Functions of Random Variables - math.usask.ca
Transcript of Functions of Random Variables - math.usask.ca
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Functions of Random Variables
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Methods for determining the distribution of
functions of Random Variables
1. Distribution function method
2. Moment generating function method
3. Transformation method
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Distribution function method
Let X, Y, Z …. have joint density f(x,y,z, …)
Let W = h( X, Y, Z, …)
First step
Find the distribution function of W
G(w) = P[W ≤ w] = P[h( X, Y, Z, …) ≤ w]
Second step
Find the density function of W
g(w) = G'(w).
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Example 1
Let X have a normal distribution with mean 0, and
variance 1. (standard normal distribution)
Let W = X2.
Find the distribution of W.
2
21
2
x
f x e
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First step
Find the distribution function of W
G(w) = P[W ≤ w] = P[ X2 ≤ w]
if 0P w X w w
2
21
2
w x
w
e dx
F w F w
where 2
21
2
x
F x f x e
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d wd w
F w F wdw dw
Second step
Find the density function of W
g(w) = G'(w).
1 1
2 2 2 21 1 1 1
2 22 2
w w
e w e w
1 1
2 21 1
2 2f w w f w w
1
2 21
if 0.2
w
w e w
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Thus if X has a standard Normal distribution then
W = X2
has density
1
2 21
if 0.2
w
g w w e w
This distribution is the Gamma distribution with a = ½
and l = ½.
This distribution is also the c2 distribution with n = 1
degree of freedom.
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Example 2
Suppose that X and Y are independent random
variables each having an exponential distribution
with parameter l (mean 1/l)
Let W = X + Y.
Find the distribution of W.
1 for 0xf x e xll
2 for 0yf y e yll
1 2,f x y f x f y
2 for 0, 0x y
e x yl
l
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First step
Find the distribution function of W = X + Y
G(w) = P[W ≤ w] = P[ X + Y ≤ w]
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1 2
0 0
w w x
P X Y w f x f y dydx
2
0 0
w w xx y
e dydxl
l
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1 2
0 0
w w x
P X Y w f x f y dydx
2
0 0
w w xx y
e dydxl
l
2
0 0
w w x
x ye e dy dxl ll
2
0 0
w xw yx e
e dxl
lll
02
0
w w x
x e ee dx
lll
l
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P X Y w 0
2
0
w w x
x e ee dx
lll
l
0
w
x we e dxl ll
0
wx
wexe
lll
l
0wwe e
wel
lll l
1 w we wel ll
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Second step
Find the density function of W
g(w) = G'(w).
1 w wde we
dw
l ll
ww wdw de
e e wdw dw
ll ll l
2w w we e wel l ll l l
2 for 0wwe wll
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Hence if X and Y are independent random variables each having an exponential distribution with parameter l then W has density
2 for 0wg w we wll
This distribution can be recognized to be the Gamma distribution with parameters a = 2 and l.
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Example: Student’s t distribution
Let Z and U be two independent random
variables with:
1. Z having a Standard Normal distribution
and
2. U having a c2 distribution with n degrees
of freedom
Find the distribution of Z
tUn
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The density of Z is:
2
21
2
z
f z e
The density of U is:
2
12 2
1
2
2
u
h u u e
n
n
n
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Therefore the joint density of Z and U is:
The distribution function of T is:
Z t
G t P T t P t P Z UU nn
2
2
12 2
1
2,
22
z u
f z u f z h u u e
n
n
n
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Therefore:
t
G t P T t P Z Un
2
2
12 2
0
1
2
22
tu
z u
u e dzdu
n
nn
n
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Illustration of limits
U
U
z z
t > 0 t < 0
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Now:
2
2
12 2
0
1
2( )
22
tu
z u
G t u e dzdu
n
nn
n
and:
2
2
12 2
0
1
2( )
22
tu
z ud
g t G t u e dz dudt
n
nn
n
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Using:
b
a
b
a
dxtxFdt
ddxtxF
dt
d),(),(
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Using the fundamental theorem of calculus:
( )
x
a
F x f t dt
then
2
2
12 2
0
1
2
22
tu
z ud
g t u e dz dudt
n
nn
n
If then ( )F x f x
2
2
122 2
0
1
2
22
t uuu
u e e du
n
n
n
n n
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Hence
221
1
2 2
0
1
2( )
22
tu
g t u e du
n
nn
n n
Using
1
0
xx e dxa l
a
a
l
1
0
1 xx e dxa
a ll
a
or
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Hence 2 1
1 21
2 21
20 2
12
2
1
tu
u e du
t
n
nn
n
n
n
and 1
212
2 2
1 12
2 2( ) 1
22
tg t
nn
nn
n n n
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or
1 12 22 2
1
2( ) 1 1
2
t tg t K
n nn
n n nn
1
2
2
K
n
nn
where
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Student’s t distribution
12 2
( ) 1t
g t K
n
n
1
2
2
K
n
nn
where
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Student – W.W. Gosset
Worked for a distillery
Not allowed to publish
Published under the
pseudonym “Student
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t distribution
standard normal distribution
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Functions of Random Variables
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Methods for determining the distribution of
functions of Random Variables
1. Distribution function method
2. Moment generating function method
3. Transformation method
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Distribution function method
Let X, Y, Z …. have joint density f(x,y,z, …)
Let W = h( X, Y, Z, …)
First step
Find the distribution function of W
G(w) = P[W ≤ w] = P[h( X, Y, Z, …) ≤ w]
Second step
Find the density function of W
g(w) = G'(w).
