MOMENT GENERATING FUNCTION AND

STATISTICAL DISTRIBUTIONS

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MOMENT GENERATING FUNCTION

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xall

tx

xall

tx

tXX

discreteisXif)x(fe

.contisXifdx)x(fe

)e(E)t(M

The m.g.f. of random variable X is defined as

for t Є (-h,h) for some h>0.

Properties of m.g.f.

• M(0)=E[1]=1

• If a r.v. X has m.g.f. M(t), then Y=aX+b has a m.g.f.

•

• M.g.f does not always exists (e.g. Cauchy distribution)

3

)at(Mebt

.derivativektheisMwhere)0(M)X(E th)k()k(k

Example

• Suppose that X has the following p.d.f.

Find the m.g.f; expectation and variance.

4

0xforxe)x(f x

CHARACTERISTIC FUNCTION

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xall

itx

xall

itx

itXX

discreteisXifxfe

contisXifdxxfe

eEt)(

.)(

)()(

The c.h.f. of random variable X is defined as

for all real numbers t. 1,12 ii

C.h.f. always exists.

Uniqueness

Theorem:

1.If two r.v.s have mg.f.s that exist and are equal, then they have the same distribution.

2.If two r,v,s have the same distribution, then they have the same m.g.f. (if they exist)

Similar statements are true for c.h.f.

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Problem

• It is sometimes the case that exact values of random variables (Y1, Y2, …) cannot be observed, but we can observe they are greater than some fixed value. Let Y1, Y2, … be i.i.d. r.v.s. Let a be a fixed number on real line. For i=1,2,… define,

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aYif

aYifX

i

ii 0

1

Problem, cont.

For example, if a manufacturing process produces parts with strength Yi that are tested to see if they can withstand stress a, then Xi denotes whether the strength is at least a or it is less than a. In such a case, we cannot directly observe the strength Yi of the ith part, but we can observe whether it breaks in stress test.

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Problem, cont.

• Define p=P(Y1≥a) and q=1-p, Sn=X1+X2+…+Xn. Note that, Sn is the number of Y1, …, Yn that exceed a.

i) Define the characteristic function, say , of a r.v. X

ii) Find

iii)Find

iv)Find P(Sn=j)

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)(tX

)(1

tX

)(tnS

Other generating functions

• logM(t) is called cumulant generating function.

• is factorial moment generating function.

• Note:

there is a simple relation between m.g.f. and f.m.g.f.

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)()( XX tEt

)()()()(ln )ln(ln XttX tEeEeEtMX

Other generating functions

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))1)...(1((|)(

...

))1((|)(

)(|)(|)(

1

12

2

11

1

kXXXEtdt

d

XXEtdt

d

XEXtEtdt

d

tXk

k

tX

tX

tX

Example

• Suppose X has the following p.m.f.

Find the expectation and variance of X.

• Solution: Let’s use factorial m.g.f.

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0,...2,1,0,!

)(

xx

exXP

x

Example

13

22

2

22

21

2)1(

1)1(

)1(

0

))(()())1((

))(()()(

))1((|)1(

)(|)1(

!

)()()(

XEXEXXE

XEXEXVar

XXEe

XEe

eeex

ettEt

tt

X

tt

X

tt

x

XX

X

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STATISTICAL DISTRIBUTIONS

Recall

• Random variable: A function defined on the sample space S that associates a real number with each outcome in S.

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Example

• Toss three coins• Sample space S={s1=HHH,s2=HHT,

…,s6=THT,s7=TTH,s8=TTT}

• Define X=number of heads: X(s1)=3,X(s6)=1,X(s8)=0

• Define Y=number of tails before first head: Y(s1)=0, Y(s6)=1, Y(s8)=3

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Random variables

• A random variable is continuous if its CDF, F(x)=P(X≤x), is continuous.

• A random variable is discrete if its CDF, F(x)=P(X≤x), is a step function.

• It is possible for a CDF to have continuous pieces and steps, but we will mostly concentrate on the previous two bullets in this course.

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SOME DISCRETE PROBABILITY

DISTRIBUTIONSDegenerate, Uniform, Bernoulli,

Binomial, Poisson, Negative Binomial, Geometric,

Hypergeometric

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DEGENERATE DISTRIBUTION

• An rv X is degenerate at point k if

1,

0, . .

X kP X x

o w

The cdf:

0,

1,

X kF x P X x

X k

UNIFORM DISTRIBUTION

• A finite number of equally spaced values are equally likely to be observed.

