One Function of Two Random Variables One Function

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 ? Practical Viewpoint § A receiver output signal usually consists of the desired signal buried in noise § the above formulation in that case reduces to Z = X + Y.

§ 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 expressed in two different ways. and

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

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

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

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 Solution § Clearly, § There are two cases of z for which the shaded areas are quite different in shape and they should be considered separately.

Example – Continued

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

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 Solution and hence If X and Y are independent, which represents the convolution of with

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

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

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

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 Solution Using and the fact that we obtain

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

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

Example X and Y are independent normal r. vs with zero Mean and common variance Determine for Solution Direct substitution with gives where we have used the substitution 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 Solution This corresponds to a circle with radius Thus * ( ) If X and Y are independent Gaussian as in the previous example, 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

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!

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

Application § Fading multipath § situation where there is Line of sight signal (constant) Multipath/Gaussian noise - 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.

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

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

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

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 Solution § We have § and hence § But and § Thus min ( X, Y ) is also exponential with parameter 2. so that

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)

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

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

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.