Distributions of functions of two random variables

14.23 Distributions of functions of two random variables

We turn now to the following problem: If X and Y are one-dimensional random variables with known distributions, how do we find the distributionof new random variables such as X + Y, XY, or

This section describes a method that helps to answer questions like this. Two new random variables

are defined by equations of the form

where Y) or Y) is the particular combination in which we are interested. From

a knowledge of the joint distribution of the two-dimensional random variable (X, Y) we calculate the joint distribution

Once is known, the individual distribu- tions of

of

and V are easily found.

To describe the method in detail, we consider a one-to-one mapping of the xy-plane onto the

defined by the pair of equations

v=

Let the inverse mapping be given by

and assume that Q and R have continuous partial derivatives. If T denotes a region in the xy-plane, let T’ denote its image in the uv-plane, as suggested by Figure 14.14. Let X and

Y be two one-dimensional continuous random variables having a continuous joint distribu- tion and assume (X, Y) has a probability density

Define new random variables and V by writing

Y). To determine a probability density of the random variable (U, V) we proceed as follows: The random variables X and Y are associated with a sample space S. For each

in S we have

Since the mapping is one-to-one, the two sets

Y(w)] and V(w ) =

and

are equal. Therefore we have (14.37)

V)

the density function of (X, Y) we can write

Using (14.37) and the formula for transforming a double integral we rewrite (14.38) as follows :

Y) T'] =

du dv.

I,I

Distributions of functions of two random variables

F IGURE 14.14 A one-to-one mapping of a region Tin the xy-plane onto a region in

the uv-plane.

Since this is valid for every region T’ in the uu-plane a density of is given by the integrand on the right; that is, we have

The densities and can now be obtained by the integration formulas

of two random variables. Given two one-dimensional random variables

EXAMPLE 1. The sum and

and Y with joint density determine density functions for the random variables

Yand

X- Y.

We use the mapping given by = x + , v = . This is a nonsingular linear transformation whose inverse is given by

The Jacobian determinant is

Calculus

of probabilities

Applying Equation (14.39) we see that a joint density g of is given by the formula

To obtain a density

we integrate with respect to and find

The change of variable x =

dx = du, transforms this to

Similarly, we find

An important special case occurs when and Y are independent. In this case the joint probability density factors into a product,

and the integrals for

and

become

EXAMPLE 2. The sum of two exponential distributions. Suppose now that each of and Y has an exponential distribution, say

= = 0 for < 0, and

for

Determine the density of

+ Y when

and Y are independent.

If < 0 the integral for is 0 since the factor = 0 for x < 0, and the factor

Solution.

x) = 0 for x 0. If 0 the integral for

becomes

dx =

dx .

To evaluate the last integral we consider two cases, = and 1. If = the integral has the value and we obtain

for

If we obtain for

Exercises

3. The maximum and minimum of two independent random variables. Let and Y be two independent one-dimensional random variables with densities

EX AM PLE

and

and

corresponding distribution functions

be the random variables

Y},

Y}.

That is, for each in the sample space,

is the minimum of the two numbers X(o), Y(o). The mapping = max

is the maximum and

v = min {x, is not to-one, so the procedure used to deduce Equation (14.39) is not applicable. However, in this case we can obtain the distribution functions of

directly from first principles. First we note that

and

. Therefore t) = Y

if, and only if,

and Y

By independence this is equal to t)P( Y t) = Thus, we have

At each point of continuity

and

we can differentiate this relation to obtain

Similarly, we have > if and only if > and Y > Therefore =

> t)P( Y > t)

At points of continuity of

and

we differentiate this relation to obtain