13 s241 functions of random variables

7/25/2019 13 S241 Functions of Random Variables

1/122


2/122

Methods for determining the distribution of

functions of Random Variables

1. Distribution function method

2. Moment generating function method

3. Transformation method


3/122

Distribution function method

Let X, , ! ". ha#e $oint densit%f&x,y,z, '

Let W ( h&X, Y, Z, '

First stepFind the distribution function of W

G&w' (P)W * w+ (P)h&X, Y, Z, '* w+

Second stepFind the densit% function of W

g&w' ( G'&w'.


4/122

-amle 1

Let X ha#e a normal distribution /ith mean 0, and

#ariance 1. &standard normal distribution'

Let W (X2.

Find the distribution of W.

( )

2

21

2

x

f x e

=


5/122

First step

Find the distribution function of WG&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=

/here ( ) ( )

2

21

2

x

F x f x e

= =


6/122

( ) ( ) ( )d wd w

F w F wdw dw

=

Second step

Find the densit% function of W

g&w' ( G'&w'.

1 1

2 2 2 2

1 1 1 1

2 22 2

w w

e w e w

= +

( ) ( )1 1

2 21 12 2

f w w f w w = +

1

2 21

if 0.

2

w

w e w

=


7/122

Thus ifX has a standard ormal distribution then

W = X2

has densit%

( )1

2 21

if 0.

2

w

g w w e w

=

This distribution is the amma distribution /ith =

and ( .

This distribution is also the2distribution /ith = 1

degree of freedom.


8/122

-amle 2

4uose thatXand Y are indeendent random#ariables each ha#ing an e-onential distribution

/ith arameter &mean 15'

Let W (X6 Y.

Find the distribution of W.

( )1 for 0x

f x e x

= ( )2 for 0

yf y e y =

( ) ( ) ( )1 2,f x y f x f y=( )2 for 0, 0x y

e x y

+=


9/122

First step

Find the distribution function of W = X 6 YG&w' (P)W * w+ (P)X6 Y * w+


10/122


11/122

[ ] ( ) ( )1 20 0

w w x

P X Y w f x f y dydx

+ =

( )2

0 0

w w xx y

e dydx

+=


12/122

[ ] ( ) ( )1 20 0

w w x

P X Y w f x f y dydx

+ =

( )2

0 0

w w xx y

e dydx

+=

2

0 0

w w xx ye e dy dx =

2

0 0

w xw yx e

e dx

= ( ) 0

2

0

w w x

x e ee dx

=


13/122

[ ]P X Y w+ ( ) 0

2

0

w w x

x e ee dx

=

0

w

x we e dx =

0

wx

we xe

=

0wwe ewe

=

1 w we we =


14/122

Second step

Find the densit% function of W

g&w' ( G'&w'.

1 w wd

e wedw

=

ww wdw dee e w

dw dw

= +

2w w we e we = +

2 for 0wwe w =


15/122

7ence ifX and Y are indeendent random #ariableseach ha#ing an e-onential distribution /ith arameter

then W has densit%

( ) 2 for 0wg w we w =

This distribution can be recogni8ed to be the Gammadistribution /ith arameters = 2 and .


16/122

-amle9 4tudent:s t distribution

LetZ and Ube t/o indeendent random#ariables /ith9

1. Z ha#ing a 4tandard ormal distribution

and

2. U ha#ing a2distribution /ith degreesof freedom

Find the distribution ofZ

tU

=


17/122

The densit% ofZ is9

( )

2

2

1

2

z

f z e

=

The densit% of U is9

( )

2

12 2

12

2

u

h u u e

=


18/122

Therefore the $oint densit% ofZ and Uis9

The distribution function of Tis9

( ) [ ]Z t

G t P T t P t P Z U U

= = =

( ) ( ) ( )

2

2

12 2

1

2,

2 2

z u

f z u f z h u u e

+

= =


19/122

Therefore9

( ) [ ] tG t P T t P Z U = = =

22

12 2

0

12

2

2

t uz u

u e dzdu

+

=


20/122

;llustration of limits

U

U

zz

t > 0 t > 0


21/122

o/9

22

12 2

0

12

& '

22

t uz u

G t u e dzdu

+

=

and9

( )

2

2

12 2

0

1

2& '

2

2

t

u z ud

g t G t u e dz dudt

+

= =


22/122


23/122


24/122

7ence

221

1

2 2

0

12

& '

