One Function of Two Random Variables

One Function of Two Random Variables X and Y : Two random variables g(x,y): a function We form a new random variable Z as Given the joint p.d.f how does one obtain the p.d.f of Z ? A receiver output signal usually consists of the desired signal buried in noise the above formulation in that case reduces to Z = X + Y. Practical Viewpoint

We have: Where in the XY plane represents the region where . need not be simply connected First find the region for every z, Then evaluate the integral there.

Example Z = X + Y. Find Integrating over all horizontal strips(like the one in figure) along the x-axis We can find by differentiating directly.

Recall - Leibnitz Differentiation Rule Suppose Then Using the above,

Example – Alternate Method of Integration If X and Y are independent, then This integral is the standard convolution of the functions and expressed in two different ways.

As a special case, suppose that Example – Conclusion If two r.vs are independent, then the density of their sum equals the convolution of their density functions. As a special case, suppose that for then using the figure we can determine the new limits for

On the other hand, by considering vertical strips first, we get Example – Conclusion In that case or On the other hand, by considering vertical strips first, we get if X and Y are independent random variables.

X and Y are independent exponential r.vs Example X and Y are independent exponential r.vs with common parameter , let Z = X + Y. Determine Solution

X and Y are independent uniform r.vs in the common interval (0,1). Example This example shows that care should be taken in using the convolution formula for r.vs with finite range. X and Y are independent uniform r.vs in the common interval (0,1). let Z = X + Y. Determine Clearly, There are two cases of z for which the shaded areas are quite different in shape and they should be considered separately. Solution

Example – Continued

By direct convolution of and we obtain the same result. In fact, for Example – Continued So, we obtain By direct convolution of and we obtain the same result. In fact, for

Example – Continued and for This figure shows which agrees with the convolution of two rectangular waveforms as well.

Example Let Determine its p.d.f and hence If X and Y are independent, which represents the convolution of with Solution

After differentiation, this gives Example – a special case Suppose For and for After differentiation, this gives

Given Z = X / Y, obtain its density function. Example Given Z = X / Y, obtain its density function. if Since by the partition theorem, we have and hence by the mutually exclusive property of the later two events Solution

Integrating over these two regions, we get Example – Continued Integrating over these two regions, we get Differentiation with respect to z gives

Example – Continued If X and Y are nonnegative random variables, then the area of integration reduces to that shown in this figure: This gives or

X and Y are jointly normal random variables with zero mean so that Example X and Y are jointly normal random variables with zero mean so that Show that the ratio Z = X / Y has a Cauchy density function centered at Using and the fact that we obtain Solution

which represents a Cauchy r.v centered at Example – Continued where Thus which represents a Cauchy r.v centered at Integrating the above from to z, we obtain the corresponding distribution function to be

Example Obtain We have But, represents the area of a circle with radius and hence This gives after repeated differentiation Solution

Example X and Y are independent normal r.vs with zero Mean and common variance Determine for Direct substitution with gives where we have used the substitution Solution Result - If X and Y are independent zero mean Gaussian r.vs with common variance then is an exponential r.vs with parameter

Example Let Find This corresponds to a circle with radius Thus If X and Y are independent Gaussian as in the previous example, Solution (*) Rayleigh distribution

Example – Conclusion If where X and Y are real, independent normal r.vs with zero mean and equal variance, then the r.v has a Rayleigh density. W is said to be a complex Gaussian r.v with zero mean, whose real and imaginary parts are independent r.vs. As we saw its magnitude has Rayleigh distribution. What about its phase

U has a Cauchy distribution with Example – Conclusion Let U has a Cauchy distribution with As a result The magnitude and phase of a zero mean complex Gaussian r.v has Rayleigh and uniform distributions respectively. We will show later, these two derived r.vs are also independent of each other!

What if X and Y have nonzero means and respectively? Example What if X and Y have nonzero means and respectively? Since Solution substituting this into (*), and letting we get Rician p.d.f. the modified Bessel function of the first kind and zeroth order where

Application Fading multipath situation where there is Line of sight signal (constant) Multipath/Gaussian noise Fading multipath situation where there is a dominant constant component (mean) a zero mean Gaussian r.v. The constant component may be the line of sight signal and the zero mean Gaussian r.v part could be due to random multipath components adding up incoherently (see diagram below). The envelope of such a signal is said to have a Rician p.d.f.

special cases of the more general order statistics. Example Determine nonlinear operators special cases of the more general order statistics. We can arrange any n-tuple such that: Solution

represent the set of order statistics among n random variables. Example - continued represent r.vs, The function that takes on the value in each possible sequence is known as the k-th order statistic. represent the set of order statistics among n random variables. represents the range, and when n = 2, we have the max and min statistics.

since and form a partition. Example - continued Since we have since and form a partition.

From the rightmost figure, If X and Y are independent, then Example - continued From the rightmost figure, If X and Y are independent, then and hence Similarly Thus

Example - continued (a) (b) (c)

Example X and Y are independent exponential r.vs with common parameter . Determine for We have and hence But and so that Thus min ( X, Y ) is also exponential with parameter 2. Solution

Example X and Y are independent exponential r.vs with common parameter . Define Determine Solution We solve it by partitioning the whole space. Since X and Y are both positive random variables in this case, we have

Example - continued (a) (b)

Let Determine the p.m.f of Z. Example – Discrete Case Let X and Y be independent Poisson random variables with parameters and respectively. Let Determine the p.m.f of Z. Z takes integer values For any gives only a finite number of options for X and Y. The event is the union of (n + 1) mutually exclusive events given by Solution

If X and Y are also independent, then Example – continued As a result If X and Y are also independent, then and hence

Thus Z represents a Poisson random variable with parameter Example – conclusion Thus Z represents a Poisson random variable with parameter Sum of independent Poisson random variables is also a Poisson random variable whose parameter is the sum of the parameters of the original random variables. Such a procedure for determining the p.m.f of functions of discrete random variables is somewhat tedious. As we shall see, the joint characteristic function can be used in this context to solve problems of this type in an easier fashion.