of eldquoxrdquot

Ex34-7 If the moments of X are E(Xr) = 08 r=123hellip Then M(t) can be determined as

Therefore P(X=0)=02 and P(X=1)=08

2020

Poisson Process Def35-1 An approximate Poisson process with parameter λgt0

The numbers of changes occurring in non-overlapping intervals are independent

The probability of exactly one change in a sufficiently short interval of length h is approximately λh

The probability of two or more changes in a sufficiently short interval is essentially zero

Determine pmf During the unit interval of length 1 there are x changes For nraquox we partition the unit interval into n subintervals of length 1n The probability of x changes in the unit interval equivThe probability of one

change in each of exactly x of these n subintervals The probability of one change in each subinterval is roughly λ(1n) The probability of two or more changes in each subinterval is essentially 0 The change occurrence or not in each subinterval becomes a Bernoulli trial

Thus for a sequence of n Bernoulli trials with probability p = λnP(X=x) can be approximated by (binomial)

2121

Poisson DistributionPoisson Distribution

2222

ExamplesExamples Ex35-1 X has a Poisson distribution with a mean of λ=5Ex35-1 X has a Poisson distribution with a mean of λ=5

Table III on p 652 lists selected values of the distributionTable III on p 652 lists selected values of the distribution P(Xle6) = 0762P(Xle6) = 0762 P(Xgt5) = 1-P(Xle5) = 1-0616 =0384P(Xgt5) = 1-P(Xle5) = 1-0616 =0384 P(X=6) = P(Xle6)-P(Xle5) = 0762-0616 = 0146P(X=6) = P(Xle6)-P(Xle5) = 0762-0616 = 0146

Ex35-2 The Poisson probability histogramsEx35-2 The Poisson probability histograms

232310

)()(

)(

xx

etxf

tx

9900559

9936192

1)1(

1

1

21

12100559 2

)(

n

iinn

n

iin xxsx

More ExamplesMore Examples Empirical data vs Theoretical formula (Ex35-3)Empirical data vs Theoretical formula (Ex35-3)

X is the number of αparticles emittedX is the number of αparticles emitted by barium-133 in 1 sec and counted by barium-133 in 1 sec and counted by a Geiger counterby a Geiger counter

100 observations are made100 observations are made

Generally with unit interval of Generally with unit interval of length t the Poisson pmf islength t the Poisson pmf is

10486101048610 Ex35-4 Assume a tape flaw is a Poisson distribution with a Ex35-4 Assume a tape flaw is a Poisson distribution with a

mean of λ= 11200 flaw per feetmean of λ= 11200 flaw per feet What is the distribution of X the number of flaws in a 4800-foot rollWhat is the distribution of X the number of flaws in a 4800-foot roll E(X)=48001200=4E(X)=48001200=4 P(X=0) = eP(X=0) = e-4-4= 0018= 0018 By Table III on p 652 P(Xle4) By Table III on p 652 P(Xle4) =0 629=0 629

10

)4()(

)4(

xx

exf

x

2424

8570

)2()3( 210020

3

0

)2(

x

x

x

eXP

Poisson with λ=np can simulate Binomial for large n aPoisson with λ=np can simulate Binomial for large n and small pnd small p (μ σ(μ σ22)=(λ λ) asymp(np npq))=(λ λ) asymp(np npq)

Ex35-6 Bulbs with 2 defective rate Ex35-6 Bulbs with 2 defective rate The probability that a box of 100 bulbs contains at most 3 The probability that a box of 100 bulbs contains at most 3 defdef

ective bulbs isective bulbs is

By Binomial the tedious computation will result inBy Binomial the tedious computation will result in

Ex35-7 p160 the comparisons of Binomial and PoisEx35-7 p160 the comparisons of Binomial and Poisson distributionsson distributions

When Poisson asymp BinomialWhen Poisson asymp Binomial

3

0

100100 8590)980()020()3(x

xxxCXP

2525

xnxqpx

nxXPxf

)()(

Ex35-8 Among a lot of 1000 parts n=100 parts are taken at randoEx35-8 Among a lot of 1000 parts n=100 parts are taken at random wo replacement The lot is accepted if no more than 2 of100 parts m wo replacement The lot is accepted if no more than 2 of100 parts taken are defective Assume p is the defective ratetaken are defective Assume p is the defective rate Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=

2)2)HypergeometricHypergeometric

NN11=1000p=1000pN=1000 N=1000 NN22=N-N=N-N11

Since N is large it makes little difference if sampling is done with or withSince N is large it makes little difference if sampling is done with or without replacement simulated by BinomrArrout replacement simulated by BinomrArr ialial

n Bernoulli trialsn Bernoulli trials For small p p=0~1 simulated by PoissorArrFor small p p=0~1 simulated by PoissorArr nn

OC(001)=092OC(001)=092OC(002)=0677OC(002)=0677OC(003)=0423OC(003)=0423OC(005)=0125OC(005)=0125OC(010)=0003OC(010)=0003

When Poisson asymp Binomial Binomial When Poisson asymp Binomial Binomial asymp Hypergeometric Hypergeometricasymp Hypergeometric Hypergeometric

21 )()(21

21

NxnNxnxxXPxfn

NN

xn

N

x

N

2

0

)100(

)100()2( 100

x

px

x

epXPpnp

1717

Negative Binomial Distribution Let X be the number of Bernoulli trials to observe the rth succ

ess X has a negative binomial distribution

Its pmf g(x) is

If r=1 X has a geometric distribution with

1818

Geometric Distribution X has a geometric distribution with the pmf

P(X gt k) P(X lek)

