§3.5 Distribution of Special Functions

• Sum and difference Z=X+Y, Z=X-YProduct and quotient

Z=XY, Z=Y\X

Functions of discrete random vectors

Suppose that

(X, Y) ～ P(X ＝ xi, Y ＝ yj) ＝ pij ， i, j ＝ 1, 2, …

then Z ＝ g(X, Y) ～ P{Z ＝ zk} ＝ ＝ pk ,

k ＝ 1, 2, …

kji zyxgkiijp

),(:,

(X,Y) (x1,y1) (x1,y2) … (xi,yj) …

pij p11 p12 pij

Z=g(X,Y) g(x1,y1) g(x1,y2) g(xi,yj)

or

EX Suppose that X and Y are independent, and the pmfs of X and Y are

X

XP

31

7.03.0

Y

YP

42

4.06.0

Find the pmf of Z=X+Y.

},{}{},{ jiji yYPxXPyYxXP Y

X 42

13

18.0 12.0

42.0 28.0

Because X and Y are independent, soSolution

),( YX

)4,3(

)2,3(

)4,1(

)2,1(

P

18.012.042.0

28.0

YXZ

35

57

thenYXZ

P

3 5 7

18.0 54.0 28.0

YX 42

13

18.0 12.0

42.0 28.0

EX Suppose that X and Y are independent and both are uniformly distributed on 0-1 with law X 0 1

P q p Try to determine the distribution law of (1)W ＝ X ＋ Y ; (2) V ＝ max(X, Y) ；(3) U ＝ min(X, Y);(4)The joint distribution law of w and V .

(X,Y) (0,0) (0,1) (1,0) (1,1)

pij

W ＝ X ＋YV ＝ max(X, Y)

U ＝ min(X, Y)

2q pq pq 2p

0 1 1 2

0 1 1 1

0 0 0 1

VW

0 1

0 1 2

2q 0 0

0 pq22p

Example Suppose and are independent of each other, then

1 2~ ( ), ~ ( )X P Y P

1 2~ ( ).Z X Y P

Example Suppose and are independent of each other , then

~ ( , ), ~ ( , )X b n p Y b m p

~ ( , ).Z X Y b m n p

Proof

r

i

irYiXPrZP0

),()(

then

i = 0 , 1 , 2 , …

j = 0 , 1 , 2 , …

!)(

i

eiXP

i1

1

!)(

j

ejYP

j2

2

Example Suppose and are independent of each other, then

1 2~ ( ), ~ ( )X P Y P

1 2~ ( ).Z X Y P

0

( ) ( , )r

i

P Z r P X i Y r i

1 2

i r-i- -1 2

0

e ei! (r - i)!

r

i

1 2( )

i r-i1 2

0

r!

! i!(r - i)!

r

i

e

r

1 2( )

1 2( ) ,!

re

r

r = 0 , 1 , …

then 1 2~ ( ).Z X Y P

a) Sum and difference

Let X and Y be continuous r.v.s with joint pdf f (x, y), Let Z=X+Y ,Find the pdf of Z=X+Y.

D

dxdyyxf ),(

D={(x, y): x+y ≤z}

The cdf of Z is :

ZF z P Z z

P X Y z

x y z

x

y

0

zyx

Z dxdyyxfzF ),()(

yz

Z dydxyxfzF ]),([)(

Let x=u-y, then

z

Z dyduyyufzF ]),([)(

zdudyyyuf ]),([

x y z

x

y

0

y

Then the pdf of Z=X+Y is :

Or

dyyyzfzFzf ZZ ),()()( '

dxxzxfzFzf ZZ ),()()( '

z

Z dudyyyufzF ]),([)(

In particular, if X and Y are independent, then

dyyfyzfzf YXZ )()()(

dxxzfxfzf YXZ )()()(

which is called the convolution of fx(.) and fy(.).

Example 1 If X and Y are independent with the same pdf

Find the pdf of Z=X+Y.

1, 0 1( )

0,

xf x

el se

dxxzfxfzf YXZ )()()(

Solution

10

10

xz

xi.e.

zxz

x

1

10

z x

z

xOz

1z x

2

1

1z

z 1z

0.Zf z So when or ,0z 2z

0

z

Zf z dxwhen ,0 1z

1 2z when ,

z

1

1Z zf z dx

2 z

so

, 0 1,

2 ,1 2,

0 , .Z

z z

f z z z

else

dxxzfxfzf YXZ )()()(

Example 2 If X and Y are independent with th

e same distribution N(0,1) , Find the pdf of Z=X

+Y.

dxxzfxfzf YXZ )()()(

Solution

22

2 21

2

z xx

e e dxπ

22( )

4 21

2

z zx

e e dxπ

2

2( )21

2

zx zxe e dx

π

22( )

4 21

2

z zx

e e dxπ

let ,2

zt x then

Zf z 2

24

1

2

zte e dt

π

2

41

2

z

e ππ

2

22 21

2 2

z

eπ

Z=X+Y ～ N(0,2).

二、 Z=Y\X, Z=XY

( ) ( , ) .Y Xf z x f x xz dx

1( ) ( , ) .XY

zf z f x dx

x x

let f(x,y) is the joint pdf of (X,Y), then the pdf of Z=Y\X is

When X and Y are independent,

1( ) ( ) ( ) .XY X Y

zf z f x f dx

x x

( ) ( ) ( ) .Y X X Yf z x f x f xz dx