1

Lecture 8

! Functions of a Random Variable

! Functions of Two or More Random Variables.

2

Functions of A Random Variable

! Suppose X has a discrete distribution withp.f. f, and Y=r(X), a function of X. The p.f. ofY is:

å=

=

====

yxrxxfyXryYyg

)(:)())(Pr()Pr()(

3

Variable with A Continuous Distribution! Suppose X has a continuous distribution with p.d.f. f,

and Y=r(X), a function of X. The d.f. G of Y can bederived as:

If Y also has a continuous distribution, its p.d.f. g canbe obtained by

at any y where G is differentiable.

ò £=

£=£=

})(:{)(

))(Pr()Pr()(

yxrxdxxf

yXryYyG

dyydGyg )()( =

4

Example! Suppose X has a uniform distribution on (-1,1),

What is the p.d.f for ?

îíì <<-

=otherwise

xforxf

0,112/1

)(

2XY =

5

Solution

1/2

1/2

2

1/2 1/2

1/2

1/2

1/2

1/2

For 0 1,( ) Pr( ) Pr( )Pr( )

12

0, 0So ( ) , 0 1

1, 1

For 0 1,( ) 1( )

21 , 0 1

2So ( )0, otherwise

y

y

yG y Y y X y

y X y

dx y

yG y y y

y

ydG yg ydy y

yyg y

-

£ <

= £ = £

= - £ £

= =

<ìï= £ <íï ³î

< <

= =

ì < <ï= íïî

ò

6

Direct Derivation of p.d.f.! Suppose Y=r(X) where r is continuous, and X lies in a

certain interval (a,b) over which the function r(x) isstrictly increasing.• Then r is a one-to-one function which maps (a,b) to (a,b) . Ithas an inverse function X=s(Y).

• For any y such that a<y<b,

• Suppose s is differentiable over (a,b), then)]([))(Pr())(Pr()Pr()(

ysFysXyXryYyG

=£=£=£=

dyydsysf

dyysdF

dyydGyg

)()]([

))(()()(

=

==

7

! Suppose Y=r(X) where r is continuous, and X lies in acertain interval (a,b) over which the function r(x) isstrictly decreasing.• Then r is a one-to-one function which maps (a,b) to (a,b) . Ithas an inverse function X=s(Y).

• For any y such that a<y<b,

• Suppose s is differentiable over (a,b), then)]([1))(Pr())(Pr()Pr()(ysFysXyXryYyG

-=³=£=£=

dyydsysf

dyysdF

dyydGyg

)()]([

))(()()(

-=

-==

8

Theorem. Let X be a random variable for which thep.d.f. is f and Pr(a<X<b)=1. Let Y=r(X), and supposethat r(x) is continuous and either strictly increasing orstrictly decreasing for a<x<b. Suppose also that r(x)maps a<x<b to a<y<b, and let X=s(Y) be the inversefunction for a <Y<b. Then the p.d.f. of Y is specifiedby

ïî

ïíì

<<=otherwise

yfordyydsysfyg

0

,)()]([)( ba

9

Example! Suppose X has a p.d.f.

What is the p.d.f. of ?îíì <<

=otherwise

xforxxf

0,103

)(2

21 XY -=

10

Solution (1)2

2

2

1 2

1

3/ 2

3/ 2

1 , 0<X<1

For 0 1,( ) Pr( )

Pr[1 ] Pr[1 ]

= Pr[ 1 ]

= 3

1- (1- ) 0, 0

So ( ) 1- (1- ) , 0 11, 1

y

Y X

yG y Y y

X yX y

X y

x dx

yy

G y y yy

-

= -

< <= £

= - £

= - £

³ -

=

£ìï= < <íï ³î

ò

1/ 23 (1 ) , 0 1( )( ) 20, otherwise

y ydG yg ydy

ì - < <ï= = íïî

11

Solution (2)! Y is a continuous, strictly decreasing function for

0<X<1, with range 0<Y<1. The inverse function isfor 0<Y<1.2/1)1()( YYsX -==

1/ 2

1/ 21/ 2

( ) 12(1 )

1 33(1 ) (1 ) , 0 12(1 ) 2( )

0, otherwise

ds ydy y

y y yyg y

= --

ì - = - < <ï -= íïî

12

Functions of Two or More Random Variables

! Suppose X1,...,Xn have a discrete joint distribution withp.f. f, and m functions Y1,...,Ym of these n randomvariables are:

• For any given values y1,...,ym, let A denote the set ofall points (x1,...,xn) such that

• Then the joint p.f. g of Y1,...,Ym is:

1 1 1

1

( , , )

( , , )

n

m m n

Y r X X

Y r X X

=

=

LM M

L

1 1 1

1

( , , )

( , , )

n

m n m

r x x y

r x x y

=

=

LM ML

1

1 1( , , )