Memory-less (EX3412)

Ex34-4 Fruit fliesrsquo eyes with frac14white and frac34red The probability of checking at least 4 flies to observe a white eye is

P(Xge4) = P(Xgt3) = (frac34)3= 04219 The probability of checking at most 4 flies to observe a white eye

isP(Xle4) = 1-(frac34)3=06836 The probability of finding the first white eye on the 4th fly checked is

P(X=4) = pq4-1= 01055 lt= P(Xle4) -P(Xle3)

Ex34-4 For a basketball player with 80 free throw X is the minimum number of throws for a total of 10 free throws Its pmf is μ= rp= 1008 = 125 σ2= rqp2= 10(02)(08)2= 3125

1919

mgf rArr pdf By Maclaurinrsquosseries expansion (ref p632)

If the moments of X E(Xr) = M(r)(0) are known M(t) is thus determined

rArr pdf can be obtained by rewriting M(t) as the weighted sum of eldquoxrdquot

Ex34-7 If the moments of X are E(Xr) = 08 r=123hellip Then M(t) can be determined as

Therefore P(X=0)=02 and P(X=1)=08

2020

Poisson Process Def35-1 An approximate Poisson process with parameter λgt0

The numbers of changes occurring in non-overlapping intervals are independent

The probability of exactly one change in a sufficiently short interval of length h is approximately λh

The probability of two or more changes in a sufficiently short interval is essentially zero

Determine pmf During the unit interval of length 1 there are x changes For nraquox we partition the unit interval into n subintervals of length 1n The probability of x changes in the unit interval equivThe probability of one

change in each of exactly x of these n subintervals The probability of one change in each subinterval is roughly λ(1n) The probability of two or more changes in each subinterval is essentially 0 The change occurrence or not in each subinterval becomes a Bernoulli trial

Thus for a sequence of n Bernoulli trials with probability p = λnP(X=x) can be approximated by (binomial)

2121

Poisson DistributionPoisson Distribution

2222

ExamplesExamples Ex35-1 X has a Poisson distribution with a mean of λ=5Ex35-1 X has a Poisson distribution with a mean of λ=5

Table III on p 652 lists selected values of the distributionTable III on p 652 lists selected values of the distribution P(Xle6) = 0762P(Xle6) = 0762 P(Xgt5) = 1-P(Xle5) = 1-0616 =0384P(Xgt5) = 1-P(Xle5) = 1-0616 =0384 P(X=6) = P(Xle6)-P(Xle5) = 0762-0616 = 0146P(X=6) = P(Xle6)-P(Xle5) = 0762-0616 = 0146

Ex35-2 The Poisson probability histogramsEx35-2 The Poisson probability histograms

232310

)()(

)(

xx

etxf

tx

9900559

9936192

1)1(

1

1

21

12100559 2

)(

n

iinn

n

iin xxsx

More ExamplesMore Examples Empirical data vs Theoretical formula (Ex35-3)Empirical data vs Theoretical formula (Ex35-3)

X is the number of αparticles emittedX is the number of αparticles emitted by barium-133 in 1 sec and counted by barium-133 in 1 sec and counted by a Geiger counterby a Geiger counter

100 observations are made100 observations are made

Generally with unit interval of Generally with unit interval of length t the Poisson pmf islength t the Poisson pmf is

10486101048610 Ex35-4 Assume a tape flaw is a Poisson distribution with a Ex35-4 Assume a tape flaw is a Poisson distribution with a

mean of λ= 11200 flaw per feetmean of λ= 11200 flaw per feet What is the distribution of X the number of flaws in a 4800-foot rollWhat is the distribution of X the number of flaws in a 4800-foot roll E(X)=48001200=4E(X)=48001200=4 P(X=0) = eP(X=0) = e-4-4= 0018= 0018 By Table III on p 652 P(Xle4) By Table III on p 652 P(Xle4) =0 629=0 629

10

)4()(

)4(

xx

exf

x

2424

8570

)2()3( 210020

3

0

)2(

x

x

x

eXP

Poisson with λ=np can simulate Binomial for large n aPoisson with λ=np can simulate Binomial for large n and small pnd small p (μ σ(μ σ22)=(λ λ) asymp(np npq))=(λ λ) asymp(np npq)

Ex35-6 Bulbs with 2 defective rate Ex35-6 Bulbs with 2 defective rate The probability that a box of 100 bulbs contains at most 3 The probability that a box of 100 bulbs contains at most 3 defdef

ective bulbs isective bulbs is

By Binomial the tedious computation will result inBy Binomial the tedious computation will result in

Ex35-7 p160 the comparisons of Binomial and PoisEx35-7 p160 the comparisons of Binomial and Poisson distributionsson distributions

When Poisson asymp BinomialWhen Poisson asymp Binomial

3

0

100100 8590)980()020()3(x

xxxCXP

2525

xnxqpx

nxXPxf

)()(

Ex35-8 Among a lot of 1000 parts n=100 parts are taken at randoEx35-8 Among a lot of 1000 parts n=100 parts are taken at random wo replacement The lot is accepted if no more than 2 of100 parts m wo replacement The lot is accepted if no more than 2 of100 parts taken are defective Assume p is the defective ratetaken are defective Assume p is the defective rate Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=

2)2)HypergeometricHypergeometric

NN11=1000p=1000pN=1000 N=1000 NN22=N-N=N-N11

Since N is large it makes little difference if sampling is done with or withSince N is large it makes little difference if sampling is done with or without replacement simulated by BinomrArrout replacement simulated by BinomrArr ialial

n Bernoulli trialsn Bernoulli trials For small p p=0~1 simulated by PoissorArrFor small p p=0~1 simulated by PoissorArr nn

