# Section Method of Moment Generating Functions by mikesanye

VIEWS: 20 PAGES: 13

• pg 1
```									Chapter 6, Section 5
Transformations of Variables

Method of Moment-Generating
Functions

 John J Currano, 04/16/2010   1
Method of Moment-Generating Functions

The crucial theorem is:

Theorem 6.1 (p. 318): If X and Y are random variables which

both have moment-generating functions, and if

mX(t) = mY(t) for all t in some interval around t = 0,

then X and Y have the same probability distribution.

2
Method of Moment-Generating Functions

Some other useful facts:

1. If U = aY + b, then

                             
m U (t )  E (e tU )  E e aYt  bt  e bt E eY (at )  e bt m Y (at ).

2. If Y1, Y2, … , Yn are independent and U = Y1 + Y2 +  + Yn, then

(
m U (t )  E (e tU )  E e tY1               ) E e
 tY2 L tYn
(     tY1 tY2
e       Le     )
tYn

( )( ) ( )
 E etY1 E etY2 L E e tYn                  by independence

 mY (t ) mY (t ) L mY (t )
1       2         n

3
Method of Moment-Generating Functions

Some other useful facts:
n
3. If Y1, Y2, … , Yn are independent and U   aiYi ,
i 1
n
then m U (t )   m Y (ai t ).
i
i 1

4. If Y1, Y2, … , Yn are independent and identically distributed (iid)

with common distribution Y, i.e., a Random Sample from Y, and

[ ].
U = Y1 + Y2 +  + Yn, then m U (t )  m Y (t )   n

4
Method of Moment-Generating Functions

Example 1. Suppose that Y1, Y2, … , Ym are independent binomial RVs
with Yi ~ bin(ni, p) [same p]. Then for i = 1, 2, … , m,


m Yi (t )  q  pet      ni

m
Let Y   Yi . Then by Property 2 on slide 3,
i 1

( )              n1  n2 L nm
m
mY (t )   m Yi (t )  q  pe t                       ,
i 1

m
so Y has a binomial distribution with n   ni trials and probability of
i 1
success, p. Thus, the sum of independent binomials with the same
probability of success, p, is also binomial.

5
Example 2. Let             Y ~ NegBin(r, p),
X0 = 0, and
Xi = # of trial on which i th success occurs.
Then
Yi = Xi  Xi 1 ~ Geom(p) for i = 1, 2, … , r, and
Y1, Y2, … , Yr are independent.
Also,
r          r
Yi   X i  X i 1  X r  X 0  X r  0  Y .
i 1       i 1
telescoping sum

r
r            pe t      
Thus, m Y (t )   m Y (t )                    by Property 2 on slide 3.
  t       
 1 qe      
i
i 1

We have found the moment-generating functions of the negative
binomial distributions using those of the geometric distributions.
6
r
 pe                t
Example 2. Y ~ NegBin(r, p), m Y (t )            
 1  qe t  .
          
We can now use the mgf of Y to find its mean and variance.
For example,

 pe 0 
E (Y )  mY (0)  r           
r 1

pe 1  qe   pe qe 
0        0         0   0

 1  qe 0 
                                  1  qe 
0 2

r 1
( ) )
 p                  p (  q  pq
1
r      
1 q                  (1  q)2

r 1
 p
r                  (
p pq      )
p                  p2

r
      .
p

7
Example 3. Let Y ~ N( ,  2) and Z  Y    1 Y   .
          
             
Then m Y (t )  exp  t  1  2 t 2 ,
2

         1 
so m Z (t )  exp   t  m Y  t  by Property 1 on slide 3
          

          1             1  
2
 exp   t  exp    t   2  2  t  
1
                         
                        
      t                2 t 2      
 exp  
 

t


         1
2

 2




  exp


( ) 1
2
t2

So Z ~ N(0, 1 ) : Z has a standard normal distribution. We proved this
fact in Chapter 4 using the Distribution Function method – this is simpler.

Transforming a random variable Y by subtracting its mean and dividing
by its standard deviation is called standardizing Y.
8
Example 4. Let Z ~ N(0,1) and Y =  +  Z .

Then m Z (t )  exp      1
2

t2 ,

so m Y (t )  exp  t  m Z  t                by Property 1 on slide 3

 exp  t  exp    (      1
2      )
 t 2

(
 exp  t             1
2
2   t)
2

So Y ~ N( ,  2) : Y has a normal distribution with mean  and
variance  2. We also proved this in Chapter 4 using the
Distribution Function method – again this method is simpler.

Transforming a standard normal random variable Z in this fashion
is a way of simulating other normal random variables.

9
Example 5. If Yi ~ N( i ,  i 2) are independent for i = 1, 2, . . . , n,

i

then m Y (t )  exp  i t    1
2
 i2 t 2    for         1 i  n. Let Y   ai Yi .
m

i 1

Then by the Property 3 on slide 4,
m

i 1
m

i 1
(
m Y (t )   m Yi ( ai t )   exp  i a i t                      1
2      (a t) .
 i2   i)  2

 m                                m                 
 exp  t  a i  i         1
t   2
      ai2  i2   ,
 i 1                 2
i 1              

m             m         
so   Y ~ N   a i  i ,  ai2  i2  . Thus, a linear combination of
 i 1        i 1       

independent normal random variables is also normal.

10
Special Case. Suppose Y1, Y2, . . . , Yn are a random sample from
n
a N(  ,    2)   - distribution, and Y  1    Yi , the sample mean.
n   i 1


Then by Example 5, Y ~ N  ,  2 n .      

Thus the sample mean of a random sample of size n from a normal
distribution also has a normal distribution.

Its mean is the same as that of the original distribution and its
variance is smaller   2/n instead of  2.

This will be used often next semester.

11
Example 6. (p. 319, Example 6.11)

1    z2 2
Let Z ~ N(0,1), so f ( z )           e       , and let Y  Z 2 . Then
2


m Y (t )  E e   tZ2
   
 etz
2   1
2
e z
2
2
dz



1    (1 2t ) z 2
          e                 2
dz
   2


 
1
e                   )dz
 z 2 ( (1 2t ) 1
2

   2

12
Example 6. (p. 319, Example 6.11) Let Z ~ N(0,1) and Y  Z 2 .

Then m Y (t )  
1
e  z2   2(12t )1  dz .
    2

The integrand is proportional to the density function of the normal

distribution with  = 0 and  2 = (1 – 2t)1. Thus, if we multiply by
1
 1  2t     inside the integral and by   1  2t 1 2 outside,
12

we obtain
m Y (t )  1  2t          1  1  2t 
1 2                 1 2
.

Therefore, Y  Z 2 has a c 2 (1) - distribution.

We proved this in Chapter 4 using the Distribution Function method.
Once again, this method is simpler.
13

```
To top