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Distribution of the Max and Min
Statistics
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Let x1, x2, … , xn denote a sample of size n from
the density f(x).
Let M = max(xi) then determine the distribution
of M.
Repeat this computation for m = min(xi)
Assume that the density is the uniform density
from 0 to q.
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Hence
10
( )
elsewhere
xf x
and the distribution function
0 0
( ) 0
1
x
xF x P X x x
x
q
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Finding the distribution function of M.
( ) max iG t P M t P x t
1 , , nP x t x t
1 nP x t P x t
0 0
0
1
n
t
tt
t
q
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Differentiating we find the density function of M.
1
0
0 otherwise
n
n
ntt
g t G tq
q
0
0.1
0.2
0.3
0.4
0.5
0.6
0 2 4 6 8 10
0
0.02
0.04
0.06
0.08
0.1
0.12
0 2 4 6 8 10
f(x) g(t)
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Finding the distribution function of m.
( ) min iG t P m t P x t
11 , , nP x t x t
11 nP x t P x t
0 0
1 1 0
1
n
t
tt
t
q
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0
0.1
0.2
0.3
0.4
0.5
0.6
0 2 4 6 8 10
Differentiating we find the density function of m.
1
1 0
0 otherwise
nn t
tg t G t
qq q
0
0.02
0.04
0.06
0.08
0.1
0.12
0 2 4 6 8 10
f(x) g(t)
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The probability integral transformation
This transformation allows one to convert
observations that come from a uniform
distribution from 0 to 1 to observations that
come from an arbitrary distribution.
Let U denote an observation having a uniform
distribution from 0 to 1.
1 0 1
( )elsewhere
ug u
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Find the distribution of X.
1( )X F ULet
Let f(x) denote an arbitrary density function and
F(x) its corresponding cumulative distribution
function.
1( )G x P X x P F U x
P U F x
F x
Hence.
g x G x F x f x
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has density f(x).
1( )X F U
Thus if U has a uniform distribution from 0 to 1.
Then
U
1( )X F U
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The Transformation Method
Theorem
Let X denote a random variable with
probability density function f(x) and U = h(X).
Assume that h(x) is either strictly increasing
(or decreasing) then the probability density of
U is:
1
1 ( )( )
dh u dxg u f h u f x
du du
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Proof
Use the distribution function method.
Step 1 Find the distribution function, G(u)
Step 2 Differentiate G (u ) to find the
probability density function g(u)
G u P U u P h X u
1
1
( ) strictly increasing
( ) strictly decreasing
P X h u h
P X h u h
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1
1
( ) strictly increasing
1 ( ) strictly decreasing
F h u h
F h u h
hence
g u G u
1
1
1
1
strictly increasing
strictly decreasing
dh uF h u h
du
dh uF h u h
du
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or
1
1 ( )( )
dh u dxg u f h u f x
du du
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Example
Suppose that X has a Normal distribution
with mean m and variance s2.
Find the distribution of U = h(x) = eX.