2

2

t u

g t u e du

+ =


25/122

7ence2 1

1 21

2 21

20 2

12

2

1

tu

u e du

t

+ +

+

+ =

+

and1

212

2 2

1 12

2 2& ' 1

22

tg t

++

+

= +


26/122

or

1 12 22 2

12

& ' 1 1

2

t tg t K

+ + + = + = +

1

2

2

K

+

=

/here


27/122

Students t distribution

12 2

& ' 1t

g t K

+

= +

1

2

2

K

+

=

/here


28/122

4tudent = >.>. osset

>or?ed for a distiller%

ot allo/ed to ublish

@ublished under the

seudon%m A4tudent


29/122

t distribution

standard normal distribution


30/122

Distribution of the Ma- and Min

4tatistics


31/122

Letx1,x2, " ,xdenote a samle of si8e from

the densit%f&x'.

Let!( ma-&x"' then determine the distribution

of!.

Reeat this comutation for # ( min&x"'

Bssume that the densit% is the uniform densit%

from 0 to .


32/122

7ence

10& '

else/here

xf x

=

0

and the distribution function

[ ]

0 0

& ' 0

1

x

xF x P X x x

x


33/122

Finding the distribution function of!.

[ ] ( )& ' ma- "G t P ! t P x t = =

[ ]1 , , P x t x t= L

[ ] [ ]1 P x t P x t= L

0 0

0

1

t

tt

t


34/122

Differentiating /e find the densit% function of!.

( ) ( )

1

0

0 other/ise

tt

g t G t

= =

f&x' g&t'


35/122

Finding the distribution function of #.

[ ] ( )& ' min "G t P # t P x t = =

[ ]11 , , P x t x t= > >L

[ ] [ ]11 P x t P x t= > >L

0 0

1 1 0

1

t

tt

t


36/122

Differentiating /e find the densit% function of #.

( ) ( )

1

1 0

0 other/ise

t

tg t G t

= =

f&x' g&t'


37/122

The robabilit% integral transformation

This transformation allo/s one to con#ert

obser#ations that come from a uniform

distribution from 0 to 1 to obser#ations that

come from an arbitrar% distribution.

Let U denote an obser#ation ha#ing a uniform

distribution from 0 to 1.

1 0 1& '

else/here

ug u

=0


38/122

Find the distribution ofX.

1& 'X F U=Let

Letf&x$denote an arbitrar% densit% function and

F%x' its corresonding cumulati#e distribution

function.

( ) [ ] 1& 'G x P X x P F U x = =

( )P U F x =

( )F x=7ence.

( ) ( ) ( ) ( )g x G x F x f x = = =


39/122


40/122

The Transformation Method

TheoremLet X denote a random #ariable /ith

robabilit% densit% functionf&x' and U (

h&X'.

Bssume that h&x' is either strictl% increasing

&or decreasing' then the robabilit% densit% of

U is9

( ) ( ) ( )1

1 & '& ' dh u dx

g u f h u f xdu du

= =


41/122

Proof


42/122

( )

( )

1

1

& ' strictl% increasing

1 & ' strictl% decreasing

F h u h

F h u h

=

hence

( ) ( )g u G u=

( )( ) ( )

( )( ) ( )

1

1

1

1

strictl% increasing

strictl% decreasing

dh u

F h u hdu

dh uF h u h

du

=


43/122

or

( ) ( ) ( )1

1 & '& ' dh u dx

g u f h u f xdu du

= =


44/122

-amle

4uose that X has a ormal distribution/ith meanand #ariance 2.

Find the distribution of U ( h&x' ( eX.

Solution:

( )( )

2

221

2

x

f x e

=

( ) ( ) ( ) ( )11 ln 1ln and

dh u d uh u u

du du u

= = =


45/122

hence

( ) ( ) ( )1

1 & '& ' dh u dx

g u f h u f xdu du

= =

( )( )2

2

ln

21 1

for 02

u

e uu

= >

This distribution is called the logCnormal

distribution


46/122

logCnormal distribution

The Transfomation Method


47/122

The Transfomation Method

&man% #ariables'Theorem

Let x1,x2,",xdenote random #ariables /ith

$oint robabilit% densit% function

f&x1,x2,",x'

Let u1( h1&-1,x2,",x'.

u2( h2&-1,x2,",x'.