( , , ) ( , , )n

m nx x A

g y y f x xÎ

= åL

L L

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Variables with A Continuous Joint Distribution

! Suppose the joint p.d.f. of X1,...,Xn is f(x1,...,xn) andY=r(X1,...,Xn). For any given value y, let Ay be thesubset of containing all points (x1,...,xn) such that

. Then

If the distribution of Y is also continuous, then thep.d.f. of Y can be found by differentiating the d.f.G(y).

nR

1( , , )nr x x y£L

1

1 1

( ) Pr( ) Pr[ ( , , ) ]

( , , )y

n

n nA

G y Y y r x x y

f x x dx dx

= £ = £

= ò òL

L L L

14

The Distribution of Maximum and Minimum Values in a Random Sample

! Suppose X1,...,Xn form a random sample ofsize n from a distribution with p.d.f. f and d.f.F. Consider

! (1) d.f. and p.d.f of ?

1max( , , )n nY X X= L

1max( , , )n nY X X= L

15

Solution! X1,...,Xn form a random sample of size n from

a distribution with p.d.f. f and d.f. F.

[ ]

[ ]

1

1

1

1

max( , , )

( ) Pr( ) Pr( , , )Pr( ) Pr( )

( ) ( ) ( ) , for -

( )( ) ( ) ( ), for -

n n

n n n

nn

nnn

Y X X

G y Y y X y X yX y X y

F y F y F y y

dG yg y n F y f y ydy

-

=

= £ = £ £

= £ £

= = ¥ < < ¥

= = ¥ < < ¥

L

LL

L

16

1 1min( , , )nY X X= L! (2) d.f. and p.d.f of ?

17

! Solution:

[ ] [ ][ ]

[ ]

1 1

1 1 1

1

1

111

min( , , )

( ) Pr( ) 1 Pr( )1 Pr( , , )1 Pr( ) Pr( )1 1 ( ) 1 ( )

1 1 ( ) , for -

( )( ) 1 ( ) ( ), for -

n

n

n

n

n

Y X X

G y Y y Y yX y X yX y X yF y F y

F y y

dG yg y n F y f y ydy

-

=

= £ = - >= - > >= - > >

= - - -

= - - ¥ < < ¥

= = - ¥ < < ¥

L

LLL

18

1 1min( , , )nY X X= L! (3) Joint d.f. and joint p.d.f of and ? 1max( , , )n nY X X= L

19

! Suppose we want to find the joint distribution of and 1 1min( , , )nY X X= L 1max( , , )n nY X X= L

[ ][ ] [ ]

1

1 1 1

1 1

1 1 1

11

1

1

21

11

For - ,( , ) Pr( )Pr( ) Pr( )Pr( ) Pr( , , )

( ) Pr( )

( ) ( ) ( )

( ) ( ) ( )

( , )( , )

n

n n n

n n n n

n n n n nn

n n i ni

nn n n

n nn n

nn

n

y yG y y Y y and Y y

Y y Y y and Y yY y y X y y X y

G y y X y

G y F y F y

F y F y F y

G y yg y yy y

=

¥ < < < ¥

= £ £ =

= £ - £ >

= £ - < £ < £

= - < £

= - -

= - -

¶=

¶ ¶

Õ

L

[ ] 21 1 1( 1) ( ) ( ) ( ) ( ), -

0, otherwise

nn n nn n F y F y f y f y y y-ì - - ¥ < < < ¥ï= í

ïî

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Transformation of A Multivariate p.d.f.! Suppose X1,...,Xn have a continuous joint

distribution with joint p.d.f. f, and n newrandom variables Y1,...,Yn are defined by:

Suppose is the support for X1,...,Xn, theimage under the transformation is T. Assumethat the transformation from S to T is a one-to-one transformation.

1 1 1

1

( , , )

( , , )

n

n n n

Y r X X

Y r X X

=

=

LM M

L

nRS Ì

21

! We can get the inverse of the transformation:

! Suppose each partial derivative exists atevery point . The Jacobian of thetransformation can be constructed:

! The joint p.d.f. g of Y1,...,Yn can be derived:

1 1 1

1

( , , )

( , , )

n

n n n

X s Y Y

X s Y Y

=

=

LM M

L

ji ys ¶¶ /

1( , , )ny y TÎL

1 1

1

1

detn

n n

n

s sy y

Js sy y

¶ ¶é ùê ú¶ ¶ê ú

= ê úê ú¶ ¶ê úê ú¶ ¶ë û

L

M M

L

1 1 1 11

[ ( , , ), , ( , , )] | | ( , , )( , , )

0n n n n

n

f s y y s y y J for y y Tg y y

otherwiseÎì

= íî

L L L LL

22

Example

! Suppose X1 and X2 have a continuous jointdistribution with p.d.f.