Solution:
2
221
2
x
f x e
m
s
s
1
1ln 1
ln and dh u d u
h u udu du u
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hence
1
1 ( )( )
dh u dxg u f h u f x
du du
2
2
ln
21 1
for 02
u
e uu
m
s
s
This distribution is called the log-normal
distribution
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log-normal distribution
0
0.02
0.04
0.06
0.08
0.1
0 10 20 30 40
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The Transfomation Method
(many variables) Theorem
Let x1, x2,…, xn denote random variables
with joint probability density function
f(x1, x2,…, xn )
Let u1 = h1(x1, x2,…, xn).
u2 = h2(x1, x2,…, xn).
un = hn(x1, x2,…, xn).
define an invertible transformation from the x’s to the u’s
⁞
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Then the joint probability density function of
u1, u2,…, un is given by:
1
1 1
1
, ,, , , ,
, ,
n
n n
n
d x xg u u f x x
d u u
1, , nf x x J
where
1
1
, ,
, ,
n
n
d x xJ
d u u
Jacobian of the transformation
1 1
1
1
det
n
n n
n
dx dx
du du
dx dx
du du
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Example Suppose that x1, x2 are independent with density
functions f1 (x1) and f2(x2)
Find the distribution of
u1 = x1+ x2
u2 = x1 - x2
Solving for x1 and x2 we get the inverse transformation
1 21
2
u ux
1 22
2
u ux
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1 2
1 2
,
,
d x xJ
d u u
The Jacobian of the transformation
1 1
1 2
2 2
1 2
det
dx dx
du du
dx dx
du du
1 1
1 1 1 1 12 2det
1 1 2 2 2 2 2
2 2
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The joint density of x1, x2 is
f(x1, x2) = f1 (x1) f2(x2)
Hence the joint density of u1 and u2 is:
1 2 1 21 2
1
2 2 2
u u u uf f
1 2 1 2, ,g u u f x x J
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From
1 2 1 21 2 1 2
1,
2 2 2
u u u ug u u f f
We can determine the distribution of u1= x1 + x2
1 1 1 2 2,g u g u u du
1 2 1 21 2 2
1
2 2 2
u u u uf f du
1 2 1 21
2
1put then ,
2 2 2
u u u u dvv u v
du
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Hence
1 2 1 21 1 1 2 2
1
2 2 2
u u u ug u f f du
1 2 1f v f u v dv
This is called the convolution of the two
densities f1 and f2.
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Example: The ex-Gaussian distribution
1. X has an exponential distribution with
parameter l.
2. Y has a normal (Gaussian) distribution with
mean m and standard deviation s.
Let X and Y be two independent random
variables such that:
Find the distribution of U = X + Y.
This distribution is used in psychology as a model
for response time to perform a task.
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Now 1
0
0 0
xe xf x
x
ll
1 2g u f v f u v dv
The density of U = X + Y is :.
2
222
1
2
x
f y e
m
s
s
2
22
0
1
2
u v
ve e dv
m
l sls
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or
2
22
02
u vv
g u e dv
ml
sl
s
2 2
2
2
2
02
u v v
e dv
m s l
sl
s
22 2
2
2 2
2
02
v u v u v
e dv
m m s l
sl
s
2 22
2 2
2
2 2
02
v u vu
e e dv
m s lm
s sl
s
![Page 60: Functions of Random Variables - math.usask.ca](https://reader033.fdocuments.us/reader033/viewer/2022060104/62927c1c6860af3a1471c547/html5/thumbnails/60.jpg)
or 2 22
2 2
2
2 2
02
v u vu
e e dv
m s lm
s sl
s
2 22 2 2 2 2
2 2
2
2 2
02
u u v u v u
e e dv
m m s l m s l m s l
s sl
s
2 22 2 2 2 2
2 2
2
2 2
0
1
2
u u v u v u
e e dv
m m s l m s l m s l
s sls
22 2
22 0
u u
e P V
m m s l
sl
![Page 61: Functions of Random Variables - math.usask.ca](https://reader033.fdocuments.us/reader033/viewer/2022060104/62927c1c6860af3a1471c547/html5/thumbnails/61.jpg)
Where V has a Normal distribution with mean
22
2
21
u ug u e
s ll m m s l
ls
2
V um m s l
and variance s2.
Hence
Where (z) is the cdf of the standard Normal
distribution
![Page 62: Functions of Random Variables - math.usask.ca](https://reader033.fdocuments.us/reader033/viewer/2022060104/62927c1c6860af3a1471c547/html5/thumbnails/62.jpg)
0
0.03
0.06
0.09
0 10 20 30
g(u)
The ex-Gaussian distribution
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Functions of Random Variables
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Methods for determining the distribution of
functions of Random Variables
1. Distribution function method
2. Moment generating function method
3. Transformation method
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Distribution function method
Let X, Y, Z …. have joint density f(x,y,z, …)
Let W = h( X, Y, Z, …)
First step
Find the distribution function of W
G(w) = P[W ≤ w] = P[h( X, Y, Z, …) ≤ w]
Second step
Find the density function of W
g(w) = G'(w).
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The Transformation Method
Theorem
Let X denote a random variable with
probability density function f(x) and U = h(X).