u( h&-1,x2,",x'.

define an in#ertible transformation from thex:s to the u:s


48/122

Then the $oint robabilit% densit% function of

u1, u2,",uis gi#en b%9

( ) ( ) ( )

( )1

1 1

1

, ,, , , ,

, ,

d x xg u u f x x

d u u=

LL L

L

( )1, , f x x &= L

/here( )

( )

1

1

, ,

, ,

d x x&

d u u=

L

L

acobian of the transformation

1 1

1

1

det

dx dx

du du

dx dx

du du

=

L

M M

K


49/122

-amle4uose that x1,x2 are indeendent /ith densit%

functionsf1 &x1' andf2&x2'

Find the distribution of

u1(x16x2u2(x1Cx2

4ol#ing forx1and x /e get the in#erse transformation

1 21

2u ux +=

1 22

2

u ux

=


50/122

( )

( )1 2

1 2

,

,

d x x&

d u u=

The acobian of the transformation

1 1

1 2

2 2

1 2

det

dx dxdu du

dx dx

du du

=

1 1

1 1 1 1 12 2det 1 1 2 2 2 2 2

2 2

= = =


51/122

The $oint densit% ofx1,x2 is

f&x1,x2' (f1 &x1'f2&x2'

7ence the $oint densit% of u1and u2is9

1 2 1 21 2

1

2 2 2

u u u uf f

+ =

( ) ( )1 2 1 2, ,g u u f x x &=


52/122

From

( )

1 2 1 2

1 2 1 2

1,

2 2 2

u u u ug u u f f

+

=

>e can determine the distribution of u1(x16x2

( ) ( )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

1ut then ,

2 2 2

u u u u d(( u (

du

+ = = =


53/122

7ence

( ) 1 2 1 21 1 1 2 21

2 2 2

u u u ug u f f du

+ =

( ) ( )1 2 1f ( f u ( d(

=

This is called the con#olution of the t/o

densitiesf1andf2.


54/122

Example: The e-Caussian distribution

1. X has an e-onential distribution /ith

arameter .

2. Y has a normal &aussian' distribution /ith

meanand standard de#iation .

LetX and Ybe t/o indeendent random

#ariables such that9

Find the distribution of U (X 6 Y.

This distribution is used in s%cholog% as a

model for resonse time to erform a tas?.

0x


55/122

o/ ( )10

0 0

xe xf x

x

=


56/122

or

( )( )

2

22

02

u ((

g u e d(

=

( )

2 2

2

2

2

02

u ( (

e d(

+ =

( ) ( )22 2

2

2 2

2

02

( u ( u (

e d(

+ + =

( ) ( )2 22

2 2

2

2 2

02

( u (u

e e d(

=


57/122

or ( ) ( )2 22

2 2

2

2 2

02

( u (u

e e d(

=

( ) ( ) ( ) ( )2 22 2 2 2 2

2 2

2

2 2

02

u u ( u ( u

e e d(

+ =

( ) ( ) ( ) ( )

2 22 2 2 2 2

2 2

2

2 2

0

1

2

u u ( u ( u

e e d(

+ =

( ) ( )

[ ]

22 2

22 0

u u

e P +

=


58/122

>here + has a ormal distribution /ith mean

( )( )

22

2

21

u ug u e

+ =

( )2

+ u = +and #ariance 2.

7ence

>here &z' is the cdf of the standard ormaldistribution


59/122

g&u'

The e-Caussian distribution


60/122


61/122


62/122

The distribution of a random #ariableX is described b%

either

1. The densit% functionf&x' ifX continuous &robabilit% mass function&x' if

X discrete', or

2. The cumulati#e distribution functionF&x', or

3. The moment generating function #X&t'

P ti


63/122

Properties

1. #X&0' ( 1

( ) ( ) ( )0 deri#ati#e of at 0. th

X X# . # t t = =2.

( )

.

. - X= =

( ) 2 33211 .2E 3E E

X# t t t t t

.

= + + + + + +L L3.

( ) ( )

( )

continuous

discrete

.

.

. .

x f x dx X- X

x , x X

= =

Let X be a random ariable ith moment


64/122

. LetXbe a random #ariable /ith moment

generating function #X&t'. Let Y( bX 6 a

Then #Y&t' ( #bX 6 a&t'(-&e )bX 6 a+t' ( eat-&e X) bt +'

( eat#X &bt'

G. LetXand Ybe t/o indeendent random

#ariables /ith moment generating function

#X&t' and #Y&t' .

