One Function of Two Random Variables

1. One Function of Two Random Variables

2. 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

3. 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.

4. 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.

5. Recall - LeibnitzDifferentiation Rule Suppose Then Using the above,

6. 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.

7. 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 • for • then using the figure we can determine the new limits for

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

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

10. 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

11. Example – Continued

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

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

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

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

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

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

18. 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

19. Example Xand 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

20. 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

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

22. Example X and Y are independent normal r.vs with zero Mean and common variance Determine for Direct substitutionwith 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

23. 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

24. 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

25. 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!

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

27. Multipath/Gaussian noise Line of sight signal (constant) Application • 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.

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

29. 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.

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

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

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

33. 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

34. 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

35. (a) (b) Example - continued

36. 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

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

38. 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.