Assume that h(x) is either strictly increasing
(or decreasing) then the probability density of
U is:
1
1 ( )( )
dh u dxg u f h u f x
du du
![Page 67: Functions of Random Variables - math.usask.ca](https://reader033.fdocuments.us/reader033/viewer/2022060104/62927c1c6860af3a1471c547/html5/thumbnails/67.jpg)
The Transfomation Method
(many variables) Theorem
Let x1, x2,…, xn denote random variables
with joint probability density function
f(x1, x2,…, xn )
Let u1 = h1(x1, x2,…, xn).
u2 = h2(x1, x2,…, xn).
un = hn(x1, x2,…, xn).
define an invertible transformation from the x’s to the u’s
⁞
![Page 68: Functions of Random Variables - math.usask.ca](https://reader033.fdocuments.us/reader033/viewer/2022060104/62927c1c6860af3a1471c547/html5/thumbnails/68.jpg)
Then the joint probability density function of
u1, u2,…, un is given by:
1
1 1
1
, ,, , , ,
, ,
n
n n
n
d x xg u u f x x
d u u
1, , nf x x J
where
1
1
, ,
, ,
n
n
d x xJ
d u u
Jacobian of the transformation
1 1
1
1
det
n
n n
n
dx dx
du du
dx dx
du du
![Page 69: Functions of Random Variables - math.usask.ca](https://reader033.fdocuments.us/reader033/viewer/2022060104/62927c1c6860af3a1471c547/html5/thumbnails/69.jpg)
Use of moment generating
functions
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Definition
Let 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
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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)
![Page 72: Functions of Random Variables - math.usask.ca](https://reader033.fdocuments.us/reader033/viewer/2022060104/62927c1c6860af3a1471c547/html5/thumbnails/72.jpg)
Properties
1. mX(0) = 1
0 derivative of at 0.k th
X Xm k m t t 2.
k
k E Xm
2 33211 .
2! 3! !
kkXm t t t t t
k
m mmm 3.
continuous
discrete
k
k
k k
x f x dx XE X
x p x Xm
![Page 73: Functions of Random Variables - math.usask.ca](https://reader033.fdocuments.us/reader033/viewer/2022060104/62927c1c6860af3a1471c547/html5/thumbnails/73.jpg)
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) = eatE(e X[ bt ])
= 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) = E(e [X + Y]t) = E(e Xt e Yt)
= E(e Xt) E(e Yt)
= mX (t) mY (t)
![Page 74: Functions of Random Variables - math.usask.ca](https://reader033.fdocuments.us/reader033/viewer/2022060104/62927c1c6860af3a1471c547/html5/thumbnails/74.jpg)
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
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M. G. F.’s - Continuous distributions
Name
Moment generating
function MX(t)
Continuous
Uniform
ebt-eat
[b-a]t
Exponential l
l t
for t < l
Gamma l
l t
a
for t < l
c2
nd.f.
1
1-2t
n/2
for t < 1/2
Normal etm+(1/2)t2s2
![Page 76: Functions of Random Variables - math.usask.ca](https://reader033.fdocuments.us/reader033/viewer/2022060104/62927c1c6860af3a1471c547/html5/thumbnails/76.jpg)
M. G. F.’s - Discrete distributions
Name
Moment
generating
function MX(t)
Discrete
Uniform
et
N etN-1
et-1
Bernoulli q + pet
Binomial (q + pet)N
Geometric pet
1-qet
Negative
Binomial
pet
1-qet k
Poisson el(et-1)
![Page 77: Functions of Random Variables - math.usask.ca](https://reader033.fdocuments.us/reader033/viewer/2022060104/62927c1c6860af3a1471c547/html5/thumbnails/77.jpg)
Moment generating function of the
gamma distribution
tX tx
Xm t E e e f x dx
1 0
0 0
xx e xf x
x
aa ll
a
where
![Page 78: Functions of Random Variables - math.usask.ca](https://reader033.fdocuments.us/reader033/viewer/2022060104/62927c1c6860af3a1471c547/html5/thumbnails/78.jpg)
tX tx
Xm t E e e f x dx
1
0
tx xe x e dxa
a ll
a
using
1
0
t xx e dx
alal
a
1
0
1a
a bxbx e dx
a
1
0
a bx
a
ax e dx
b
or
![Page 79: Functions of Random Variables - math.usask.ca](https://reader033.fdocuments.us/reader033/viewer/2022060104/62927c1c6860af3a1471c547/html5/thumbnails/79.jpg)
then
1
0
t x
Xm t x e dxa
lal
a
t
a
a
al
a l
tt
al
ll
![Page 80: Functions of Random Variables - math.usask.ca](https://reader033.fdocuments.us/reader033/viewer/2022060104/62927c1c6860af3a1471c547/html5/thumbnails/80.jpg)
Moment generating function of the
Standard Normal distribution
tX tx
Xm t E e e f x dx
2
21
2
x
f x e
where
thus
2 2
2 21 1
2 2
x xtx
tx
