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§3.5 Distribution of Special Functions. Sum and difference Z=X+Y, Z=X-Y Product and quotient Z=XY, Z=Y\X. Functions of discrete random vectors. Suppose that (X, Y) ~ P(X = x i , Y = y j ) = p ij , i, j = 1, 2, … - PowerPoint PPT Presentation
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§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