Download Joint Probability Density Function of two functions of two

Joint Probability Density Function of two functions of two random variables
We consider the transformation ( g1 , g 2 ) : R 222 2  R 2 . We have to find out the joint
probability density function f Z1 ,Z2  z1 , z2  where Z1  g1  X , Y  and Z2  g2  X , Y  . We
hve to find out the joint probability density function f Z1 ,Z2  z1 , z2  where z1  g1  x, y 
and z2  g2  x, y  . Suppose the inverse mapping relation is
x  h1 ( z1 , z2 ) and y  h2 ( z1 , z2 )
Consider a differential region of area dz1 dz2 at point ( z1 , z2 ) in the Z1  Z 2 plane.
Let us see how the corners of the differential region are mapped to the X  Y plane.
Observe that
h
h
h1 ( z1  dz1 , z2 )  h1 ( z1 , z2 )  1 dz1  x  1 dz1
 z1
 z1
h
h
h2 ( z1  dz1 , z2 )  h1 ( z1 , z2 )  2 dz1  y  2 dz1
 z1
 z1
Therefore,
h
h
The point ( z1  dz1 , z2 ) is mapped to the point ( x  1 dz1 , y  2 dz ) in the
 z1
 z1
X  Y plane.
Y
Z2
( x, y )
( z1 , z2 )
(x 
 h1
h
dz1 , y  2 dz )
 z1
 z1
( z1  dz1 , z2 )
X
Z1
We can similarly find the points in the X  Y plane corresponding to ( z1 , z2  dz2 ) and
( z1  dz1 , z2  dz2 ). The mapping is shown in Fig. We notice that each differential region
in the X  Y plane is a parallelogram. It can be shown that the differential parallelogram
at ( x, y ) has a area
J ( z1 , z2 ) dz1dz2 where J ( z1 , z2 ) is the Jacobian of the
transformation defined as the determinant
 h1
 z1
J ( z1 , z2 ) 
 h2
 z1
 h1
 z2
 h2
 z2
Further, it can be shown that the absolute values of the Jacobians of the forward and the
inverse transform are inverse of each other so that
1
J ( z1 , z2 ) 
J ( x, y)
where
 g1
x
J ( x, y ) 
 g2
x
 g1
y
 g2
y
Therefore, the differential parallelogram in Fig. has an area of
dz1dz2
.
J ( x, y )
Suppose the transformation z1  g1  x, y  and z2  g2  x, y  has n roots and let
( xi , yi ), i  1, 2,..n be the roots. The inverse mapping of the differential region in the
X  Y plane will be n differential regions corresponding to n roots. The inverse
mapping is illustrated in the following figure for n  4. As these parallelograms are nonoverlapping,
n
f Z1 , Z 2  z1 , z2  dz1 dz2   f X ,Y ( x, y )
i 1

n
f X ,Y ( x, y )
i 1
J ( xi , yi )
f Z1 , Z 2  z1 , z2   
dz1 dz2
J ( xi , yi )
Remark
 If z1  g1  x, y  and z2  g2  x, y  does not have a root in ( x, y ),
then
f Z1 ,Z2  z1 , z2   0.
(x 
Y
(x 
Z2
( z1  dz1 , z2  dz2 )
( z1 , z2  dz2 )
( z1 , z2 )
x
x
y
y
dz1 
dz2 , y 
dz1 
dz )
 z1
 z2
 z1
 z2 2
x
y
dz z , y 
dz )
 z2
 z2 2
( x, y )
( z1  dz1 , z2 )
(x 
x
y
dz1 , y 
dz )
 z1
 z1 1
X
Z1
Y
Z2
( z1 , z2 )
( z1  dz1 , z2 )
Z1
Example: pdf of linear transformation
Z1  aX  bY
Z 2  cX  dY
Then
X
dz1  bz2
az  cz1
, y 2
ad  bc
ad  bc
a b
J ( x, y ) 
 ad  bc
c d
Suppose X and Y are two independent Gaussian random variables each with mean 0 and
Y
variance  2 . Given R  X 2  Y 2 and   tan 1 , find f R (r ) and f ( ) .
X
-----------------------------------------------------------------------------------------------------------Solution:
x
We have x  r cos and y  r sin  so that r 2  x 2  y 2 …………. (1)
and tan  
From (1)
r x
  cos 
x r
r y
and
  sin 
y r
From (2)

y
sin 
 2

2
x x  y
r

x
cos 
 2

2
y x  y
r
 cos 
 J  det  sin 


r
 f R, (r , ) 
sin  
1
cos   
r
r 
f X ( x, y) 

J
 x r cos
y  r sin

r
2 2
2
2
 r 2 cos2   r 2 sin 2 
2

2

e
.e
y
……………… (2)
x

2

 f R (r ) 
r
2 2
2
 r 2
e 2
f R , (r ,  )d
0

r
2
2
 r 2
e 2
0r 

f ( )   f R, (r , )dr
0

1

2 2 0

2
 r 2
re 2 dr
1
0    2
2
Rician Distribution:

X and Y are independent Gaussian variables with non zero mean  X and Y
respectively and constant variance.



We have to find the joint density function of the random variable Z  X 2  Y 2 .
Envelope of a sinusoidal + a narrow band Gaussian noise.
Received noise in a multipath situation.
Z  X 2 Y 2
Y
  tan 1
X
We have to find J(x, y) corresponding to z and  .
 z z 
 x y 

J ( x, y )  det 
   
 x y 


From
z  x2  y 2
We have,
z 2  x2  y 2
y
x
y
and tan  
x
and tan  
Therefore,
and
also
and
z x
  cos 
x z
z y
  sin 
y z

y
y
 2
  2 cos 2 
2
x
x sec 
x

y
1

 cos 2 
2
y x sec  x
 cos 
 J ( x, y )  det  y 2 cos 2 


x2
cos3  y sin  cos 2 


x
x2
 x cos   y sin  
 cos 2  

x2


2
z cos  1


x2
z
sin  

1
2 
cos x

x
Consider the transformation as shown in the diagram below:
f X ,Y ( x, y ) 
1
2 2
e
2
2

 1 2   x   X   y  Y  
2 

We have to find the density at (x, y) corresponding to z and  .
From the above figure,
X   X  Z cos    cos 0
and
f X ,Y ( x, y) 
Y  Y  Z sin    sin 0
1
2 2

e
2

2
 1 2   z cos   cos0   z sin   sin  
2 

1


1 e 2 2  z2 2 z cos( 0 ) 2 
2
2
z2   2 z

cos( 0 )
1
2 2 e 2

e
2 2