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## Utah State University ECE 6010 Stochastic Processes Homework # 3 Solutions

1. Show that .

Therefore,

2. Suppose . Use the ch.f. of to find an expression for , .

3. Suppose and are the indicator functions of events and , respectively. Find , and show that and are independent if and only if .

So from the equation above,

are independent

independent

Therefore, and are independent.

4. Suppose is a ch.f. Show that is also a ch.f.

Let be another random variable, so and .

So, is a ch.f. for r.v. .

5. Suppose and are jointly Gaussian. Use ch.f.s to show that .

Now,

6. Suppose and are jointly continuous. (a) Show that

and thus that

For and jointly continuous,

Now,

(b) Suppose . Show that

Now,

therefore,

substituting this in the first equation we get

7. Suppose and are independent continuous r.v.s with c.d.f.s and , respectively. Suppose further that for all . Show that

 (because independent)

8. Prove Jensen's inequality for the case of simple-function r.v.'s

First, we prove that the convexity idea generalizes to multiple points. For a convex function we know that

The general result is:

 (***)

where and .

We'll do it for three points, from which the induction to points should be straightforward. Let . Consider

Factor out of the first two terms the quantity :

Since , convexity applies to the two terms in the square brackets:

 (*)

Now letting and , we see that we have obtained

for which convexity applies again:

 (**)

Combining (*) and (**) we obtain

Now to Jensen's inequality. It, too, is proved by induction. We will demonstrate explicitly the first couple of steps. Suppose (a simple function involving a single set . Then takes on two values: , with probability and 0 , with probability . Then

and for a convex function

where the inequality follows since is convex, and .

Now consider a simple function involving two disjoint sets:

Then takes on three values, , , and 0, with probabilities and , respectively. Then

And

by the convexity in (***).

9. Prove the Schwartz inequality.

Consider the quantity , which is for all values of the real constant . Expanding, we have

 (*)

Now find the value of that minimizes the right-hand side by differentiating with respect to :

so the minimizing . Substitution into (*) we have

Simplifying, we obtain the expression