Then #X/Y&t' (-&e)X 6 Y+t' (-&e Xt e Yt'

(-&e Xt'-&e Yt'

( #X &t' #Y &t'


65/122

H. LetXand Ybe t/o random #ariables /ith

moment generating function #X&t' and #Y&t'

and t/o distribution functionsFX&x' and

FY&y' resecti#el%.

Let #X &t' ( #Y &t' thenFX&x' (FY&x'.

This ensures that the distribution of a random

#ariable can be identified b% its moment

generating function

M F : I ti di t ib ti


66/122

M. . F.:s C Iontinuous distributions

M F : Di t di t ib ti


67/122

M. . F.:s C Discrete distributions

Moment generating function of the


68/122


gamma distribution

( ) ( ) ( )tX txX# t - e e f x dx

= =

( ) ( )1 0

0 0

xx e xf x

x

=


69/122

( ) ( ) ( )tX txX# t - e e f x dx

= =

( )1

0

tx xe x e dx

=

using

( )( )1

0

t xx e dx

=

( )1

0

1

a

a bxb

x e dxa

=( )1

0

a bx

a

ax e dx

b

=

or

then


70/122

then

( ) ( )( )1

0

t x

X# t x e dx

=

( )

( )

( )t

=

tt

= <



71/122


4tandard ormal distribution

( ) ( ) ( )tX txX# t - e e f x dx

= =

( )

2

21

2

x

f x e

=

/here

thus

( )

2 2

2 21 1

2 2

x xtx

tx

X# t e e dx e dx

+

= =

( )2

1x a


72/122

>e /ill use( )

22

0

11

2

x a

be dxb

=

( )

2

21

2

xtx

X# t e dx

+

= 2

221

2

x tx

e dx

= ( )

22 2 2 22

2 2 2 2

1 1

2 2

x tx tx t t t

e e dx e e dx

+

= =

2

2

t

e= 2 3


73/122

ote9

( )

2

2 32 2

2

22 2

12 2E 3E

t

X

t t

t# t e

= = + + + +L

2 3

12E 3E FE

x x x xe x= + + + + +L

2 H 2

2 31

2 2 2E 2 3E 2 E

#

#

t t t t

#= + + + + + +L L

Blso

( ) 2 332112E 3E

X# t t t t

= + + + +L

2 3 x x x


74/122

ote9

( )

2

2 32 2

2

22 2

12 2E 3E

t

X

t t

t# t e

= = + + + +L

12E 3E FE

x x x xe x= + + + + +L

2 H 2

2 31

2 2 2E 2 3E 2 E

#

#

t t t t

#= + + + + + +L L

Blso ( ) 2 332112E 3E

X# t t t t = + + + +L

( )momentth x f x dx

= =


75/122

Juating coefficients of t, /e get

( )

21for 2 then2 E 2 E

#

#. #

# #

= =

0 if is odd and . =

1 2 3 hence 0, 1, 0, 3 = = = =


76/122


77/122

Thus Y = aX / b has a normal distribution /ith


78/122

ThusZ has a standard normal distribution .

Special Case:thez transformation

1XZ X aX b

= = + = +

10Z a b

= + = + =

2

2 2 2 21 1Z a

= = =

ThusY = aX / b has a normal distribution /ith

mean a/ band #ariance a22.

-amle


79/122

-amle4uose thatXand Yare indeendent eachha#ing a

normal distribution /ith meansX andY , standardde#iations X and Y

Find the distribution of = X / Y

( )

2 2

2XX

tt

X# t e

+=

Solution:

( )

2 2

2

YY

tt

Y# t e

+=

( ) ( ) ( )

2 2 2 2

2 2

X YX Y

t tt t

X Y X Y# t # t # t e e

+ +

+ = =

o/

or


80/122

or

( ) ( )

( )2 2 2

2

X Y

X Y

tt

X Y# t e

++ +

+ =

( the moment generating function of thenormal distribution /ith meanX /Yand

#ariance2 2

X Y +

ThusY = X / Y has a normal distribution

/ith meanX /Yand #ariance2 2

X Y

+

-amle


81/122

-amle4uose thatXand Yare indeendent eachha#ing anormal distribution /ith means

Xand

Y, standard

de#iations X and Y

Find the distribution of1 = aX / bY

( )