• Example: throw a fair die. P(X=1)=…=P(X=6)=1/6

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,...2,1N;N,...,2,1x;N

1)xX(P

12

)1N)(1N()X(Var;

2

1N)X(E

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BERNOULLI DISTRIBUTION

• A Bernoulli trial is an experiment with only two outcomes. An r.v. X has Bernoulli(p) distribution if

1 with probability ;0 1

0 with probability 1

pX p

p

1p0and;1,0xfor)p1(p)xX(P x1x

BERNOULLI DISTRIBUTION

• P(X=0)=1-p and P(X=1)=p

• E(X)=p

•

22

)1(

)1()1()0(

E(X))-E(XVar(X)22

2

pp

pppp

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BINOMIAL DISTRIBUTION• Define an rv Y by

Y = total number of successes in n Bernoulli trials.

1.There are n trials (n is finite and fixed).2. Each trial can result in a success or a failure.3. The probability p of success is the same for all

the trials.4. All the trials of the experiment are independent.

1

~ , where ~ .n

i ii

Y X Bin n p X Ber p

Let ~ , . ,independent

i iX Bin n p Then

1 21

~ , .k

i ki

X Bin n n n p

BINOMIAL DISTRIBUTION

• Example: • There are black and white balls in a box. Select

and record the color of the ball. Put it back and re-pick (sampling with replacement).

• n: number of independent and identical trials • p: probability of success (e.g. probability of

picking a black ball)• X: number of successes in n trials

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BINOMIAL THEOREM

• For any real numbers x and y and integer n>0

inin

i

n yxi

nyx

0

)(

BINOMIAL DISTRIBUTION

• If Y~Bin(n,p), then

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p)-np(1Var(Y)

npE(Y)

ntY )]p1(pe[)t(M

10,...,1,0)1()(

pnypp

y

nyYP yny

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POISSON DISTRIBUTION

• The number of occurrences in a given time interval can be modeled by the Poisson distribution.

• e.g. number of customers to arrive in a bank between 13:00 and 13:30.

• Another application is in spatial distributions. • e.g. modeling the distribution of bomb hits in an

area or the distribution of fish in a lake.

POISSON DISTRIBUTION

• If X~ Poi(λ), then

• E(X)= Var(X)=λ

•

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]}1)[exp(exp{)( ttM X

0,...2,1,0,!

)(

xx

exXP

x

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Relationship between Binomial and Poisson

~ , with mgf 1nt

XX Bin n p M t pe p

Let =np.

1

lim lim 1

1lim 1

nt

Xn n

ntte

Yn

M t pe p

ee M t

n

The mgf of Poisson()

The limiting distribution of Binomial rv is the Poisson distribution.

NEGATIVE BINOMIAL DISTRIBUTION (PASCAL OR WAITING TIME DISTRIBUTION)

• X: number of Bernoulli trials required to get a fixed number of failures before the r th success; or, alternatively,

• Y: number of Bernoulli trials required to get a fixed number of successes, such as r successes.

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NEGATIVE BINOMIAL DISTRIBUTION (PASCAL OR WAITING TIME DISTRIBUTION)

X~NB(r,p)

1p0,...;1,0x;)p1(px

1xr)xX(P xr

2p

)p1(r)X(Var

p

)p1(r)X(E

rtrX ]e)p1(1[p)t(M

NEGATIVE BINOMIAL DISTRIBUTION

• An alternative form of the pdf:

Note: Y=X+r

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1p0,...;1r,ry;)p1(p1r

1y)yY(P ryr

2p

)p1(r)X(Var)Y(Var

p

rr)X(E)Y(E

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GEOMETRIC DISTRIBUTION• Distribution of the number of Bernoulli trials

required to get the first success.• It is the special case of the Negative Binomial

Distribution r=1.

1

1 , 1,2,x

P X x p p x

X~Geometric(p)

2p

)p1()X(Var

p

1)X(E

• Example: If probability is 0.001 that a light bulb will fail on any given day, then what is the probability that it will last at least 30 days?

• Solution:

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GEOMETRIC DISTRIBUTION

97.0)999.0()001.01(001.0)30X(P 30

31x

1x

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HYPERGEOMETRIC DISTRIBUTION

• A box contains N marbles. Of these, M are red. Suppose that n marbles are drawn randomly from the box without replacement. The distribution of the number of red marbles, x is

, 0,1,...,

M N M

x n xP X x x n

N

n

It is dealing with finite population.

X~Hypergeometric(N,M,n)

HYPERGEOMETRIC DISTRIBUTION

• As N →∞, hypergeometric → binomial.

