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Use of moment generating functions. Definition. Let X denote a random variable with probability density function f ( x ) if continuous (probability mass function p ( x ) if discrete) Then m X ( t ) = the moment generating function of X. - PowerPoint PPT Presentation
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Use of moment generating functions
DefinitionLet X denote a random variable with probability density function f(x) if continuous (probability mass function p(x) if discrete)
Then
mX(t) = the moment generating function of X
tXE e
if is continuous
if is discrete
tx
tx
x
e f x dx X
e p x X
The distribution of a random variable X is described by either
1. The density function f(x) if X continuous (probability mass function p(x) if X discrete), or
2. The cumulative distribution function F(x), or
3. The moment generating function mX(t)
Properties1. mX(0) = 1
0 derivative of at 0.k thX Xm k m t t 2.
kk E X
2 33211 .
2! 3! !kk
Xm t t t t tk
3.
continuous
discrete
k
kk k
x f x dx XE X
x p x X
4. Let X be a random variable with moment generating function mX(t). Let Y = bX + a
Then mY(t) = mbX + a(t)
= E(e [bX + a]t) = eatmX (bt)
5. Let X and Y be two independent random variables with moment generating function mX(t) and mY(t) .
Then mX+Y(t) = mX (t) mY (t)
6. Let X and Y be two random variables with moment generating function mX(t) and mY(t) and two distribution functions FX(x) and FY(y) respectively.
Let mX (t) = mY (t) then FX(x) = FY(x).
This ensures that the distribution of a random variable can be identified by its moment generating function
M. G. F.’s - Continuous distributions
Name
Moment generating function MX(t)
Continuous Uniform
ebt-eat
[b-a]t
Exponential t
for t <
Gamma t
for t <
2
d.f.
1
1-2t /2
for t < 1/2
Normal et+(1/2)t22
M. G. F.’s - Discrete distributions
Name
Moment generating
function MX(t)
Discrete Uniform
et
N etN-1et-1
Bernoulli q + pet Binomial (q + pet)N
Geometric pet
1-qet
Negative Binomial
pet
1-qet k
Poisson e(et-1)
Moment generating function of the gamma distribution
tX txXm t E e e f x dx
1 0
0 0
xx e xf x
x
where
tX txXm t E e e f x dx
1
0
tx xe x e dx
using
1
0
t xx e dx
1
0
1a
a bxbx e dx
a
1
0
a bxa
ax e dx
b
or
then
1
0
t xXm t x e dx
t
tt
Moment generating function of the Standard Normal distribution
tX txXm t E e e f x dx
2
21
2
x
f x e
where
thus
2 2
2 21 1
2 2
x xtxtx
Xm t e e dx e dx
We will use 2
22
0
11
2
x a
be dxb
2
21
2
xtx
Xm t e dx
2 2
21
2
x tx
e dx
22 2 2 22
2 2 2 21 1
2 2
x tx tx t t t
e e dx e e dx
2
2
t
e
Note:
2
2 32 2
22
2 21
2 2! 3!
t
X
t t
tm t e
2 3 4
12! 3! 4!
x x x xe x
2 4 6 2
2 31
2 2 2! 2 3! 2 !
m
m
t t t t
m
Also
2 33211
2! 3!Xm t t t t
Note:
2
2 32 2
22
2 21
2 2! 3!
t
X
t t
tm t e
2 3 4
12! 3! 4!
x x x xe x
2 4 6 2
2 31
2 2 2! 2 3! 2 !
m
m
t t t t
m
Also 2 33211
2! 3!Xm t t t t
momentth kk k x f x dx
Equating coefficients of tk, we get
21
for 2 then 2 ! 2 !
mm
k mm m
0 if is odd andk k
1 2 3 4hence 0, 1, 0, 3
Using of moment generating functions to find the distribution of
functions of Random Variables
ExampleSuppose that X has a normal distribution with mean and standard deviation .
Find the distribution of Y = aX + b
2 2
2
tt
Xm t e
Solution:
22
2
atatbt bt
aX b Xm t e m at e e
2 2 2
2
a ta b t
e
= the moment generating function of the normal distribution with mean a + b and variance a22.
Thus Z has a standard normal distribution .
Special Case: the z transformation
1XZ X aX b
10Z a b
22 2 2 21
1Z a
Thus Y = aX + b has a normal distribution with mean a + b and variance a22.