2 2

2XX

tt

X# t e

+=

Solution:

( )

2 2

2

Y

Y

t

tY# t e

+=

( ) ( ) ( ) ( ) ( )aX bY aX bY X Y # t # t # t # at # bt + = =

o/

( ) ( )

( ) ( )

2 22 2

2 2

X YX Y

at bt at bt

e e

+ +

=

or


82/122

or

( ) ( )

( )2 2 2 2 2

2

X Y

X Y

a b ta b t

aX bY # t e

++ +

+ =

( the moment generating function of thenormal distribution /ith mean aX / bY

and #ariance2 2 2 2

X Ya b +

ThusY = aX / bY has a normal

distribution /ith mean aX / bYand

#ariance2 2 2 2

X Ya b +

4ecial Iase9


83/122

4ecial Iase9

ThusY = X 2 Y has a normal distribution

/ith meanX 2Yand #ariance

( ) ( )2 22 2 2 2

1 1X Y X Y

+ + = +

a ( 61 and b( C1.

-amle &-tension to indeendent RV:s'


84/122

-amle &-tension to indeendent RV s'4uose thatX1,X2, ",Xare indeendent eachha#ing a

normal distribution /ith means", standard de#iations "&for " = 1, 2, " ,'

Find the distribution of1 = a1X1/ a1X2/ / aX

( )

2 2

2""

"

tt

X# t e

+=

Solution:

( ) ( ) ( )1 1 1 1 a X a X a X a X

# t # t # t + + =L Lo/

( ) ( )

( ) ( )

22 221 1

1 12 2

a ta ta t a t

e e

+ +

= L

&for " = 1, 2, " ,'

( ) ( )1 1 X X

# a t # a t = L

or


85/122

or

( ) ( )

( )2 2 2 2 21 11 1

1 1

......

2

a a ta a t

a X a X # t e

+ ++ + +

+ + =L

( the moment generating function of thenormal distribution /ith mean

and #ariance

ThusY = a1X1/ / aXhas a normal

distribution /ith mean a1

1

/ / a

and

#ariance

1 1 ... a a + +2 2 2 2

1 1

...

a a + +

2 2 2 21 1 ... a a + +

Special case:


86/122

1 2

1a a a

= = = =L

1 2 = = = =L2 2 2 2

1 1 1 = = = =L

;n this caseX1,X2, ",Xis a samle from a

normal distribution /ith mean, and standardde#iations , and

( )1 21 1 X X X

= + + +L

the samle meanX= =

p


87/122

Thus

2 2 2 2 21 1 ...x a a = + +

and #ariance

1 1 ...x a a = + +has a normal distribution /ith mean

1 1 ... Y x a x a x= = + +

( ) ( )11 1... x x = + +

( ) ( )1 1... = + + =

2 2 2 22 2 21 1 1...

= + + = =

Summary


88/122

;fx1,x2, ",xis a samle from a normal

distribution /ith mean, and standardde#iations , then the samle meanx=

y

22

x

=

and #ariance

x =

has a normal distribution /ith mean

standard de#iation x

=


89/122

@oulation

4amling distribution

ofx

The Law of Large umbers


90/122

4uosex1,x2, ",xis a samle &indeendent

identicall% distributed = i.i.d.' from adistribution /ith mean,

the samle meanx=

g

Then

1 as for all 0P x < >

Let

Proof:@re#iousl% /e used Tcheb%che#:s Theorem.

This assumes &2' is finite.

Proof: &use moment generating functions'


91/122

>e /ill use the follo/ing fact9

Let

#1&t', #2&t', "

denote a seJuence of moment generating functions

corresonding to the seJuence of distribution

functions9F1&x' ,F2&x', "

Let #&t' be a moment generating function

corresonding to the distribution functionF&x' then

if

& g g '

( ) ( )lim for all in an inter#al about 0.""

# t # t t

=

( ) ( )lim for all .""

F x F x x

=then

Let x x denote a seJuence of indeendent


92/122

Letx1,x2, " denote a seJuence of indeendent

random #ariables coming from a distribution /ith

moment generating function #&t' and distributionfunctionF&x'.