OC(001)=092OC(001)=092OC(002)=0677OC(002)=0677OC(003)=0423OC(003)=0423OC(005)=0125OC(005)=0125OC(010)=0003OC(010)=0003

When Poisson asymp Binomial Binomial When Poisson asymp Binomial Binomial asymp Hypergeometric Hypergeometricasymp Hypergeometric Hypergeometric

21 )()(21

21

NxnNxnxxXPxfn

NN

xn

N

x

N

2

0

)100(

)100()2( 100

x

px

x

epXPpnp

1818

Geometric Distribution X has a geometric distribution with the pmf

P(X gt k) P(X lek)

Memory-less (EX3412)

Ex34-4 Fruit fliesrsquo eyes with frac14white and frac34red The probability of checking at least 4 flies to observe a white eye is

P(Xge4) = P(Xgt3) = (frac34)3= 04219 The probability of checking at most 4 flies to observe a white eye

isP(Xle4) = 1-(frac34)3=06836 The probability of finding the first white eye on the 4th fly checked is

P(X=4) = pq4-1= 01055 lt= P(Xle4) -P(Xle3)

Ex34-4 For a basketball player with 80 free throw X is the minimum number of throws for a total of 10 free throws Its pmf is μ= rp= 1008 = 125 σ2= rqp2= 10(02)(08)2= 3125

1919

mgf rArr pdf By Maclaurinrsquosseries expansion (ref p632)

If the moments of X E(Xr) = M(r)(0) are known M(t) is thus determined

rArr pdf can be obtained by rewriting M(t) as the weighted sum of eldquoxrdquot

Ex34-7 If the moments of X are E(Xr) = 08 r=123hellip Then M(t) can be determined as

Therefore P(X=0)=02 and P(X=1)=08

2020

Poisson Process Def35-1 An approximate Poisson process with parameter λgt0

The numbers of changes occurring in non-overlapping intervals are independent

The probability of exactly one change in a sufficiently short interval of length h is approximately λh

The probability of two or more changes in a sufficiently short interval is essentially zero

Determine pmf During the unit interval of length 1 there are x changes For nraquox we partition the unit interval into n subintervals of length 1n The probability of x changes in the unit interval equivThe probability of one

change in each of exactly x of these n subintervals The probability of one change in each subinterval is roughly λ(1n) The probability of two or more changes in each subinterval is essentially 0 The change occurrence or not in each subinterval becomes a Bernoulli trial

Thus for a sequence of n Bernoulli trials with probability p = λnP(X=x) can be approximated by (binomial)

2121

Poisson DistributionPoisson Distribution

2222

ExamplesExamples Ex35-1 X has a Poisson distribution with a mean of λ=5Ex35-1 X has a Poisson distribution with a mean of λ=5

Table III on p 652 lists selected values of the distributionTable III on p 652 lists selected values of the distribution P(Xle6) = 0762P(Xle6) = 0762 P(Xgt5) = 1-P(Xle5) = 1-0616 =0384P(Xgt5) = 1-P(Xle5) = 1-0616 =0384 P(X=6) = P(Xle6)-P(Xle5) = 0762-0616 = 0146P(X=6) = P(Xle6)-P(Xle5) = 0762-0616 = 0146

Ex35-2 The Poisson probability histogramsEx35-2 The Poisson probability histograms

232310

)()(

)(

xx

etxf

tx

9900559

9936192

1)1(

1

1

21

12100559 2

)(

n

iinn

n

iin xxsx

More ExamplesMore Examples Empirical data vs Theoretical formula (Ex35-3)Empirical data vs Theoretical formula (Ex35-3)

X is the number of αparticles emittedX is the number of αparticles emitted by barium-133 in 1 sec and counted by barium-133 in 1 sec and counted by a Geiger counterby a Geiger counter

100 observations are made100 observations are made

Generally with unit interval of Generally with unit interval of length t the Poisson pmf islength t the Poisson pmf is

10486101048610 Ex35-4 Assume a tape flaw is a Poisson distribution with a Ex35-4 Assume a tape flaw is a Poisson distribution with a

mean of λ= 11200 flaw per feetmean of λ= 11200 flaw per feet What is the distribution of X the number of flaws in a 4800-foot rollWhat is the distribution of X the number of flaws in a 4800-foot roll E(X)=48001200=4E(X)=48001200=4 P(X=0) = eP(X=0) = e-4-4= 0018= 0018 By Table III on p 652 P(Xle4) By Table III on p 652 P(Xle4) =0 629=0 629

10

)4()(

)4(

xx

exf

x

2424

8570

)2()3( 210020

3

0

)2(

x

x

x

eXP

Poisson with λ=np can simulate Binomial for large n aPoisson with λ=np can simulate Binomial for large n and small pnd small p (μ σ(μ σ22)=(λ λ) asymp(np npq))=(λ λ) asymp(np npq)

Ex35-6 Bulbs with 2 defective rate Ex35-6 Bulbs with 2 defective rate The probability that a box of 100 bulbs contains at most 3 The probability that a box of 100 bulbs contains at most 3 defdef

ective bulbs isective bulbs is

By Binomial the tedious computation will result inBy Binomial the tedious computation will result in

Ex35-7 p160 the comparisons of Binomial and PoisEx35-7 p160 the comparisons of Binomial and Poisson distributionsson distributions