Xm t e e dx e dx
![Page 81: Functions of Random Variables - math.usask.ca](https://reader033.fdocuments.us/reader033/viewer/2022060104/62927c1c6860af3a1471c547/html5/thumbnails/81.jpg)
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
![Page 82: Functions of Random Variables - math.usask.ca](https://reader033.fdocuments.us/reader033/viewer/2022060104/62927c1c6860af3a1471c547/html5/thumbnails/82.jpg)
Note:
2
2 32 2
2
22 2
12 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
mmm
![Page 83: Functions of Random Variables - math.usask.ca](https://reader033.fdocuments.us/reader033/viewer/2022060104/62927c1c6860af3a1471c547/html5/thumbnails/83.jpg)
Note:
2
2 32 2
2
22 2
12 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
mmm
momentth k
k k x f x dxm
![Page 84: Functions of Random Variables - math.usask.ca](https://reader033.fdocuments.us/reader033/viewer/2022060104/62927c1c6860af3a1471c547/html5/thumbnails/84.jpg)
Equating coefficients of tk, we get
21
for 2 then 2 ! 2 !
m
mk m
m m
m
0 if is odd andk km
1 2 3 4hence 0, 1, 0, 3m m m m
![Page 85: Functions of Random Variables - math.usask.ca](https://reader033.fdocuments.us/reader033/viewer/2022060104/62927c1c6860af3a1471c547/html5/thumbnails/85.jpg)
Using of moment generating
functions to find the distribution of
functions of Random Variables
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Example
Suppose that X has a normal distribution with
mean m and standard deviation s.
Find the distribution of Y = aX + b
2 2
2
tt
Xm t es
m
Solution:
22
2
atat
bt bt
aX b Xm t e m at e e
sm
2 2 2
2
a ta b t
es
m
= the moment generating function of the normal
distribution with mean am + b and variance a2s2.
![Page 87: Functions of Random Variables - math.usask.ca](https://reader033.fdocuments.us/reader033/viewer/2022060104/62927c1c6860af3a1471c547/html5/thumbnails/87.jpg)
Thus Z has a standard normal distribution .
Special Case: the z transformation
1XZ X aX b
m m
s s s
10Z a b
mm m m
s s
2
2 2 2 211Z as s s
s
Thus Y = aX + b has a normal distribution with
mean am + b and variance a2s2.
![Page 88: Functions of Random Variables - math.usask.ca](https://reader033.fdocuments.us/reader033/viewer/2022060104/62927c1c6860af3a1471c547/html5/thumbnails/88.jpg)
Example Suppose that X and Y are independent each having a normal distribution with means mX and mY , standard deviations sX and sY
Find the distribution of S = X + Y
2 2
2
XX
tt
Xm t es
m
Solution:
2 2
2
YY
tt
Ym t es
m
2 2 2 2
2 2
X YX Y
t tt t
X Y X Ym t m t m t e es s
m m
Now
![Page 89: Functions of Random Variables - math.usask.ca](https://reader033.fdocuments.us/reader033/viewer/2022060104/62927c1c6860af3a1471c547/html5/thumbnails/89.jpg)
or
2 2 2
2
X Y
X Y
tt
X Ym t e
s sm m
= the moment generating function of the
normal distribution with mean mX + mY and
variance
2 2
X Ys s
Thus Y = X + Y has a normal distribution
with mean mX + mY and variance
2 2
X Ys s
![Page 90: Functions of Random Variables - math.usask.ca](https://reader033.fdocuments.us/reader033/viewer/2022060104/62927c1c6860af3a1471c547/html5/thumbnails/90.jpg)
Example Suppose that X and Y are independent each having a normal distribution with means mX and mY , standard deviations sX and sY
Find the distribution of L = aX + bY
2 2
2
XX
tt
Xm t es
m
Solution:
2 2
2
YY
tt
Ym t es
m
aX bY aX bY X Ym t m t m t m at m bt
Now
2 22 2
2 2
X YX Y
at btat bt
e e
s sm m
![Page 91: Functions of Random Variables - math.usask.ca](https://reader033.fdocuments.us/reader033/viewer/2022060104/62927c1c6860af3a1471c547/html5/thumbnails/91.jpg)
or
2 2 2 2 2
2
X Y
X Y
a b ta b t
aX bYm t e
s sm m
= the moment generating function of the
normal distribution with mean amX + bmY
and variance
2 2 2 2
X Ya bs s
Thus L = aX + bY has a normal
distribution with mean amX + bmY and
variance
2 2 2 2
X Ya bs s
![Page 92: Functions of Random Variables - math.usask.ca](https://reader033.fdocuments.us/reader033/viewer/2022060104/62927c1c6860af3a1471c547/html5/thumbnails/92.jpg)
Special Case:
Thus Y = X - Y has a normal distribution
with mean mX - mY and variance
2 22 2 2 21 1
X Y X Ys s s s
a = +1 and b = -1.
