§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＝xi, Y＝yj)＝pij ，i, j＝1, 2, …
then Z＝g(X, Y)～P{Z＝zk}＝
ij
i ,k : g ( xi , y j ) zk
k＝1, 2, …
or
p
…
(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)
＝pk ,
…
EX Suppose that X and Y are independent, and the
pmfs of X and Y are
X
1
3
Y
2
4
PX
0.3
0.7
PY
0.6
0.4
Find the pmf of Z=X+Y.
Solution Because X and Y are independent, so
P{ X  xi ,Y  y j }  P{ X  xi }P{Y  y j },
Y
2
4
X
0.12
1 0.18
3 0.42
0.28
X
1
3
Y
P
2 4
0.18 0.12
0.42 0.28
Z  X Y
0.18
0.12
0.42
0.28
( X ,Y ) Z  X  Y
(1,2)
(1,4)
( 3,2)
( 3,4)
3
5
5
7
3
5
7
0.18
0.54
0.28
then
P
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)
pij
(0,0)
V＝max(X, Y)
U＝min(X, Y)
V
0
1
(1,0)
pq
pq
0
1
1
p
2
0
1
0
1
1
0
1
q
W＝X＋Y
W
(0,1)
2
0
0
q
0
2
(1,1)
1
2
0
0
2 pq
p
2
2
Example Suppose X ~ P(1 ), Y ~ P(2 ) and are
independent of each other, then
Z  X  Y ~ P(1  2 ).
Example Suppose X ~ b(n, p ), Y ~ b(m, p ) and are independent
of each other , then
Z  X  Y ~ b(m  n, p).
Example Suppose
X ~ P(1 ), Y ~ P(2 ) and are
independent of each other, then
Z  X  Y ~ P(1  2 ).
Proof
e  1 1i
P ( X  i) 
i=0,1,2,…
i!
 2 j
e 2
P (Y  j ) 
j=0,1,2,…
j!
then
r
P ( Z  r )   P ( X  i ,Y  r  i )
i 0
r
P ( Z  r )   P ( X  i, Y  r  i )
i 0
r
 e
- 1

i 0


then
e
i!
 ( 1  2 )
r!
e
i
1
 ( 1  2 )
e
- 2

r-i
2
(r - i)!
r
r!
i r-i


1 2
i  0 i!(r - i)!
(1  2 ) , r = 0 , 1 , …
r
r!
Z  X  Y ~ P(1  2 ).
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.
y
The cdf of Z is :
FZ  z   P  Z  z 
 P  X  Y  z
  f ( x, y)dxdy
D
D={(x, y): x+y ≤z}
0
x
x y z
 f ( x, y)dxdy
FZ ( z ) 
x  y z

z y


FZ ( z )   [ 
f ( x, y)dx ]dy
y
y
0
Let x=u-y, then

z


z



FZ ( z )   [  f ( u  y, y)du]dy
  [  f ( u  y, y)dy]du
x
x y z
z



FZ ( z )   [  f ( u  y, y)dy]du
Then the pdf of Z=X+Y is :

fZ ( z )  F ( z )   f ( z  y, y)dy
'
Z

Or

fZ ( z )  F ( z )   f ( x, z  x )dx
'
Z

In particular, if X and Y are independent, then

f Z ( z )   f X ( z  y ) fY ( y )dy


 f ( z) 
Z




f X ( x ) fY ( z  x )dx
which is called the convolution of fx(.) and fy(.).
Example 1 If X and Y are independent with
the same pdf
1, 0  x  1
f ( x)  
el se
0,
Find the pdf of Z=X+Y.
Solution

fZ ( z )   f X ( x ) fY ( z  x )dx

 0  x 1

0  z  x  1
i.e.
 0  x 1

z  1  x  z

f Z ( z )   f X ( x ) fY ( z  x )dx

when z  0 or z  2 , f Z  z   0.
So
when 0  z  1 ,
f Z  z    dx  z
0
z
when 1  z  2 ,
f Z  z    dx  2  z
1
z 1
so
 z , 0  z  1,

f Z  z    2  z ,1  z  2,
 0 , else.

z  x 1
z
2
z
1
z
O zz 1 1
zx
x
Example 2 If X and Y are independent with
the same distribution N(0,1) , Find the pdf of
Z=X+Y.
Solution

fZ ( z )   f X ( x ) fY ( z  x )dx

2
z x


x2

2
1 
2

e

e
dx

2π 
z2
1   2  ( x 2  zx )

e e
dx

2π 
1

e
2π
z2

4



e
z
 ( x  )2
2
dx
1

e
2π
z2

4



e
z
 ( x  )2
2
dx
z
let t  x  , then
2
1
fZ  z   e
2π
z2

4

1
 e dt  2π e
t2


z2
2
1
2 2 
e
2π  2
Z=X+Y ～ N(0,2).
z2

4
 π
二、Z=Y\X, Z=XY
let f(x,y) is the joint pdf of (X,Y), then the pdf
of Z=Y\X is
fY X ( z)  

f XY ( z )  



x f ( x, xz)dx.
1
z
f ( x, )dx.
x
x
When X and Y are independent,
fY X ( z )  

f XY ( z )  



x f X ( x) fY ( xz )dx.
1
z
f X ( x) fY ( )dx.
x
x