ExampleSuppose that X and Y are independent each having a normal distribution with means X and Y , standard deviations X and Y
Find the distribution of S = X + Y
2 2
2X
Xt
t
Xm t e
Solution:
2 2
2Y
Yt
t
Ym t e
2 2 2 2
2 2X Y
X Yt t
t t
X Y X Ym t m t m t e e
Now
or
2 2 2
2
X YX X
tt
X Ym t e
= the moment generating function of the normal distribution with mean X + Y and variance
2 2
X Y
Thus Y = X + Y has a normal distribution with mean X + Y and variance 2 2
X Y
ExampleSuppose that X and Y are independent each having a normal distribution with means X and Y , standard deviations X and Y
Find the distribution of L = aX + bY
2 2
2X
Xt
t
Xm t e
Solution:
2 2
2Y
Yt
t
Ym t e
aX bY aX bY X Ym t m t m t m at m bt Now
2 22 2
2 2X Y
X Yat bt
at bte e
or
2 2 2 2 2
2
X YX X
a b ta b t
aX bYm t e
= the moment generating function of the normal distribution with mean aX + bY and variance
2 2 2 2
X Ya b
Thus Y = aX + bY has a normal distribution with mean aX + BY and variance
2 2 2 2
X Ya b
Special Case:
Thus Y = X - Y has a normal distribution with mean X - Y and variance
2 22 2 2 21 1
X Y X Y
a = +1 and b = -1.
Example (Extension to n independent RV’s)Suppose that X1, X2, …, Xn are independent each having a normal distribution with means i, standard deviations i
(for i = 1, 2, … , n)
Find the distribution of L = a1X1 + a1X2 + …+ anXn
2 2
2i
i
i
tt
Xm t e
Solution:
1 1 1 1n n n na X a X a X a Xm t m t m t Now
22 221 1
1 1 2 2n n
n n
a ta ta t a t
e e
(for i = 1, 2, … , n)
1 1 nX X nm a t m a t
or
2 2 2 2 2
1 11 1
1 1
......
2
n nn n
n n
a a ta a t
a X a Xm t e
= the moment generating function of the normal distribution with mean
and variance
Thus Y = a1X1 + … + anXn has a normal distribution with mean a11 + …+ ann and variance
1 1 ... n na a 2 2 2 21 1 ... n na a
2 2 2 21 1 ... n na a
1 2
1na a a
n
1 2 n 2 2 2 21 1 1
In this case X1, X2, …, Xn is a sample from a normal distribution with mean , and standard deviations and
1 2
1nL X X X
n
the sample meanX
Special case:
Thus
2 2 2 2 21 1 ...x n na a
and variance
1 1 ...x n na a has a normal distribution with mean
1 1 ... n nY x a x a x
11 1... nx xn n
1 1...n n
2 2 2 22 2 21 1 1
... nn n n n
If x1, x2, …, xn is a sample from a normal distribution with mean , and standard deviations then the sample meanx
Summary
22x n
and variance
x has a normal distribution with mean
standard deviation xn
0
0.1
0.2
0.3
0.4
20 30 40 50 60
Population
Sampling distribution of x
If x1, x2, …, xn is a sample from a distribution with mean , and standard deviations then if n is large the sample meanx
The Central Limit theorem
22x n
and variance
x has a normal distribution with mean
standard deviation xn
We will use the following fact: Let
m1(t), m2(t), … denote a sequence of moment generating functions corresponding to the sequence of distribution functions:
F1(x) , F2(x), … Let m(t) be a moment generating function corresponding to the distribution function F(x) then if
Proof: (use moment generating functions)
lim for all in an interval about 0.ii
m t m t t
lim for all .ii
F x F x x
then
Let x1, x2, … denote a sequence of independent random variables coming from a distribution with moment generating function m(t) and distribution function F(x).
1 2 1 2
=n n nS x x x x x xm t m t m t m t m t
Let Sn = x1 + x2 + … + xn then
=n
m t
1 2now n nx x x Sx
n n
1or n
n
n
x SS
n
t tm t m t m m
n n
Let x n n
z x
n
then
nn n
t t
z x
nt ntm t e m e m
n
and ln lnz
n tm t t n m
n
Then ln lnz
n tm t t n m
n
2 2
2 2 2ln
t tm u
u u
2
2 2Let or and
t t tu n n
u un
2
2 2
ln m u ut
u
0
Now lim ln lim lnz zn u
m t m t
2
2 20
lnlimu
m u ut
u
2
2 0lim using L'Hopital's rule
2u
m u
m ut
u
2
22
2 0lim using L'Hopital's rule again
2u
m u m u m u
m ut
2
22
2 0lim using L'Hopital's rule again
2u
m u m u m u
m ut
22
2
0 0
2
m mt
222 2
2 2 2
i iE x E xt t
222thus lim ln and lim
2
t
z zn n
tm t m t e
2
2Now t
m t e
Is the moment generating function of the standard normal distribution
Thus the limiting distribution of z is the standard normal distribution
2
21
i.e. lim2
x u
zn
F x e du
Q.E.D.