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Unit 5: Functions of Random Variables

2
there corresponds a point (x , x ) in R given by the above transformation and conversely to Notes
1 2
every point (z , z ) there corresponds a unique point (x , x ) such that
1 2 l 2
z = g (x , x )
1 1 1 2
z = g (x , x )
2 2 1 2
For example suppose that Z = X + X . Then we can choose Z = X – X . You can easily see that
1 1 2 2 1 2
2
the transformation (x , x )  (z , z ) from R to R is one-one and in this case we have
2
1 2 1 2
Z  Z Z  Z
X  1 2 and X  1 2
1 2
2 2
So, in general, one can assume that we can express (X , X ) in terms of (Z , Z ) uniquely.
1 2 1 2
That means that there is exist real valued functions h and h such that
l 2
X = h (Z , Z )
1 1 1 2
X = h (Z , Z )
2 2 1 2
Let us further assume that h and h have continuous partial derivatives with respect to Z , Z .
1 2 1 2
Consider the Jacobian of the tranformation (Z , Z ) (X , X )
1 2 1 2
 h 1  h 2
z  1 z  2   h  h 2   h  h 2
1
2
 h 2  h 2 z  1 z  2 z  2 z  2
z  2 z  2
Recall that you. have seen ‘Jacobians’ in Unit 9, Block 3 of MTE - 07 we denote this Jacobian by
 (x ,x )
J  1 2 . Assume that J is not zero for all (z , z ). Then, it can be shown, by using the change
 (z ,z ) l 2
1 2
of variable formula for double integrals [see MTE-07, Unit 11, Block 41 we can show that the
random vector (Z , Z ) has a density and the density function  (z , z ) of (Z , Z ) is
1 2 l 2 1 2
 (z , z ) = f [ h (z , z ), h (z , z ) ] | J | if (z , z ) E B ...(2)
1 2 1 1 2 2 1 2 1 2
= 0 otherwise
where B = { (z , z ) : z = g (x , x ), z = g (x , x ) for some (x , x ) }.
1 2 1 1 1 2 2 1 2 1 2
From the joint density of (Z , Z ) obtained above, the marginal density of Z can be derived and
1 2 1
it is given by

 2 (z )   f(z ,z ) dz 2
1
2
1
–
Let us now compute the density function of Z = X + X where X and X are independent and
1 1 2 1 2
identically distributed standard normal variables. We have seen that in this cask Z – X + X and
2 1 2
we can write
Z + Z Z – Z
X  1 2 and X = 1 2 . Let us now calculate the Jacobian of the transformation. It is
1
2
2 2
given by
 x  x
1 1 1 1
z  z  1
1 2  2 2   .
 x  x 1 1 2
2 2 
z  z  2 2
1 2
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