![Page 93: Functions of Random Variables - math.usask.ca](https://reader033.fdocuments.us/reader033/viewer/2022060104/62927c1c6860af3a1471c547/html5/thumbnails/93.jpg)
Example (Extension to n independent RV’s)
Suppose that X1, X2, …, Xn are independent each having a normal distribution with means mi, standard deviations si
(for i = 1, 2, … , n)
Find the distribution of L = a1X1 + a2X2 + …+ anXn
2 2
2
ii
i
tt
Xm t es
m
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 12 2
n nn n
a ta ta t a t
e e
ssm m
(for i = 1, 2, … , n)
1 1 nX X nm a t m a t
![Page 94: Functions of Random Variables - math.usask.ca](https://reader033.fdocuments.us/reader033/viewer/2022060104/62927c1c6860af3a1471c547/html5/thumbnails/94.jpg)
or
2 2 2 2 21 1
1 1
1 1
......
2
n n
n n
n n
a a ta a t
a X a Xm t e
s sm m
= the moment generating function of the
normal distribution with mean
and variance
Thus Y = a1X1 + … + anXn has a normal
distribution with mean a1m1 + …+ anmn
and variance
1 1 ... n na am m 2 2 2 2
1 1 ... n na as s
2 2 2 2
1 1 ... n na as s
![Page 95: Functions of Random Variables - math.usask.ca](https://reader033.fdocuments.us/reader033/viewer/2022060104/62927c1c6860af3a1471c547/html5/thumbnails/95.jpg)
1 2
1na a a
n
1 2 nm m m m
2 2 2 2
1 2 ns s s s
In this case X1, X2, …, Xn is a sample from a
normal distribution with mean m, and standard
deviations s,and
1 2
1nL X X X
n
the sample meanX
Special case:
![Page 96: Functions of Random Variables - math.usask.ca](https://reader033.fdocuments.us/reader033/viewer/2022060104/62927c1c6860af3a1471c547/html5/thumbnails/96.jpg)
Thus
2 2 2 2 2
1 1 ...x n na as s s
and variance
1 1 ...x n na am m m
has a normal distribution with mean
1 1 ... n nY x a x a x
11 1... nx x
n n
1 1...n nm m m
2 2 2 22 2 21 1 1
... nn n n n
ss s s
![Page 97: Functions of Random Variables - math.usask.ca](https://reader033.fdocuments.us/reader033/viewer/2022060104/62927c1c6860af3a1471c547/html5/thumbnails/97.jpg)
If x1, x2, …, xn is a sample from a normal
distribution with mean m, and standard
deviations s,then the sample meanx
Summary
22
xn
ss
and variance
xm m
has a normal distribution with mean
standard deviation xn
ss
![Page 98: Functions of Random Variables - math.usask.ca](https://reader033.fdocuments.us/reader033/viewer/2022060104/62927c1c6860af3a1471c547/html5/thumbnails/98.jpg)
0
0.1
0.2
0.3
0.4
20 30 40 50 60
Population
Sampling distribution
of x
![Page 99: Functions of Random Variables - math.usask.ca](https://reader033.fdocuments.us/reader033/viewer/2022060104/62927c1c6860af3a1471c547/html5/thumbnails/99.jpg)
Suppose x1, x2, …, xn is a sample (independent
identically distributed – i.i.d.) from a
distribution with mean m,
the sample meanx
The Law of Large Numbers
Then
1 as for all 0P x nm
Let
Proof: Previously we used Tchebychev’s Theorem.
This assumes s(s2) is finite.
