ECE 534 Homework Assignment 3 for Spring 2008 - Problems and Solutions - Prof. Venugopal V, Assignments of Electrical and Electronics Engineering

The third homework assignment for the ece 534 course during the spring 2008 semester. The assignment includes several problems related to gaussian random variables, conditional expectations, and conditional covariance matrices. Students are required to read chapter 3 of the textbook and the solutions to the even-numbered problems. The document also provides a solution for problem 3.17, which deals with the joint distribution of gaussian vectors.

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ECE 534 Spring 2008
February 20, 2008
Homework Assignment 3
Due Date: Monday, March 3 (in class)
Reading: Chapter 3 of text. Also read the solutions to the even numbered problems of Chapter 3
given at the end of the book.
1. Problem 3.1.
2. Problem 3.3. Hint for (b): Can E[X|Y] be negative.
3. Problem 3.5.
4. Problem 3.7. Note: Regarding the uniqueness in part (b), the answer is non-unique. You may
want to think of an example to show that the answer is non-unique, but you don’t have to do it
as part of the homework.
5. Using the joint characteristic function show that if X1,X2,X3,X4are jointly Gaussian (correlated)
random variables with zero mean, then
E[X1X2X3X4] = E[X1X2]E[X3X4] + E[X1X3]E[X2X4] + E[X1X4]E[X2X3]
6. Problem 3.9.
7. Problem 3.15.
8. (Problem 3.17 rewritten with better notation.) Assume that Xand Zare independent Gaussian
vectors with means µX,µZand covariance matrices KXand KZ. Let Y=X+Z. Then clearly
Xand Yare jointly Gaussian.
(a) Show that
E[X|Y] = µX+KX(KX+KZ)1(Y(µX+µZ))
=KZ(KX+KZ)1µX+KX(KX+KZ)1(YµZ)
(b) Show that the conditional covariance matrix of X, given Yis given by
Cov(e) = KXKX(KX+KZ)1KX=KX(KX+KZ)1KZ= (K1
X+K1
Z)1
where e=XE[X|Y]
c
V. V. Veeravalli, 2008 1

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ECE 534 Spring 2008 February 20, 2008

Homework Assignment 3

Due Date: Monday, March 3 (in class)

Reading: Chapter 3 of text. Also read the solutions to the even numbered problems of Chapter 3 given at the end of the book.

  1. Problem 3.1.
  2. Problem 3.3. Hint for (b): Can E[X|Y ] be negative.
  3. Problem 3.5.
  4. Problem 3.7. Note: Regarding the uniqueness in part (b), the answer is non-unique. You may want to think of an example to show that the answer is non-unique, but you don’t have to do it as part of the homework.
  5. Using the joint characteristic function show that if X 1 , X 2 , X 3 , X 4 are jointly Gaussian (correlated) random variables with zero mean, then E[X 1 X 2 X 3 X 4 ] = E[X 1 X 2 ]E[X 3 X 4 ] + E[X 1 X 3 ]E[X 2 X 4 ] + E[X 1 X 4 ]E[X 2 X 3 ]
  6. Problem 3.9.
  7. Problem 3.15.
  8. (Problem 3.17 rewritten with better notation.) Assume that X and Z are independent Gaussian vectors with means μX , μZ and covariance matrices KX and KZ. Let Y = X + Z. Then clearly X and Y are jointly Gaussian. (a) Show that E[X|Y ] = μX + KX (KX + KZ )−^1 (Y − (μX + μZ )) = KZ (KX + KZ )−^1 μX + KX (KX + KZ )−^1 (Y − μZ ) (b) Show that the conditional covariance matrix of X, given Y is given by Cov(e) = KX − KX (KX + KZ )−^1 KX = KX (KX + KZ )−^1 KZ = (K− X^1 + K Z− 1 )−^1 where e = X − E[X|Y ]

©cV. V. Veeravalli, 2008 1