Moment Generating Functions
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Transcript Moment Generating Functions
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 E (Y k )
k!
t k y k p( y )
y
k!
(t y)k
p( y )
k!
y
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!
k 0 y
(t y)k
y k 0 k !
p( y )
May rearrange,
since finite for
|t| < b.
et y p( y ) E[ety ]
y
Moment Generating Function
• That is,
m(t ) E[ety ]
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
th
k
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)!t 1E (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
Common MGFs
• The MGFs for some of the discrete distributions
we’ve seen include:
binomial: m(t ) ( pet q)n
pet
geometric: m(t )
t
1 qe
Poisson: m(t ) e
( et 1)
Geometric MGF
1
3
e
t
t
e
• Consider the MGF m(t )
t
t
2
1 3 e 3 2e
• Use derivatives to determine the first and second
moments.
t
m(t )
3e
3 2e
t 2
And so,
E (Y ) m(0)
3e0
3 2e
0 2
3
3
1
Geometric MGF
• Since m(t )
3et
3 2e
t 2
V (Y ) E (Y 2 ) [ E (Y )]2
• We have
m(t )
t
3 2e
t 3
And so,
E (Y ) m(t )
2
15 (3)2 6
3e (3 2e )
t
3e (3 2e )
0
0
3 2e
0 3
15
Geometric MGF
1
3
e
t
t
e
• Since m(t )
t
t
2
1 3 e 3 2e
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
2
13
1 3 3 1
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[e ]
tY
mW (t ) E[et ( aY b ) ]
E[e atY ebt ]
e E[ e ]
bt
atY
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
bt
e amY (at ) bmY (at )
And so, letting t = 0,
E (W ) mW (0) e0 amY (0) bmY (0)
aE(Y ) b as expected.
V(aY+b)
• Now, based on the MGF, can you again
consider V(W) = V(aY + b).
2
mW (0) E (W ) ?
• …and so V(W) = V(aY + b) = a2V(Y).
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
P(| Y | k ) 1 2
k
Remember, it’s just
a lower bound.
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