Page 69 - DMTH404_STATISTICS

Unit 5: Functions of Random Variables

Since x > 0, it follows that 2z + z > 0, that is, z > –2z1. Notes
l l 2 2
Since x > 0, it follows that z > 0. Obviously, - < z < .
2 2 l
Further more you can check that the Jacobian of the transformation is equal to 2.
Now the joint density function is


1  (x y)
fx ,x , (x , x ) = e 2
4
1 2 1 2
Therefore from (2) the joint density function of (Z , Z ) is
1 2
4 [z , z ] = fx , x , [2z + z , z ] | 1 | provided (z , z )  B
1 2 1 2 1 2 2 l 2
= 0 , otherwise.
Therefore

1
f[z ,z ]  e  1 z  2 z if (z , z )  B
1
2
2 1 2
= 0 otherwise ...*
and the marginal density of Z is
1
 1
z (z ) =  e  1 z  2 z dz if – < z < 0
1 1 2 2 1
 2z
1
 1
=  e  1 z  2 z dz if 0  z < 
2
1
0 2
This shows that
1  |z 1 |
z (z ) = e , – < z < 
1 1 2 1
The distribution with the density function given by * is known as double exponential distribution.
An important application of the transformation approach is to determine distribution of the
sum of two independent random variables not necessarily identically distributed. Let us now
look at this problem.
Suppose X and X are independent random variables with the density functions f1 (x ) and f (x )
1 2 1 2 2
respectively and we want to determine the distribution function of X + X . Let Z = X + X . We
1 2 1 1 2
apply transformation method here. Set Z = X . Then the transformation (X , X )  (Z , Z ) is
2 2 1 2 1 2
invertible and
X = Z – Z
1 1 2
X = Z .
2 2
The Jacobian of the transformation is equal to unity. Since the joint density of (X , X ) is f (x )
1 2 1 1
f (x ), it follows from (2) that the joint density of (Z , Z ) is given by
2 2 1 2
(z , z ) = f (z – z ) f (z ), – < z , z < .
1 2 1 1 2 2 2 l 2

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