![Page 100: Functions of Random Variables - math.usask.ca](https://reader033.fdocuments.us/reader033/viewer/2022060104/62927c1c6860af3a1471c547/html5/thumbnails/100.jpg)
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
![Page 101: Functions of Random Variables - math.usask.ca](https://reader033.fdocuments.us/reader033/viewer/2022060104/62927c1c6860af3a1471c547/html5/thumbnails/101.jpg)
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
nn
n
x SS
n
t tm t m t m m
n n
![Page 102: Functions of Random Variables - math.usask.ca](https://reader033.fdocuments.us/reader033/viewer/2022060104/62927c1c6860af3a1471c547/html5/thumbnails/102.jpg)
using L’Hopitals rule
now ln ln ln
n
x
t tm t m n m
n n
ln where
t m u tu
u n
0
lnThus lim ln limx
n u
t m um t
u
0
0lim
1 0u
m ut
m u mt t
mm
![Page 103: Functions of Random Variables - math.usask.ca](https://reader033.fdocuments.us/reader033/viewer/2022060104/62927c1c6860af3a1471c547/html5/thumbnails/103.jpg)
is the moment generating function of
a random variable that takes on the value m with
probability 1.
and lim for all values of .xn
F x F x x
tm t e m
1
i.e. andx
p xx
m
m
0
and distribution function and1
xF x
x
m
m
Thus lim t
xn
m t m t e m
![Page 104: Functions of Random Variables - math.usask.ca](https://reader033.fdocuments.us/reader033/viewer/2022060104/62927c1c6860af3a1471c547/html5/thumbnails/104.jpg)
Now
0
since and1
xF x
x
m
m
P x P xm m m
x xF Fm m
1 if 0F Fm m
as n
Q.E.D.
![Page 105: Functions of Random Variables - math.usask.ca](https://reader033.fdocuments.us/reader033/viewer/2022060104/62927c1c6860af3a1471c547/html5/thumbnails/105.jpg)
If x1, x2, …, xn is a sample from a distribution
with mean m, and standard deviations s,then
if n is large the sample meanx
The Central Limit theorem
22
xn
ss
and variance
xm m
has a normal distribution with mean
standard deviation xn
ss
![Page 106: Functions of Random Variables - math.usask.ca](https://reader033.fdocuments.us/reader033/viewer/2022060104/62927c1c6860af3a1471c547/html5/thumbnails/106.jpg)
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
![Page 107: Functions of Random Variables - math.usask.ca](https://reader033.fdocuments.us/reader033/viewer/2022060104/62927c1c6860af3a1471c547/html5/thumbnails/107.jpg)
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
nn
n
x SS
n
t tm t m t m m
n n
![Page 108: Functions of Random Variables - math.usask.ca](https://reader033.fdocuments.us/reader033/viewer/2022060104/62927c1c6860af3a1471c547/html5/thumbnails/108.jpg)
Let x n n
z x
n
m m
s s s
then
nn n
t t
z x
nt ntm t e m e m
n
m m
s s
s s
and ln lnz
n tm t t n m
n
m
s s
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Then ln lnz
n tm t t n m
n
m
s s
2 2
2 2 2ln
t tm u
u u
m
s s
2
2 2Let or and
t t tu n n
u un s ss
2
2 2
ln m u ut
u
m
s
![Page 110: Functions of Random Variables - math.usask.ca](https://reader033.fdocuments.us/reader033/viewer/2022060104/62927c1c6860af3a1471c547/html5/thumbnails/110.jpg)
0
Now lim ln lim lnz zn u
m t m t
2
2 20
lnlimu
m u ut
u
m
s
2
2 0lim using L'Hopital's rule
2u
m u
m ut
u
m
s
2
22
2 0lim using L'Hopital's rule again
2u
m u m u m u
m ut
s
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2
22
2 0lim using L'Hopital's rule again
2u
m u m u m u
m ut
s
2
2
2
0 0
2
m mt
s
222 2
2 2 2
i iE x E xt t
s
2
2
2thus lim ln and lim2
t
z zn n
tm t m t e
![Page 112: Functions of Random Variables - math.usask.ca](https://reader033.fdocuments.us/reader033/viewer/2022060104/62927c1c6860af3a1471c547/html5/thumbnails/112.jpg)
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.
