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 - Leibnitz Differentiation 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. 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

8. 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.

9. 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

10. 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

11. Example – Continued

12. 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

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

16. 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

17. Integrating over these two regions, we get
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. 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

20. 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

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 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

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. 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!

26. 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

27. 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.

28. 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

29. 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.

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

31. 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

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

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. Example - continued
(a)
(b)

36. 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

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

38. 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.