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Statistics

Notes
z
=  e  x   e  x   dx
 
0
z
=  e  x   z e  z


0 
–z
= 1 – e – ze –z
Now we leave it to you to check that the density function of Z is
f (z) =  z e for z > 0
-z
2
z
= 0 otherwise
In this density function familiar to you? Recall that this function is the gamma density function
you have studied in Unit 11. Hence Example 2 says that the sum of two independent exponential
random variables has gamma distribution.
Let us consider another example.

Example 3: Suppose X and Y are independent random varihles with the same density
function I (x) and the distribution function F(x). Define Z = max(X,Y). Let us determine the
distribution function of 2.

By definition, the distribution function F is given by
z
F (z) = P [Z  z]
z
= P[max (X, Y)  z]
= P [ X  z , Y  z ]
= P [X  z] P [Y  Z] = [F (z)] 2

by the independence of X and Y and the fact that
P [X  z ] = P [Y  z] = F(z).
Since F is diflerentiable almost everywhere and the density corresponding to F is fit follows that
Z has a probability density function fz and
fz (z) = 2F (z) f(z), – < z < .
To get more practise why don’t you try some exercises now.

The examples and exercises discussed above deal with the method of obtaining the distribution
function of Z= g(X,Y) directly. This method is applicable even when (X,Y) does not have a
density function.

Next we shall discuss another method for obtaining the distribution and density functions.

5.1.2 Transformation Approach

Suppose (X , X ) is a bivariate random vector with the density function fx , x (x , x ) and we
1 2 1 2 l 2
would like to determine the distribution function of the density function of Z = g (X , X ). To
1 1 1 2
determine this, let us suppose that we can find another function Z = g (X , X ) such that the
2 2 1 2
2
transformation from (X , X ) to (Z , Z ) is one-to-one. In other words to every point (x , x ) in R ,
1 2 1 2 1 2

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