What is the joint p.d.f. of Y1 and Y2?

îíì <<<<

=otherwise

xandxforxxxxf

010 104

),( 212121

, and Let 2122

11 XXYXXY ==

23

• The inverse of the transformation is

• S={(x1,x2): 0<x1<1, 0<x2<1}T={(y1,y2): y1>0, 0<y2<1, 0<y1y2<1, y2/y1<1}

• We have

1

22122212111 ),(),(

YYYYsXYYYYsX ====

212

231

2

1

2

2

1

2

1

1

2

1

1

121

21

21

21

yyys

yy

ys

yy

ys

yy

ys

=¶¶

-=¶¶

=¶¶

=¶¶

24

! The Jacobian is

! The p.d.f. of Y1 and Y2 is:1

1

2131

2

2

1

1

2

21||

21

121

21

21

21

det

yJ

yyyy

y

yy

yy

J

=

=

úúúúú

û

ù

êêêêê

ë

é

-

=

ïî

ïíì Î=

=otherwise

TyyforyyJyysyysfyyg

0

),(2||)],(),,([),( 211

2212211

21

25

The Sum of Two Random Variables! Suppose that X1 and X2 are i.i.d. random

variables and the p.d.f. for each is:

What is the p.d.f. g of Y=X1+X2?îíì >

=-

otherwisexfore

xfx

00

)(

26

Solution (1)

1

1

11 2

1

1 2

1 2 2 10 0

1 2 2 10 0

( )2 10 0

10

For 0,( ) Pr( )

Pr[ ]

( , )

= ( ) ( )

=

= ( )

1-0,

So ( )

y y x

y y x

y y x x x

y xy

y y

yG y Y y

X X y

f x x dx dx

f x f x dx dx

e dx dx

e e dx

e yey

G y

-

-

- - +

--

- -

>= £= + £

=

- +

= -

=

ò òò òò òò

01- , 0y ye ye y- -

£ìí

- >î

, 0( )( )0, otherwise

yye ydG yg ydy

-ì >= = í

î

27

Solution (2)! X1 and X2 have a given joint p.d.f. f, and we

want to find the p.d.f for Y=X1+X2.• Let Z=X2, then the transformation from X1 and X2 to

Y and Z will be a one-to-one linear transformation.

• The inverse of the transformation is

• S={(x1,x2): x1>0, x2>0}T={(y,z): y>0, 0<z<y}

• We have

X1=Y-Z, X2=Z

11011

det =÷÷ø

öççè

æ -=J

28

• The joint p.d.f. of Y and Z is:

• The marginal p.d.f. g of Y can be obtained by:

)()(),(),(0 zfzyfzzyfzyg -=-=

( )

0

0

For 0, ( ) ( ) ( )

, 0So

0, otherwise

y y z z

y y y

y

y g y f y z f z dz

e e dz

e dz ye

ye yg(y)

¥

-¥

- - -

- -

-

> = -

=

= =

ì >= íî

òòò

29

The Range! Suppose X1,...,Xn form a random sample of size n

from a distribution with p.d.f. f and d.f. F.and .W=Yn-Y1 is called the range of the sample. What isthe p.d.f. of W?

Solution: We already derived the joint p.d.f.g(y1,yn) of Y1 and Yn. If we let Z=Y1,then the transformation from Y1 and Yn

to W and Z will be a one-to-one lineartransformation.

),,min( 11 nXXY != ),,max( 1 nn XXY !=

30

• The inverse of the transformation is

• S={(y1,yn): -∞<y1<yn< ∞}T={(w,z): w>0, -∞< z < ∞}

• We have |J|=1.

• The marginal p.d.f of W is

Y1=Z, Yn=W+Z

[ ] 2

( , ) ( , )

( 1) ( ) ( ) ( ) ( ), ( , )0, otherwise

n

h w z g z w z

n n F w z F z f z f w z w z T-

= +

ì - + - + Îï= íïî

ò¥

¥-= dzzwhwh ),()(1

31

Example! Suppose that n variables X1,…,Xn form a random

sample from a uniform distribution on the interval(0,1). What is the p.d.f. of the range of the sample?

! Solution:

F(x)=x for 0<x<1.

îíì <<

=otherwise

xforxf

0101

)(

32

• The inverse of the transformation is

• S={(y1,yn): 0<y1<yn<1}T={(w,z): 0<w<1, 0< z <1-w}

• So

• The marginal p.d.f of W is

Y1=Z, Yn=W+Z

îíì -<<<<-

=-

otherwisewzwwnn

zwhn

010,10)1(

),(2

1 2 2

01

( 1) ( 1) (1 ) 0 1( )

0

w n nn n w dz n n w w for wh w

otherwise

- - -ì - = - - < <ï= íïî

ò