1

. All six faces of a fair die have equal probabilities of appearing

. Suppose you suspect the face with the six dots does not appear with equal probability

. You toss the die 180 times trying to detect any irregularity

.a

. If the die is fair, what is the ensemble average of the number of times the six dots will appear?

(1 pt)

b

. Of course, in practice the actual number of times the six dots will appear will not be exactly equal to that ensemble average

. Write down the analytical express for the probability that a fair die will have less than 20 and and more than 40 number of six dots appearing in 180 tosess? Do not evaluate this expression

.(2 pts)

c

. Since the above analytical expression of the probability is too computationally complicated to evaluate, let us use the Central Limit Theorem to approximate this expression

. What is the numerical value for this aproximate expression? Hint: We know for a Gaussian CDF, (0) = 0

.5; (1) = 0

.8413; (2) = 0

.9772

. (3 pts)

2

. Let {X1 , X2 , X3 } be a zero-mean random sequence with an autocorrelation function

R(Xi Xj ) = E{Xi Xj } = a|ij| , i, j = 1, 2, 3, where 0 < a < 1

.a

. Under the mean-square (ms) error criterion, we vary {c1 , c2 } such that 2 (c1 , c2 ) =

E{(X3 (c1 X1 + c2 X2 ))2 } is minimized with 2 (1 , c2 ) = min{2 (c1 , c2 )} =

min c

E{(X3 (1 X1 + c2 X2 ))2 }

. Find {1 , c2 }

. (2 pts)

. Find 2 (1 , c2 )

. (2 pts)

c

c

min c

b

. Under the mean-square (ms) error criterion, we vary {c1 } such that 2 (c1 ) = E{(X2 c1 X1 )2 } is minimized with 2 (1 ) = min{2 (c1 )} = E{(X2 (1 X1 )2 }

.c

min c

Find {1 }

. (2 pts)

. Find 2 (1 )

. (2 pts)

c

c

min

3

. Let X and Y be two independent positive-valued r

.v

.s with their pdfs given by fX (x) = 0, x < 0 and fY (y) = 0, y < 0

. Dene the r

.v

. Z = X Y, where is the regular multiplication (e

.g

., 3 2 = 6

.) and is not the convolution operation

. Find

the pdf of fZ (z), 0 < z <

. Hint: Start with FZ (z)

.(6 pts)

4

. Let the random vector X = [X1 , X2 ]T have a mean vector X = [1, 2]T and a covariance

matrix

[

]

1 1

RX =

.1 2

Let Y = AX, where

[

A=

1 2

1 3

]

.Find the mean vector Y and the covariance matrix RY of Y

.