When Poisson asymp BinomialWhen Poisson asymp Binomial

3

0

100100 8590)980()020()3(x

xxxCXP

2525

xnxqpx

nxXPxf

)()(

Ex35-8 Among a lot of 1000 parts n=100 parts are taken at randoEx35-8 Among a lot of 1000 parts n=100 parts are taken at random wo replacement The lot is accepted if no more than 2 of100 parts m wo replacement The lot is accepted if no more than 2 of100 parts taken are defective Assume p is the defective ratetaken are defective Assume p is the defective rate Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=

2)2)HypergeometricHypergeometric

NN11=1000p=1000pN=1000 N=1000 NN22=N-N=N-N11

Since N is large it makes little difference if sampling is done with or withSince N is large it makes little difference if sampling is done with or without replacement simulated by BinomrArrout replacement simulated by BinomrArr ialial

n Bernoulli trialsn Bernoulli trials For small p p=0~1 simulated by PoissorArrFor small p p=0~1 simulated by PoissorArr nn

OC(001)=092OC(001)=092OC(002)=0677OC(002)=0677OC(003)=0423OC(003)=0423OC(005)=0125OC(005)=0125OC(010)=0003OC(010)=0003

When Poisson asymp Binomial Binomial When Poisson asymp Binomial Binomial asymp Hypergeometric Hypergeometricasymp Hypergeometric Hypergeometric

21 )()(21

21

NxnNxnxxXPxfn

NN

xn

N

x

N

2

0

)100(

)100()2( 100

x

px

x

epXPpnp

1919

mgf rArr pdf By Maclaurinrsquosseries expansion (ref p632)

If the moments of X E(Xr) = M(r)(0) are known M(t) is thus determined

rArr pdf can be obtained by rewriting M(t) as the weighted sum of eldquoxrdquot

Ex34-7 If the moments of X are E(Xr) = 08 r=123hellip Then M(t) can be determined as

Therefore P(X=0)=02 and P(X=1)=08

2020

Poisson Process Def35-1 An approximate Poisson process with parameter λgt0

The numbers of changes occurring in non-overlapping intervals are independent

The probability of exactly one change in a sufficiently short interval of length h is approximately λh

The probability of two or more changes in a sufficiently short interval is essentially zero

Determine pmf During the unit interval of length 1 there are x changes For nraquox we partition the unit interval into n subintervals of length 1n The probability of x changes in the unit interval equivThe probability of one

change in each of exactly x of these n subintervals The probability of one change in each subinterval is roughly λ(1n) The probability of two or more changes in each subinterval is essentially 0 The change occurrence or not in each subinterval becomes a Bernoulli trial

Thus for a sequence of n Bernoulli trials with probability p = λnP(X=x) can be approximated by (binomial)

2121

Poisson DistributionPoisson Distribution

2222

ExamplesExamples Ex35-1 X has a Poisson distribution with a mean of λ=5Ex35-1 X has a Poisson distribution with a mean of λ=5

Table III on p 652 lists selected values of the distributionTable III on p 652 lists selected values of the distribution P(Xle6) = 0762P(Xle6) = 0762 P(Xgt5) = 1-P(Xle5) = 1-0616 =0384P(Xgt5) = 1-P(Xle5) = 1-0616 =0384 P(X=6) = P(Xle6)-P(Xle5) = 0762-0616 = 0146P(X=6) = P(Xle6)-P(Xle5) = 0762-0616 = 0146

Ex35-2 The Poisson probability histogramsEx35-2 The Poisson probability histograms

232310

)()(

)(

xx

etxf

tx

9900559

9936192

1)1(

1

1

21

12100559 2

)(

n

iinn

n

iin xxsx

More ExamplesMore Examples Empirical data vs Theoretical formula (Ex35-3)Empirical data vs Theoretical formula (Ex35-3)

X is the number of αparticles emittedX is the number of αparticles emitted by barium-133 in 1 sec and counted by barium-133 in 1 sec and counted by a Geiger counterby a Geiger counter

100 observations are made100 observations are made

Generally with unit interval of Generally with unit interval of length t the Poisson pmf islength t the Poisson pmf is

10486101048610 Ex35-4 Assume a tape flaw is a Poisson distribution with a Ex35-4 Assume a tape flaw is a Poisson distribution with a

mean of λ= 11200 flaw per feetmean of λ= 11200 flaw per feet What is the distribution of X the number of flaws in a 4800-foot rollWhat is the distribution of X the number of flaws in a 4800-foot roll E(X)=48001200=4E(X)=48001200=4 P(X=0) = eP(X=0) = e-4-4= 0018= 0018 By Table III on p 652 P(Xle4) By Table III on p 652 P(Xle4) =0 629=0 629

10

)4()(

)4(

xx

exf

x

2424

8570

)2()3( 210020

3

0

)2(

x

x

x

eXP

Poisson with λ=np can simulate Binomial for large n aPoisson with λ=np can simulate Binomial for large n and small pnd small p (μ σ(μ σ22)=(λ λ) asymp(np npq))=(λ λ) asymp(np npq)

Ex35-6 Bulbs with 2 defective rate Ex35-6 Bulbs with 2 defective rate The probability that a box of 100 bulbs contains at most 3 The probability that a box of 100 bulbs contains at most 3 defdef

ective bulbs isective bulbs is

By Binomial the tedious computation will result inBy Binomial the tedious computation will result in

Ex35-7 p160 the comparisons of Binomial and PoisEx35-7 p160 the comparisons of Binomial and Poisson distributionsson distributions