( ) ( ) ( ) ( ) ( )1 2 1 2

( 0 x x x x x x

# t # t # t # t # t + + += L L

Let (x16x26 " 6xthen

( )(

# t

1 2no/ x x x x

+ + += =L

( ) ( )1or

x 00

t t# t # t # #

= = =


93/122

using L:7oitals rule

( )no/ ln ln lnxt t

# t # #

= =

( )ln /here

t # u tu

u = =

( ) ( )0ln

Thus lim ln limx u

t # u# t

u = ( )

( ) ( )

( )00

lim 1 0u

# ut

# u #

t t#

= = =


94/122

is the moment generating function of

a random #ariable that ta?es on the #alue/ithrobabilit% 1.

( ) ( )and lim for all #alues of .x

F x F x x

=

( ) t# t e=

( )1

i.e. andx

xx

==0

( )

0

and distribution function and1

x

F x x


95/122

o/

( )0

since and

1

xF x

x

as

K..D.

The Central Limit theorem


96/122

;fx1,x2, ",xis a samle from a distribution

/ith mean, and standard de#iations , thenif is large the samle meanx=

22

x

=

and #ariance

x =



=

Proof: &use moment generating functions'


97/122

>e /ill use the follo/ing fact9

Let

#1&t', #2&t', "

denote a seJuence of moment generating functions

corresonding to the seJuence of distribution

functions9F1&x' ,F2&x', "

Let #&t' be a moment generating function

corresonding to the distribution functionF&x' then

if ( ) ( )lim for all in an inter#al about 0.""

# t # t t

=

( ) ( )lim for all .""

F x F x x

=then

Let x x denote a seJuence of indeendent


98/122

Letx1,x2, " denote a seJuence of indeendent

random #ariables coming from a distribution /ith

moment generating function #&t' and distributionfunctionF&x'.

( ) ( ) ( ) ( ) ( )1 2 1 2

( 0 x x x x x x

# t # t # t # t # t + + += L L

Let (x16x26 " 6xthen

( )(

# t

1 2no/ x x x x

+ + += =L

( ) ( )1or

x 00

t t# t # t # #

= = =


99/122

Letx

z x

= =

( )then

t t

z x

t t # t e # e #

= =

( )and ln lnz t

# t t #

= +

2


100/122

( )Then ln lnz t

# t t #

= +

( )2 2

2 2 2ln

t t# u

u u

= +

2

2 2Let or and

t t tu

u u = = =

( )2

2 2

ln # u ut

u

=


101/122

( )( ) ( )( )0

o/ lim ln lim lnz z u

# t # t

=

( )2

2 20

lnlimu

# u ut

u

=

( )( )2

2 0lim using L7oitals rule

2u

# u

# ut

u

=

( ) ( ) ( )

( )

2

22

2 0lim using L7oitals rule again

2u

# u # u # u

# ut

=


102/122

( ) ( ) ( )

( )

2

2

2

2 0lim using L7oitals rule again

2u

# u # u # u

# ut

=

( ) ( )

22

20 0

2# #t

=

( ) ( )222 2

2 2 2

" "- x - xt t

= =

( )( ) ( )( )2

2

2thus lim ln and lim2

t

z z

t# t # t e

= =

2t


103/122

( )

2

2o/t

# t e=

;s the moment generating function of the standard

normal distribution

Thus the limiting distribution ofz is the standardnormal distribution

( )

2

21

i.e. lim 2

x u

z F x e du

=

Q.E.D.


104/122

The Ientral Limit theorem

illustrated

The Central Limit theorem


105/122

;fx1,x2, ",xis a samle from a distribution

/ith mean, and standard de#iations , thenif is large the samle meanx=

22

x

=

and #ariance

x =



=

The Central Limit theorem illustrated


106/122

;fx1,x2are indeendent from the uniform

distirbution from 0 to 1. Find the distributionof9 the samle meanx=

1 21 2 and 2 2

x x00 x x x += + = =

let


107/122

0


108/122

o/9 ( )122

0x 0 a0= = =

The densit% of is9x

( ) ( ) ( )2 2dh x g g xdx

= =

( )

12

12

2 0 2 1 2 0

2 2 1 2 2 2 1 1

0 other/ise 0 other/ise

x x x x

x x x x

= =

1


109/122

= 1

10

10

= 2

= 3

10


110/122

Distributions of functions of

Random Variablesamma distribution, 32distribution,

-onential distribution

Therorem


111/122

LetX andY denote a indeendent random #ariables

each ha#ing a gamma distribution /ith arameters&,1' and &,2'. Then W(X 6 Y has a gamma

distribution /ith arameters &, 1 62'.