• In that case, sampling with or without replacement does not make much difference (especially if n/N is small).

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MULTIVARIATE

DISTRIBUTIONS

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EXTENDED HYPERGEOMETRIC DISTRIBUTION

• Suppose that a collection consists of a finite number of items, N and that there are k+1 different types; M1 of type 1, M2 of type 2, and so on. Select n items at random without replacement, and let Xi be the number of items of type i that are selected. The vector X=(X1, X2,…,Xk) has an extended hypergeometric distribution and the joint pdf is

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. xand M where

},...,1,0{ ,

...

,...,,

11k

11k

1

1

2

2

1

1

21

k

ii

k

ii

iik

k

k

k

k

xnMN

Mx

n

N

x

M

x

M

x

M

x

M

xxxf

MULTINOMIAL DISTRIBUTION

• Let E1,E2,...,Ek,Ek+1 be k+1 mutually exclusive and exhaustive events which can occur on any trial of an experiment with P(Ei)=pi,i=1,2,…,k+1. On n independent trials of the experiment, let Xi be the number of occurrences of the event Ei. Then, the vector X=(X1, X2,…,Xk) has a multinomial distribution with joint pdf

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.1p and xwhere

},...,1,0{ ,...!!...!

!,...,,

11k

11k

121121

21121

k

ii

k

ii

ixk

xx

kk

pxn

nxpppxxx

nxxxf k

• Experiment involves drawing with replacement.

• Binomial is a special case of multinomial with k+1=2

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MULTINOMIAL DISTRIBUTION

MULTINOMIAL DISTRIBUTION

• Consider trinomial case for simplicity.

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},...,1,0{ ,)1()!(!!

!, 2121

21212121

21 nxppppxxnxx

nxxf i

xxnxx

ntt

n

x

xn

x

xxnxtxt

XtXt

ppepep

ppepepxxnxx

n

eEttM

)1(

)1()()()!(!!

!

)(),(

2121

0 02121

2121

21

21

1

1

2

212211

2211

MULTINOMIAL DISTRIBUTION

• M.g.f. of X1:

X1~Bin(n,p1)

Similarly, X2~Bin(n,p2)

But, Cov(X1,X2)≠0!

Cov(X1,X2)=?42

ntXt pepeEtM )1()()0,( 110

1111

• Example: Suppose we have a bowl with 10 marbles - 2 red marbles, 3 green marbles, and 5 blue marbles. We randomly select 4 marbles from the bowl, with replacement. What is the probability of selecting 2 green marbles and 2 blue marbles?

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MULTINOMIAL DISTRIBUTION

• n = 4, k+1=3, nred = 0, ngreen = 2, nblue = 2

• pred = 0.2, pgreen = 0.3, pblue = 0.5• P = [ n! / ( n1! * n2! * ... nk! ) ] * ( p1

n1 * p2

n2 * . . . * pk

nk )

P = [ 4! / ( 0! * 2! * 2! ) ] * [ (0.2)0 * (0.3)2 * (0.5)2 ]

P = 0.135

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MULTINOMIAL DISTRIBUTION

Problem

1. a) Does a distribution exist for which the m.g.f. ? If yes, find it. If

no, prove it.

b) Does a distribution exist for which the m.g.f. ? If yes, find it. If no, prove it.

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t

ttM X

1

)(

tX etM )(

Problem

2. An appliance store receives a shipment of 30 microwave ovens, 5 of which are (unknown to the manager) defective. The store manager selects 4 ovens at random, without replacement, and tests to see if they are defective. Let X=number of defectives found. Calculate the pmf and cdf of X.

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Problem

3. Let X denote the number of “do loops” in a Fortran program and Y the number of runs needed for a novice to debug the program. Assume that the joint density for (X,Y) is given in the following table.

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Problem

x/y 1 2 3 4

0 0.059 0.1 0.05 0.001

1 0.093 0.12 0.082 0.003

2 0.065 0.102 0.1 0.01

3 0.05 0.075 0.07 0.0248

Problem

a) Find the probability that a randomly selected program contains at most one “do loop” and requires at least two runs to debug the program.

b) Find E[XY].

c) Find the marginal densities for X and Y. Find the mean and variance for both X and Y.

d) Find the probability that a randomly selected program requires at least two runs to debug given that it contains exactly one “do loop”.

e) Find Cov(X,Y). Find the correlation between X and Y. Based on the observed value correlation, can you claim that X and Y are not independent? Why?

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