When Poisson asymp BinomialWhen Poisson asymp Binomial

3

0

100100 8590)980()020()3(x

xxxCXP

2525

xnxqpx

nxXPxf

)()(

Ex35-8 Among a lot of 1000 parts n=100 parts are taken at randoEx35-8 Among a lot of 1000 parts n=100 parts are taken at random wo replacement The lot is accepted if no more than 2 of100 parts m wo replacement The lot is accepted if no more than 2 of100 parts taken are defective Assume p is the defective ratetaken are defective Assume p is the defective rate Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=

2)2)HypergeometricHypergeometric

NN11=1000p=1000pN=1000 N=1000 NN22=N-N=N-N11

Since N is large it makes little difference if sampling is done with or withSince N is large it makes little difference if sampling is done with or without replacement simulated by BinomrArrout replacement simulated by BinomrArr ialial

n Bernoulli trialsn Bernoulli trials For small p p=0~1 simulated by PoissorArrFor small p p=0~1 simulated by PoissorArr nn

OC(001)=092OC(001)=092OC(002)=0677OC(002)=0677OC(003)=0423OC(003)=0423OC(005)=0125OC(005)=0125OC(010)=0003OC(010)=0003

When Poisson asymp Binomial Binomial When Poisson asymp Binomial Binomial asymp Hypergeometric Hypergeometricasymp Hypergeometric Hypergeometric

21 )()(21

21

NxnNxnxxXPxfn

NN

xn

N

x

N

2

0

)100(

)100()2( 100

x

px

x

epXPpnp

2020

Poisson Process Def35-1 An approximate Poisson process with parameter λgt0

The numbers of changes occurring in non-overlapping intervals are independent

The probability of exactly one change in a sufficiently short interval of length h is approximately λh

The probability of two or more changes in a sufficiently short interval is essentially zero

Determine pmf During the unit interval of length 1 there are x changes For nraquox we partition the unit interval into n subintervals of length 1n The probability of x changes in the unit interval equivThe probability of one

change in each of exactly x of these n subintervals The probability of one change in each subinterval is roughly λ(1n) The probability of two or more changes in each subinterval is essentially 0 The change occurrence or not in each subinterval becomes a Bernoulli trial

Thus for a sequence of n Bernoulli trials with probability p = λnP(X=x) can be approximated by (binomial)

2121

Poisson DistributionPoisson Distribution

2222

ExamplesExamples Ex35-1 X has a Poisson distribution with a mean of λ=5Ex35-1 X has a Poisson distribution with a mean of λ=5

Table III on p 652 lists selected values of the distributionTable III on p 652 lists selected values of the distribution P(Xle6) = 0762P(Xle6) = 0762 P(Xgt5) = 1-P(Xle5) = 1-0616 =0384P(Xgt5) = 1-P(Xle5) = 1-0616 =0384 P(X=6) = P(Xle6)-P(Xle5) = 0762-0616 = 0146P(X=6) = P(Xle6)-P(Xle5) = 0762-0616 = 0146

Ex35-2 The Poisson probability histogramsEx35-2 The Poisson probability histograms

232310

)()(

)(

xx

etxf

tx

9900559

9936192

1)1(

1

1

21

12100559 2

)(

n

iinn

n

iin xxsx

More ExamplesMore Examples Empirical data vs Theoretical formula (Ex35-3)Empirical data vs Theoretical formula (Ex35-3)

X is the number of αparticles emittedX is the number of αparticles emitted by barium-133 in 1 sec and counted by barium-133 in 1 sec and counted by a Geiger counterby a Geiger counter

100 observations are made100 observations are made

Generally with unit interval of Generally with unit interval of length t the Poisson pmf islength t the Poisson pmf is

10486101048610 Ex35-4 Assume a tape flaw is a Poisson distribution with a Ex35-4 Assume a tape flaw is a Poisson distribution with a

mean of λ= 11200 flaw per feetmean of λ= 11200 flaw per feet What is the distribution of X the number of flaws in a 4800-foot rollWhat is the distribution of X the number of flaws in a 4800-foot roll E(X)=48001200=4E(X)=48001200=4 P(X=0) = eP(X=0) = e-4-4= 0018= 0018 By Table III on p 652 P(Xle4) By Table III on p 652 P(Xle4) =0 629=0 629

10

)4()(

)4(

xx

exf

x

2424

8570

)2()3( 210020

3

0

)2(

x

x

x

eXP

Poisson with λ=np can simulate Binomial for large n aPoisson with λ=np can simulate Binomial for large n and small pnd small p (μ σ(μ σ22)=(λ λ) asymp(np npq))=(λ λ) asymp(np npq)

Ex35-6 Bulbs with 2 defective rate Ex35-6 Bulbs with 2 defective rate The probability that a box of 100 bulbs contains at most 3 The probability that a box of 100 bulbs contains at most 3 defdef

ective bulbs isective bulbs is

By Binomial the tedious computation will result inBy Binomial the tedious computation will result in

Ex35-7 p160 the comparisons of Binomial and PoisEx35-7 p160 the comparisons of Binomial and Poisson distributionsson distributions

When Poisson asymp BinomialWhen Poisson asymp Binomial

3

0

100100 8590)980()020()3(x

xxxCXP

2525

xnxqpx

nxXPxf

)()(

Ex35-8 Among a lot of 1000 parts n=100 parts are taken at randoEx35-8 Among a lot of 1000 parts n=100 parts are taken at random wo replacement The lot is accepted if no more than 2 of100 parts m wo replacement The lot is accepted if no more than 2 of100 parts taken are defective Assume p is the defective ratetaken are defective Assume p is the defective rate Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=

