# Probability

Document Sample

```					Generating Functions
The Moments of Y
• We have referred to E(Y) and E(Y2) as the first and
second moments of Y, respectively. In general,
E(Yk) is the kth moment of Y.
• Consider the polynomial where the moments of Y
are incorporated into the coefficients
   k                     2    2    3    3
t                       t E (Y ) t E (Y )
 k ! E (Y )  1  t E (Y )  2!  3! 
k 0
k
Moment Generating Function
• If the sum converges for all t in some interval |t| < b,
the polynomial is called the moment-generating
function, m(t), for the random variable Y.
t 2 E (Y 2 ) t 3 E (Y 3 )
m(t )  1  t E (Y )                           
2!           3!
• And we may note that for each k,
t k  y k p( y )
t k E (Y k )                           (t y )k
      y
         p( y )
k!                 k!           y   k!
Moment Generating Function
• Hence, the moment-generating function is given by
t 2 E (Y 2 ) t 3 E (Y 3 )
m(t )  1  t E (Y )                           
2!           3!

(t y ) k
                p( y )
k 0 y    k!
  (t y)k             May rearrange,
                 p( y )   since finite for
y  k 0 k !                  |t| < b.

  et y p( y )  E[ety ]
y
Moment Generating Function
• That is,
m(t )  E[ety ]   ety p ( y )
y

t 2 E (Y 2 ) t 3 E (Y 3 )
 1  t E (Y )                           
2!           3!
is the polynomial whose coefficients involve the
moments of Y.
The      k th   moment
• To retrieve the kth moment from the MGF,
evaluate the kth derivative at t = 0.
d k [m(t )] k !t 0 E (Y k ) (k  1)!t1E (Y k 1 ) t 2 E (Y k 2 )
                                                    
dt            k!            (k  1)!                2!

• And so, letting t = 0:
k
d [m(t )]
 E (Y k )
dt     t 0
Geometric MGF
• For the geometric distribution,
m(t )  E[e ]   e ( q
ty            ty        y 1
p)
y

 et p  e 2t qp  e3t q 2 p  e 4t q 3 p 
 e p 1  e q  e q  e q 
t         t            2t    2          3t   3

 1          pet
 e p
t
t 

 1  e q  1  qet
Common MGFs
• The MGFs for some of the discrete distributions
we’ve seen include:
binomial: m(t )  ( pe  q)t       n

t
pe
geometric: m(t ) 
1  qet
r
 pe  t
negative binomial: m(t )          t 
 1  qe 
 ( et 1)
Poisson: m(t )  e
Recognize the distribution
• Identify the distribution having the moment
generating function
20
 e 3
t
m(t )       
 4 
• Give the mean and variance for this distribution.
• Could use the derivatives, but is that necessary?
Geometric MGF
1       t          t
e    e
• Consider the MGF m(t )             3
1  3 e 3  2e
2  t       t

• Use derivatives to determine the first and second
moments.                 t
3e
m(t ) 
3  2e 
t 2

And so,
3e0                3
E (Y )  m(0)                            3
3  2e  0 2          1
Geometric MGF
3et
• Since m(t ) 
3  2e  t 2

V (Y )  E (Y 2 )  [ E (Y )]2
• We have
3e (3  2e )
t           t                  15  (3)  6
2

m(t ) 
 3  2e 
t 3

And so,
3e (3  2e )
0           0
E (Y )  m(t ) 
2
 15
 3  2e 
0 3
Geometric MGF
1       t   t
e   e
• Since m(t )     3

1  3 e 3  2e
2  t       t

is for a geometric random variable with p = 1/3,
our prior results tell us
E(Y) = 1/p and V(Y) = (1 – p)/p2.

1                  1 1 3 2  9 
E (Y )      3 and V (Y )           6
1 3 3  1 
2
13

which do agree with our current results.
All the moments
• Although the mean and variance help to describe a
distribution, they alone do not uniquely describe a
distribution.
• All the moments are necessary to uniquely
describe a probability distribution.
• That is, if two random variables have equal MGFs,
(i.e., mY(t) = mZ(t) for |t| < b ),
then they have the same probability distribution.
m(aY+b)?
• For the random variable Y with MGF m(t),
consider W = aY + b.
m(t )  mY (t )  E[etY ]
mW (t )  E[et ( aY b ) ]   Construct the MGF for
the random variable
 E[e atY ebt ]     W= 2Y + 3, where Y is
a geometric random
 e E[e ]
bt     atY
variable with p = 4/5.
 e mY (at )
bt
E(aY+b)
• Now, based on the MGF, we could again
consider E(W) = E(aY + b).
d bt
                            
mW (t )  e mY (at )  ebt mY (at )(a)  mY (at )bebt
dt
 e  amY (at )  bmY (at ) 
bt

And so, letting t = 0,
E (W )  mW (0)  e0  amY (0)  bmY (0) 
              
 aE (Y )  b as expected.
Tchebysheff’s Theorem
• For “bell-shaped” distributions, the empirical rule
gave us a 68-95-99.7% rule for probability a value
falls within 1, 2, or 3 standard deviations from the
mean, respectively.
• When the distribution is not so bell-shaped,
Tchebysheff tells use the probability of being
within k standard deviations of the mean is
at least 1 – 1/k2, for k > 0.
1   Remember, it’s just
P(| Y   |  k )  1  2   a lower bound.
k
A Skewed Distribution
• Consider a binomial experiment with n = 10
and p = 0.1.

P(| Y  1|  2(0.95))
1
 1  2  0.75
2
A Skewed Distribution
• Verify Tchebysheff’s lower bound for k = 2:

P(| Y  1|  2(0.95))
 P(0.9  Y  2.9)
1
 1  2  0.75
2

P(0.9  Y  2.9)  0.34868  0.38742  0.19371  0.93

```
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
 views: 8 posted: 10/5/2012 language: English pages: 18