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Chapter 3, Section 9 Discrete Random Variables. Moment-Generating Functions.  John J Currano, 12/15/2008. æ. ö. k. (  c ) k  j E [ Y j ]. E [ ( Y – c ) k ]. k. =. å. ç. ÷. j. è. ø. =. j. 0. E [ ( Y – E ( Y ) ) k ] = E [ ( Y –  ) k ]. m. =. k.

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Chapter 3 section 9 discrete random variables

Chapter 3, Section 9Discrete Random Variables

Moment-Generating Functions

 John J Currano,12/15/2008


æ

ö

k

(c)kj E [Y j]

E[(Y – c)k]

k

=

å

ç

÷

j

è

ø

=

j

0

E[(Y – E(Y))k] = E[(Y – )k]

m

=

k

Definitions. Let Y be a discrete random variable with probability function, p(y), and let k = 1, 2, 3, . Then:

is the kth moment of Y (about the origin).

is the kth moment of Y about c.

is the kth central moment of Y.

Notes.=E(Y) = the first moment of Y ( =1)

 2=V(Y) = the second central moment of Y ( =2 )

=E(Y2) – [E(Y)]2=2 – (1)2.

What is the first central moment of Y ?


Definition. If Y is a discrete random variable, its moment-generating function (mgf) is the function

provided this function of t exists (converges) in some interval around 0.

Note. The mgf, when it exists, gives a better characterization of the distribution than the mean and variance – it completely determines the distribution in the sense that if two random variables have the same mgf, then they have the same distribution.

So to find the distribution of a random variable, we can find its mgf and compare it to a list of known mgfs; if the mgf is in the list, we have found the distribution.

We shall also see that the mgf gives us an easy way to find the distribution’s mean and variance.


(

)

tY

ty

å

=

=

m

(

t

)

E

e

e

p

(

y

)

y

k

¥

(

ty

)

å

å

=

p

(

y

)

k

!

=

y

k

0

k

¥

(

ty

)

å

å

=

p

(

y

)

k

!

=

k

0

y

k

k

¥

t

t

k

å

å

=

factoring out

y

p

(

y

)

k

!

k

!

=

k

0

y

(

)

k

¥

t

k

å

=

E

Y

k

!

=

k

0

Why it is called the moment-generating function:

Using the Maclaurin Series for ex at the right, we derive

by Theorem 3.2

using the Maclaurin series for ety

interchanging the summations

by Theorem 3.2


-

k

2

¥

(

)

(

)

(

)

-

×

k

(

k

1

)

t

2

1

k

2

2

¢

¢

¢

¢

=

=

=

å

m

(

t

)

E

Y

;

m

(

0

)

E

Y

E

Y

;

k

!

2

!

=

k

2

Why it is called the moment-generating function:

Thus,

et cetera.


t

2

t

3

t

3

1

2

=

+

+

m

(

t

)

e

e

e

6

6

6

t

2

t

3

t

¢

9

1

4

14

7

¢

=

+

+

=

=

=

m

(

t

)

e

e

e

,

so

E

(

Y

)

m

(

0

)

.

6

6

6

6

3

¢

¢

=

m

(

t

)

(

)

t

2

t

3

t

2

8

36

1

27

¢

¢

+

+

=

=

=

e

e

e

,

so

E

Y

m

(

0

)

6

.

6

6

6

6

(

)

2

2

=

=

5

7

-

6

3

9

Theorem. for k = 1, 2, 3, . . . where mY(t) is the mgf of Y (provided it exists). Also, mY(0) = E(e0Y) = E(1) = 1 for any RV, Y.

Example(p. 142 #3.155). Given that

find: (a) E(Y); (b) V(Y); (c) the distribution of Y.