2)2)HypergeometricHypergeometric

NN11=1000p=1000pN=1000 N=1000 NN22=N-N=N-N11

Since N is large it makes little difference if sampling is done with or withSince N is large it makes little difference if sampling is done with or without replacement simulated by BinomrArrout replacement simulated by BinomrArr ialial

n Bernoulli trialsn Bernoulli trials For small p p=0~1 simulated by PoissorArrFor small p p=0~1 simulated by PoissorArr nn

OC(001)=092OC(001)=092OC(002)=0677OC(002)=0677OC(003)=0423OC(003)=0423OC(005)=0125OC(005)=0125OC(010)=0003OC(010)=0003

When Poisson asymp Binomial Binomial When Poisson asymp Binomial Binomial asymp Hypergeometric Hypergeometricasymp Hypergeometric Hypergeometric

21 )()(21

21

NxnNxnxxXPxfn

NN

xn

N

x

N

2

0

)100(

)100()2( 100

x

px

x

epXPpnp

2121

Poisson DistributionPoisson Distribution

2222

ExamplesExamples Ex35-1 X has a Poisson distribution with a mean of λ=5Ex35-1 X has a Poisson distribution with a mean of λ=5

Table III on p 652 lists selected values of the distributionTable III on p 652 lists selected values of the distribution P(Xle6) = 0762P(Xle6) = 0762 P(Xgt5) = 1-P(Xle5) = 1-0616 =0384P(Xgt5) = 1-P(Xle5) = 1-0616 =0384 P(X=6) = P(Xle6)-P(Xle5) = 0762-0616 = 0146P(X=6) = P(Xle6)-P(Xle5) = 0762-0616 = 0146

Ex35-2 The Poisson probability histogramsEx35-2 The Poisson probability histograms

232310

)()(

)(

xx

etxf

tx

9900559

9936192

1)1(

1

1

21

12100559 2

)(

n

iinn

n

iin xxsx

More ExamplesMore Examples Empirical data vs Theoretical formula (Ex35-3)Empirical data vs Theoretical formula (Ex35-3)

X is the number of αparticles emittedX is the number of αparticles emitted by barium-133 in 1 sec and counted by barium-133 in 1 sec and counted by a Geiger counterby a Geiger counter

100 observations are made100 observations are made

Generally with unit interval of Generally with unit interval of length t the Poisson pmf islength t the Poisson pmf is

10486101048610 Ex35-4 Assume a tape flaw is a Poisson distribution with a Ex35-4 Assume a tape flaw is a Poisson distribution with a

mean of λ= 11200 flaw per feetmean of λ= 11200 flaw per feet What is the distribution of X the number of flaws in a 4800-foot rollWhat is the distribution of X the number of flaws in a 4800-foot roll E(X)=48001200=4E(X)=48001200=4 P(X=0) = eP(X=0) = e-4-4= 0018= 0018 By Table III on p 652 P(Xle4) By Table III on p 652 P(Xle4) =0 629=0 629

10

)4()(

)4(

xx

exf

x

2424

8570

)2()3( 210020

3

0

)2(

x

x

x

eXP

Poisson with λ=np can simulate Binomial for large n aPoisson with λ=np can simulate Binomial for large n and small pnd small p (μ σ(μ σ22)=(λ λ) asymp(np npq))=(λ λ) asymp(np npq)

Ex35-6 Bulbs with 2 defective rate Ex35-6 Bulbs with 2 defective rate The probability that a box of 100 bulbs contains at most 3 The probability that a box of 100 bulbs contains at most 3 defdef

ective bulbs isective bulbs is

By Binomial the tedious computation will result inBy Binomial the tedious computation will result in

Ex35-7 p160 the comparisons of Binomial and PoisEx35-7 p160 the comparisons of Binomial and Poisson distributionsson distributions

When Poisson asymp BinomialWhen Poisson asymp Binomial

3

0

100100 8590)980()020()3(x

xxxCXP

2525

xnxqpx

nxXPxf

)()(

Ex35-8 Among a lot of 1000 parts n=100 parts are taken at randoEx35-8 Among a lot of 1000 parts n=100 parts are taken at random wo replacement The lot is accepted if no more than 2 of100 parts m wo replacement The lot is accepted if no more than 2 of100 parts taken are defective Assume p is the defective ratetaken are defective Assume p is the defective rate Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=

2)2)HypergeometricHypergeometric

NN11=1000p=1000pN=1000 N=1000 NN22=N-N=N-N11

Since N is large it makes little difference if sampling is done with or withSince N is large it makes little difference if sampling is done with or without replacement simulated by BinomrArrout replacement simulated by BinomrArr ialial

n Bernoulli trialsn Bernoulli trials For small p p=0~1 simulated by PoissorArrFor small p p=0~1 simulated by PoissorArr nn

OC(001)=092OC(001)=092OC(002)=0677OC(002)=0677OC(003)=0423OC(003)=0423OC(005)=0125OC(005)=0125OC(010)=0003OC(010)=0003

When Poisson asymp Binomial Binomial When Poisson asymp Binomial Binomial asymp Hypergeometric Hypergeometricasymp Hypergeometric Hypergeometric

21 )()(21

21

NxnNxnxxXPxfn

NN

xn

N

x

N

2

0

)100(

)100()2( 100

x

px

x

epXPpnp

2222

ExamplesExamples Ex35-1 X has a Poisson distribution with a mean of λ=5Ex35-1 X has a Poisson distribution with a mean of λ=5

Table III on p 652 lists selected values of the distributionTable III on p 652 lists selected values of the distribution P(Xle6) = 0762P(Xle6) = 0762 P(Xgt5) = 1-P(Xle5) = 1-0616 =0384P(Xgt5) = 1-P(Xle5) = 1-0616 =0384 P(X=6) = P(Xle6)-P(Xle5) = 0762-0616 = 0146P(X=6) = P(Xle6)-P(Xle5) = 0762-0616 = 0146

