Page 68 - DMTH404_STATISTICS

Statistics

Notes Now since X and X are independent, we have
1 2
1 1 2 1  1 2 2 x
fx , x (x , x ) = e  x e 2 ,  x ,x   .
2 2
l 2 1 2 2 1 1 2
Hence by (1) the joint probability density function of (Z , Z ) is
1 2
 z  z z  z 2 
(z , z ) = f 1 2 , 1 |J|,  z , z  
1 2   1 2
 2 2 
1   1 z  z 2  1 z  z 2  2  


= exp   1    1  

4   2  2  2  2   
2
1   z  2 z  
= exp  1  2   , –   z ,z  
 
2
1
4  4 4  
 
Then the marginal density of Z is given by
1
2
 1 1 z
 z (z )   f(z1,z2) z 2 e ,  z   .

4
1
1
2
– 4
Note that we can calculate the marginal density of Z2 also. It is given by
2
 1 1 z
4

 z (z )    (z ,z ) z 2 e ,  z   .
2
2
1
2
2
– 4
In other words Z has N(0, 2) and Z has N(0, 2) as their distribution functions. In fact Z and Z
1 2 1 2
are independent random variables since
(z , z ) = z (z ) z (z )
1 2 1 1 2 2
for all z and z .
l 2
We shall illustrate this method with one more example.
Example 4 : Suppose X and X are independent random variables with common density function
1 2
1 x /2
f (x) = e  for 0 < x < 
2
= 0 otherwise.
1
Let us find the distribution function of Z = (X – X ).
1 2 1 2
Here it is convenient to choose Z = X . Note that the transformation
2 2
(X , X )  (Z , Z ) gives a one-to-one mapping from the set A = { (x , x ) : 0 < x < w, 0 < x < } onto
1 2 1 2 1 2 1 2
the set
B = { (z , z ) : z > 0, - < x <  and z > –2z }. The inverse transformation is
1 2 2 2 2 l
X = 2Z + Z
1 1 2
and
X = Z .
2 2

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