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## Utah State University ECE 6010 Stochastic Processes Homework # 4 Solutions

1. Suppose .
1. Show that and .

Using characteristic functions:

Taking the gradient with respect to we have

Now evaluating at and dividing by we obtain

Taking the gradient again with respect to we have

and evaluating at we have

Dividing by we obtain , from which it follows that .

2. Show that .

We note that . Then

But we can compute the expected value using what we know from :

So

From the form of the characteristic function we identify that is Gaussian with

3. Suppose and write . Show that .

If we have, by applying the previous result

2. Suppose you have a random number generator which is capable of generating random numbers distributed as . Describe how to generate random vectors .

Generate -dimensional random vectors by generating realizations of the scalar random variable . Then .

Write (the Cholesky factorization). Let . Then

and

3. Suppose and are r.v.s. Show that is minimized over all functions be the function

Assume .

1. First approach: We will assume (only for convenience) that the r.v.s are continuous. We can write

To minimize this, we can minimize the inner integral

for each value of . Since is a fixed value for the inner integral, is a fixed value for each value of , and we can take the derivative with respect to and equate to zero to minimize:

The integral on the left is equal to 1 (since it integrates over the entire set of ), and the integral on the right is equal to .

2. Approach 2: This one is more appealing, because it is not expressed in terms of integrals (making it immediately applicable to all types of distributions), it does not require interchanging limiting operations (i.e., taking derivatives inside integrals), and does not even require derivatives with respect to functions (which was only partially justified in the first approach). This presentation is due to Ross.

The proof is accomplished by establishing the following inequality:

This is accomplished as follows:

as may be verified by expanding out the second line. However, given the term (which has the underbrace), being a function of , can be treated as a constant and factored out of the expectation. Thus

We thus obtain

or

Taking expectations of both sides we obtain

and equality holds if .

4. Let . Let , where

Determine the relationship between and and .

First observe that . Write down two expressions for the density, first in terms of the separate parameters,

and second in terms of the matrix parameters,

Equating the constants in front we must have

It is clear that and are the first and second elements of . Let us assume only for convenience that . Expanding the exponent of the second form of the pdf and equating to terms in the first form of the pdf we find

Comparing all of these, we have

5. Suppose where

1. The value is measured. Determine the best estimate for .

We can write for the mean of the unmeasured variables. The estimation formula is

where is the covariance of the unmeasured values with the measured value,

and . We obtain

2. In a separate problem, the values and are measured. Determine the best estimate of .

where

and

Thus

3. Determine a random vector which is a whitened version of .

Using M ATLAB ,

                S = [4 2 1; 2 6 3; 1 3 8];
R = chol(S);
C = R';
C*C'   % check the result

we determine that

Now let . Then is whitened.