Ex35-2 The Poisson probability histogramsEx35-2 The Poisson probability histograms

232310

)()(

)(

xx

etxf

tx

9900559

9936192

1)1(

1

1

21

12100559 2

)(

n

iinn

n

iin xxsx

More ExamplesMore Examples Empirical data vs Theoretical formula (Ex35-3)Empirical data vs Theoretical formula (Ex35-3)

X is the number of αparticles emittedX is the number of αparticles emitted by barium-133 in 1 sec and counted by barium-133 in 1 sec and counted by a Geiger counterby a Geiger counter

100 observations are made100 observations are made

Generally with unit interval of Generally with unit interval of length t the Poisson pmf islength t the Poisson pmf is

10486101048610 Ex35-4 Assume a tape flaw is a Poisson distribution with a Ex35-4 Assume a tape flaw is a Poisson distribution with a

mean of λ= 11200 flaw per feetmean of λ= 11200 flaw per feet What is the distribution of X the number of flaws in a 4800-foot rollWhat is the distribution of X the number of flaws in a 4800-foot roll E(X)=48001200=4E(X)=48001200=4 P(X=0) = eP(X=0) = e-4-4= 0018= 0018 By Table III on p 652 P(Xle4) By Table III on p 652 P(Xle4) =0 629=0 629

10

)4()(

)4(

xx

exf

x

2424

8570

)2()3( 210020

3

0

)2(

x

x

x

eXP

Poisson with λ=np can simulate Binomial for large n aPoisson with λ=np can simulate Binomial for large n and small pnd small p (μ σ(μ σ22)=(λ λ) asymp(np npq))=(λ λ) asymp(np npq)

Ex35-6 Bulbs with 2 defective rate Ex35-6 Bulbs with 2 defective rate The probability that a box of 100 bulbs contains at most 3 The probability that a box of 100 bulbs contains at most 3 defdef

ective bulbs isective bulbs is

By Binomial the tedious computation will result inBy Binomial the tedious computation will result in

Ex35-7 p160 the comparisons of Binomial and PoisEx35-7 p160 the comparisons of Binomial and Poisson distributionsson distributions

When Poisson asymp BinomialWhen Poisson asymp Binomial

3

0

100100 8590)980()020()3(x

xxxCXP

2525

xnxqpx

nxXPxf

)()(

Ex35-8 Among a lot of 1000 parts n=100 parts are taken at randoEx35-8 Among a lot of 1000 parts n=100 parts are taken at random wo replacement The lot is accepted if no more than 2 of100 parts m wo replacement The lot is accepted if no more than 2 of100 parts taken are defective Assume p is the defective ratetaken are defective Assume p is the defective rate Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=

2)2)HypergeometricHypergeometric

NN11=1000p=1000pN=1000 N=1000 NN22=N-N=N-N11

Since N is large it makes little difference if sampling is done with or withSince N is large it makes little difference if sampling is done with or without replacement simulated by BinomrArrout replacement simulated by BinomrArr ialial

n Bernoulli trialsn Bernoulli trials For small p p=0~1 simulated by PoissorArrFor small p p=0~1 simulated by PoissorArr nn

OC(001)=092OC(001)=092OC(002)=0677OC(002)=0677OC(003)=0423OC(003)=0423OC(005)=0125OC(005)=0125OC(010)=0003OC(010)=0003

When Poisson asymp Binomial Binomial When Poisson asymp Binomial Binomial asymp Hypergeometric Hypergeometricasymp Hypergeometric Hypergeometric

21 )()(21

21

NxnNxnxxXPxfn

NN

xn

N

x

N

2

0

)100(

)100()2( 100

x

px

x

epXPpnp

232310

)()(

)(

xx

etxf

tx

9900559

9936192

1)1(

1

1

21

12100559 2

)(

n

iinn

n

iin xxsx

More ExamplesMore Examples Empirical data vs Theoretical formula (Ex35-3)Empirical data vs Theoretical formula (Ex35-3)

X is the number of αparticles emittedX is the number of αparticles emitted by barium-133 in 1 sec and counted by barium-133 in 1 sec and counted by a Geiger counterby a Geiger counter

100 observations are made100 observations are made

Generally with unit interval of Generally with unit interval of length t the Poisson pmf islength t the Poisson pmf is

10486101048610 Ex35-4 Assume a tape flaw is a Poisson distribution with a Ex35-4 Assume a tape flaw is a Poisson distribution with a

mean of λ= 11200 flaw per feetmean of λ= 11200 flaw per feet What is the distribution of X the number of flaws in a 4800-foot rollWhat is the distribution of X the number of flaws in a 4800-foot roll E(X)=48001200=4E(X)=48001200=4 P(X=0) = eP(X=0) = e-4-4= 0018= 0018 By Table III on p 652 P(Xle4) By Table III on p 652 P(Xle4) =0 629=0 629

10

)4()(

)4(

xx

exf

x

2424

8570

)2()3( 210020

3

0

)2(

x

x

x

eXP

Poisson with λ=np can simulate Binomial for large n aPoisson with λ=np can simulate Binomial for large n and small pnd small p (μ σ(μ σ22)=(λ λ) asymp(np npq))=(λ λ) asymp(np npq)

Ex35-6 Bulbs with 2 defective rate Ex35-6 Bulbs with 2 defective rate The probability that a box of 100 bulbs contains at most 3 The probability that a box of 100 bulbs contains at most 3 defdef

ective bulbs isective bulbs is

By Binomial the tedious computation will result inBy Binomial the tedious computation will result in