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The Central Limit theorem
illustrated
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If x1, x2, …, xn is a sample from a distribution
with mean m, and standard deviations s,then
if n is large the sample meanx
The Central Limit theorem
22
xn
ss
and variance
xm m
has a normal distribution with mean
standard deviation xn
ss
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If x1, x2 are independent from the uniform
distirbution from 0 to 1. Find the distribution
of: the sample meanx
The Central Limit theorem illustrated
1 21 2 and
2 2
x xSS x x x
let
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1 2G s P S s P x x s Now
2
2
0 0
0 12
21 1 2
2
1 1
s
s s
ss
s
0 1
2 1 2
0 otherwise
s s
g s G s s s
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Now: 12
2
Sx S aS
The density of is:x
2 2dS
h x g S g xdx
12
12
2 0 2 1 2 0
2 2 1 2 2 2 1 1
0 otherwise 0 otherwise
x x x x
x x x x
![Page 118: Functions of Random Variables - math.usask.ca](https://reader033.fdocuments.us/reader033/viewer/2022060104/62927c1c6860af3a1471c547/html5/thumbnails/118.jpg)
n = 1
1 0
1 0
n = 2
n = 3
1 0
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Distributions of functions of
Random Variables
Gamma distribution, c2 distribution,
Exponential distribution
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Therorem
Let X and Y denote a independent random variables
each having a gamma distribution with parameters
(l,a1) and (l,a2). Then W = X + Y has a gamma
distribution with parameters (l, a1 + a2).
Proof:
1 2
and X Ym t m tt t
a al l
l l
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1 2 1 2
t t t
a a a al l l
l l l
Therefore X Y X Ym t m t m t
Recognizing that this is the moment generating
function of the gamma distribution with parameters
(l, a1 + a2) we conclude that W = X + Y has a
gamma distribution with parameters (l, a1 + a2).
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Therorem (extension to n RV’s)
Let x1, x2, … , xn denote n independent random
variables each having a gamma distribution with
parameters (l,ai), i = 1, 2, …, n.
Then W = x1 + x2 + … + xn has a gamma distribution
with parameters (l, a1 + a2 +… + an).
Proof:
1,2...,i
ixm t i nt
al
l
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1 2 1 2 ...
...n n
t t t t
a a a a a al l l l
l l l l
1 2 1 2... ...
n nx x x x x xm t m t m t m t
Recognizing that this is the moment generating
function of the gamma distribution with parameters
(l, a1 + a2 +…+ an) we conclude that
W = x1 + x2 + … + xn has a gamma distribution with
parameters (l, a1 + a2 +…+ an).
Therefore
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Therorem
Suppose that x is a random variable having a
gamma distribution with parameters (l,a).
Then W = ax has a gamma distribution with
parameters (l/a, a).
Proof:
xm tt
al
l
then ax xam t m at
at ta
aa l
l
ll
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1. Let X and Y be independent random variables
having an exponential distribution with parameter
l then X + Y has a gamma distribution with a= 2
and l
Special Cases
2. Let x1, x2,…, xn, be independent random variables
having a exponential distribution with parameter l
then S = x1+ x2 +…+ xn has a gamma distribution
with a= n and l3. Let x1, x2,…, xn, be independent random variables
having a exponential distribution with parameter l
then
has a gamma distribution with a= n and nl
1 nx xSx
n n
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0
0.1
0.2
0.3
0.4
0.5
0.6
0 5 10 15 20
pop'n
n = 4
n = 10
n = 15
n = 20
Distribution of
population – Exponential distribution
x
Another illustration of the central limit theorem
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4. Let X and Y be independent random variables
having a c2 distribution with n1 and n2 degrees of
freedom respectively then X + Y has a c2
distribution with degrees of freedom n1 + n2.
Special Cases -continued
5. Let x1, x2,…, xn, be independent random variables
having a c2 distribution with n1 , n2 ,…, nn degrees
of freedom respectively then x1+ x2 +…+ xn has a
c2 distribution with degrees of freedom n1 +…+ nn.
Both of these properties follow from the fact that a
c2 random variable with n degrees of freedom is a
random variable with l= ½ and a = n/2.
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If z has a Standard Normal distribution then z2 has a
c2 distribution with 1 degree of freedom.
Recall
Thus if z1, z2,…, zn are independent random variables
each having Standard Normal distribution then
has a c2 distribution with n degrees of freedom.
2 2 2
1 2 ...U z z zn
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Therorem
Suppose that U1 and U2 are independent random
variables and that U = U1 + U2 Suppose that U1
and U have a c2 distribution with degrees of
freedom n1andn respectively. (n1 < n)
Then U2 has a c2 distribution with degrees of
freedom n2 =n -n1
Proof:
12
1
12
12
Now
v
Um tt
212
12
and
v
Um tt
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1 2
Also U U Um t m t m t
2
12 2
12
12
1122
11 22
12
v
vv
v
t
t
t
2
1
Hence U
U
U
m tm t
m t
Q.E.D.
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Tables for Standard Normal
distribution