(a)

(b)

Thus, V(Y) = E(Y2) [E(Y)]


t

2

t

3

t

3

1

2

=

+

+

m

(

t

)

e

e

e

6

6

6

(

)

tY

ty

t

2

t

3

t

3

1

2

=

=

=

+

+

å

m

(

t

)

E

e

e

p

(

y

)

e

e

e

6

6

6

y

3

1

2

,

,

,

6

6

6

Theorem. for k = 1, 2, 3, . . . where mY(t) is the mgf of Y (provided it exists). Also, mY(0) = E(e0Y) = E(1) = 1 for any RV, Y.

Example(p. 142 #3.155). Given that

find: (a) E(Y); (b) V(Y); (c) the distribution of Y.

(c) Since and the

support of Y consists of the coefficients of t in the exponents of the powers of e in the nonzero terms of the power series, Y has support

{1, 2, 3}. The probability that Y assumes each of these values is the coefficient of the exponential in the corresponding term, so

respectively.

Y = 1, 2, 3, with probabilities


æ

ö

n

n

-

ty

y

n

y

ç

÷

å

e

p

q

ç

÷

y

è

ø

=

y

0

(

)

æ

ö

n

n

y

-

t

n

y

ç

÷

å

=

pe

q

ç

÷

y

è

ø

=

y

0

Example. Find the moment-generating function of Y ~ bin(n, p) and use it to find E(Y) and V(Y).

Solution.

Then,

by Theorem 3.2

= (pet+ q)nby the Binomial Theorem


Theorem. If Y ~ bin(n, p), then its mgf is

)

(

-

n

1

t

t

+

×

n

pe

q

pe

-

-

n

1

n

1

¢

Þ

=

=

+

×

×

=

×

×

=

E

(

Y

)

m

(

0

)

n

(

p

q

)

p

1

n

(

1

)

p

1

np

(

)

(

)

-

-

n

1

n

2

é

ù

t

t

t

t

t

+

×

-

+

×

×

+

n

pe

q

pe

n

(

n

1

)

pe

q

pe

pe

¢

¢

=

m

(

t

)

ë

û

(

)

2

¢

¢

Þ

=

=

E

Y

m

(

0

)

-

-

n

2

n

1

-

×

×

×

×

+

×

×

n

(

n

1

)

(

1

)

p

1

p

1

n

(

1

)

p

1

2

2

2

2

=

-

+

=

-

+

n

(

n

1

)

p

np

n

p

np

np

2

2

2

2

2

Þ

=

-

+

-

V(Y) = E(Y2)  [E(Y)]

n

p

np

np

(

np

)

2

=

-

+

=

-

=

np

np

np

(

1

p

)

npq

.

Now differentiate to find E(Y) and V(Y):

(a)

(b)


(

)

2

5,

bin

3

2

3

3

2

æ

ö

æ

ö

5

5

æ

ö

æ

ö

æ

ö

æ

ö

2

1

2

1

ç

÷

ç

÷

+

ç

÷

ç

÷

ç

÷

ç

÷

ç

÷

ç

÷

3

3

3

3

è

ø

è

ø

è

ø

è

ø

2

3

è

ø

è

ø

Example. If the moment-generating function of a random variable, Y,

, find Pr(Y = 2 or 3).

Solution. From the form of the moment-generating function,

, so that

we know that Y ~

P(Y = 2 or 3) =

P(Y = 2) + P(Y = 3) =


Homework Problem Results (pp. 142-143)

3.147 If Y ~ Geom(p), then

3.158 If Y is a random variable with moment-generating function, mY(t), and W = aY + b where a and b are constants, then

mW (t) = maY+b (t) = et bmY (at).


(

)

t

l

-

e

1

=

m

(

t

)

e

.

Y

Other Moment-Generating Functions

1. If Y ~ NegBin(r, p), then

This is most easily proved using results in Chapter 6.

Exercise. Use mY(t) to find E(Y) and V(Y).

2. If Y ~ Poisson(), then

This is Example 3.23 on p. 140, where it is proved and then used (in Example 3.24) to find E(Y) and V(Y).


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