Ex35-7 p160 the comparisons of Binomial and PoisEx35-7 p160 the comparisons of Binomial and Poisson distributionsson distributions

When Poisson asymp BinomialWhen Poisson asymp Binomial

3

0

100100 8590)980()020()3(x

xxxCXP

2525

xnxqpx

nxXPxf

)()(

Ex35-8 Among a lot of 1000 parts n=100 parts are taken at randoEx35-8 Among a lot of 1000 parts n=100 parts are taken at random wo replacement The lot is accepted if no more than 2 of100 parts m wo replacement The lot is accepted if no more than 2 of100 parts taken are defective Assume p is the defective ratetaken are defective Assume p is the defective rate Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=

2)2)HypergeometricHypergeometric

NN11=1000p=1000pN=1000 N=1000 NN22=N-N=N-N11

Since N is large it makes little difference if sampling is done with or withSince N is large it makes little difference if sampling is done with or without replacement simulated by BinomrArrout replacement simulated by BinomrArr ialial

n Bernoulli trialsn Bernoulli trials For small p p=0~1 simulated by PoissorArrFor small p p=0~1 simulated by PoissorArr nn

OC(001)=092OC(001)=092OC(002)=0677OC(002)=0677OC(003)=0423OC(003)=0423OC(005)=0125OC(005)=0125OC(010)=0003OC(010)=0003

When Poisson asymp Binomial Binomial When Poisson asymp Binomial Binomial asymp Hypergeometric Hypergeometricasymp Hypergeometric Hypergeometric

21 )()(21

21

NxnNxnxxXPxfn

NN

xn

N

x

N

2

0

)100(

)100()2( 100

x

px

x

epXPpnp

2424

8570

)2()3( 210020

3

0

)2(

x

x

x

eXP

Poisson with λ=np can simulate Binomial for large n aPoisson with λ=np can simulate Binomial for large n and small pnd small p (μ σ(μ σ22)=(λ λ) asymp(np npq))=(λ λ) asymp(np npq)

Ex35-6 Bulbs with 2 defective rate Ex35-6 Bulbs with 2 defective rate The probability that a box of 100 bulbs contains at most 3 The probability that a box of 100 bulbs contains at most 3 defdef

ective bulbs isective bulbs is

By Binomial the tedious computation will result inBy Binomial the tedious computation will result in

Ex35-7 p160 the comparisons of Binomial and PoisEx35-7 p160 the comparisons of Binomial and Poisson distributionsson distributions

When Poisson asymp BinomialWhen Poisson asymp Binomial

3

0

100100 8590)980()020()3(x

xxxCXP

2525

xnxqpx

nxXPxf

)()(

Ex35-8 Among a lot of 1000 parts n=100 parts are taken at randoEx35-8 Among a lot of 1000 parts n=100 parts are taken at random wo replacement The lot is accepted if no more than 2 of100 parts m wo replacement The lot is accepted if no more than 2 of100 parts taken are defective Assume p is the defective ratetaken are defective Assume p is the defective rate Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=

2)2)HypergeometricHypergeometric

NN11=1000p=1000pN=1000 N=1000 NN22=N-N=N-N11

Since N is large it makes little difference if sampling is done with or withSince N is large it makes little difference if sampling is done with or without replacement simulated by BinomrArrout replacement simulated by BinomrArr ialial

n Bernoulli trialsn Bernoulli trials For small p p=0~1 simulated by PoissorArrFor small p p=0~1 simulated by PoissorArr nn

OC(001)=092OC(001)=092OC(002)=0677OC(002)=0677OC(003)=0423OC(003)=0423OC(005)=0125OC(005)=0125OC(010)=0003OC(010)=0003

When Poisson asymp Binomial Binomial When Poisson asymp Binomial Binomial asymp Hypergeometric Hypergeometricasymp Hypergeometric Hypergeometric

21 )()(21

21

NxnNxnxxXPxfn

NN

xn

N

x

N

2

0

)100(

)100()2( 100

x

px

x

epXPpnp

2525

xnxqpx

nxXPxf

)()(

Ex35-8 Among a lot of 1000 parts n=100 parts are taken at randoEx35-8 Among a lot of 1000 parts n=100 parts are taken at random wo replacement The lot is accepted if no more than 2 of100 parts m wo replacement The lot is accepted if no more than 2 of100 parts taken are defective Assume p is the defective ratetaken are defective Assume p is the defective rate Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=Operating characteristic curve OC(p) = P(Xle2) = P(X=0)+P(X=1)+P(X=

2)2)HypergeometricHypergeometric

NN11=1000p=1000pN=1000 N=1000 NN22=N-N=N-N11

Since N is large it makes little difference if sampling is done with or withSince N is large it makes little difference if sampling is done with or without replacement simulated by BinomrArrout replacement simulated by BinomrArr ialial

n Bernoulli trialsn Bernoulli trials For small p p=0~1 simulated by PoissorArrFor small p p=0~1 simulated by PoissorArr nn

OC(001)=092OC(001)=092OC(002)=0677OC(002)=0677OC(003)=0423OC(003)=0423OC(005)=0125OC(005)=0125OC(010)=0003OC(010)=0003

When Poisson asymp Binomial Binomial When Poisson asymp Binomial Binomial asymp Hypergeometric Hypergeometricasymp Hypergeometric Hypergeometric

21 )()(21

21

NxnNxnxxXPxfn

NN

xn

N

x

N

2

0

)100(

)100()2( 100

x

px

x

epXPpnp