Proof:

( ) ( )

1 2

andX Y# t # t t t

= =


112/122

1 2 1 2

t t t

+ = =

( ) ( ) ( )Therefore X Y X Y# t # t # t + =

Recogni8ing that this is the moment generating

function of the gamma distribution /ith arameters

&, 16 2' /e conclude that W(X 6 Y has a

gamma distribution /ith arameters &, 16 2'.

Therorem&e-tension to RV:s'


113/122

Letx1,x2, " ,xdenote indeendent random #ariables each

ha#ing a gamma distribution /ith arameters &,"', " ( 1, 2, ",.

Then W(x16x26 " 6xhas a gamma distribution /ith

arameters &, 1 62 6" 6 '.

Proof:

( ) 1, 2...,

"

"x# t " t

= =

Therefore


114/122

1 2 1 2 ...

...

t t t t

+ + + = =

( ) ( ) ( ) ( )1 2 1 2

... ...

x x x x x x# t # t # t # t + + + =

Recogni8ing that this is the moment generating

function of the gamma distribution /ith arameters

&, 16 2 6"6 n' /e conclude that

W(x16x2/ / xhas a gamma distribution /ith

arameters &, 16 2 6"6 '.

Therorem


115/122

4uose thatxis a random #ariable ha#ing a

gamma distribution /ith arameters &,'.Then W( ax has a gamma distribution /ith

arameters &5a, '.

Proof:( )x# t

t

=

( ) ( )then ax x a# t # at at t

a

= = =

Special Cases


116/122

1. LetX and Ybe indeendent random #ariables

ha#ing an e-onential distribution /ith arameter

then X 6 Y has a gamma distribution /ith ( 2

and 2. Letx1,x2,",x,be indeendent random #ariables

ha#ing a e-onential distribution /ith arameter then (x16x26"6xhas a gamma distribution

/ith ( and 3. Letx1,x2,",x,be indeendent random #ariables

ha#ing a e-onential distribution /ith arameter then

has a gamma distribution /ith ( and

1 x x0x

+ += = K

Distribution of

l i i l di ib i

x


117/122

oulation = -onential distribution

Bnother illustration of the central limit theorem

L d b i d d d i bl

Special Cases !continued


118/122

. LetX and Ybe indeendent random #ariables

ha#ing a2 distribution /ith 1 and 2 degrees of

freedom resecti#el% then X 6 Y has a2

distribution /ith degrees of freedom 1 6 2.

G. Letx1,x2,",x,be indeendent random #ariables

ha#ing a2 distribution /ith 1 ,2 ,", degreesof freedom resecti#el% thenx16x26"6xhas a

2 distribution /ith degrees of freedom 1 6"6 .

oth of these roerties follo/ from the fact that a2 random #ariable /ith degrees of freedom is a

random #ariable /ith ( and ( 52.

;f h 4 d d l di ib i h h

Recall


119/122

;f z has a 4tandard ormal distribution thenz2 has a

2 distribution /ith 1 degree of freedom.

Thus ifz1,z2,",zare indeendent random #ariables

each ha#ing 4tandard ormal distribution then

has a2 distribution /ith degrees of freedom.

2 2 21 2 ...U z z z = + + +

Therorem


120/122

4uose that U1and U2are indeendent random #ariables and

that U ( U16 U24uose that U1and Uha#e a2

distribution/ith degrees of freedom 1andresecti#el%. &1N '

Then U2has a2distribution /ith degrees of freedom 2(C1

Proof:

( )12

1

12

12

o/

(

U# tt

=

( )21

2

12

and

(

U# tt

=

( ) ( ) ( )Blso U U U# t # t # t=


121/122

( ) ( ) ( )1 2

Blso U U U# t # t # t

2

12 2

12

12

1122

11 22

12

(

((

(

t

t

t

= =

( ) ( )( )2

1

7enceU

U

U

# t# t

# t=

Q.E.D.


122/122

Tables for 4tandard ormal distri

bution
http://14%20s241%20use%20of%20normal%20tables.ppt/http://14%20s241%20use%20of%20normal%20tables.ppt/http://14%20s241%20use%20of%20normal%20tables.ppt/http://14%20s241%20use%20of%20normal%20tables.ppt/

13 s241 functions of random variables

Documents