Solutions to HW Problems Chapter 6
6.1 The rearrangement leads to Y2 X 1 , Y1 X 2 , Y4 X 3 , etc. so the sequence
X 1 , X 2 , X 3 ,., X n becomes the sequence Y2 , Y1 , Y4 , Y3 ,., Yn , Yn 1 . Since there is no size
ordering or any other functional relati
1
Chapter 10 solutions
1. (a) It was shown in the text that the m.s. limit is a linear operator, namely if [] and
[] in the m.s.-sense, then
lim ([] + []) = + (m.s.)
(1)
We are here interested in extending this relationship to derivatives, by showing:
(1
Solutions to Chapter 11
11.1. For minimum variance, we want to minimize diagonal terms of
2
(Y Y )(Y Y )T ,
where Y = AX.
2 = (AX Y )(AX Y )T = AK1 AT + K2 AK12 K21 AT .
Now write A = A0 + ; is = 0 for minimum variance?
2 = (A0 + )K1 (A0 + )T + K2 (A0 + )
ECE514 Random Processes Fall 2010Karhunen-Loeve Transform
November 30
Here is the analysis that seemed confusing in class. Recall that n (t) is a basis function (not a basis vector), it is an eigen-function from the interval [T /2, +T /2] to R. Similarly,
ECE514 Random Process Fall 2010
HW6 - Due: December 14, 2010
Contact [email protected]
Recommended reading: Please read Sections 8.18.3 and 9.1. The following problems t well into the material we covered in class. Solutions appear in the course notes, an
ECE514 Random Process Fall 2010
HW6 - Due: December 14, 2010
Contact [email protected]
Submission in groups of up to 4 students is strongly encouraged. Electronic submission to [email protected] In this homework, we evaluate a nancial investment strateg
ECE514 Random Process Fall 2010
HW4 - Due: December 2, 2010
Contact [email protected]
Recommended reading: Please read Sections 7.17.4 with emphasis on examples and proofs we went through in class. Section 7.5 can be skimmed, and read all of Section 7.6. 1. P
ECE514 Random Process Fall 2010
HW4 - Due: November 16, 2010
Contact [email protected]
Recommended reading: Please read most of Sections 4.1-4.8 and Section 4.9 pages 113-114. The following can be skipped: The proof in Sec. 4.5. Examples 4.8.1 and 4.8.4 in Se
ECE514 Random Process Fall 2010
HW4 - Due: November 16, 2010
Contact [email protected]
NOTE: an illustrative Matlab question has been added, and the submission date has been extended from November 11 to November 16.
Recommended reading: Please read most of Se
ECE514 Random Process Fall 2010
HW4 - Due: November 11, 2010
Contact [email protected]
NOTE: a Matlab question is expected to be added to this assignment; in that case the submission date will be pushed back. Please make sure to check the course webpage perio
1
Chapter 9 solutions
1. (a) [ ()] = [] = for + 1 for all integers . So the mean of the output
analog process () = a constant.
(b) Assume times 1 and 2 satisfy
1 + 1
then
2 + 1
[ (1 ) (2 )] = [ ]
= [b1 c b2 c]
where bc is the least integer function, i.e
1
Solutions to Chapter 2
1. Calculating with the binomial probability law, for the given random variable , we obtain
2
X 10
=0
5
X 10
[ = 2] =
=3
8
X 10
[ = 3] =
=6
10
X 10
[ = 4] =
[ = 1] =
=9
(3) (07)10 0383
(03) (07)10 057
(03) (07)10 00472 and
(03
1
Errata for Stark and Woods, 4th ed., 1st printing
1. p. 70, Problem 1.36: In lines 2-3, should be . Also, should be , and 0 should be 0 .
2. p. 72, Problem 1.48: Add missing lines after independent trials. Two new lines:
(a) What is the probability that
1
Solutions to Chapter 4
1. The sample mean for the set of numbers is given by
12
1 X
=
107
12
=1
The standard deviation is given by
v
u
12
u1 X
( )2 398
= t
12
=1
2. We are given that is Bernoulli distributed over cfw_0 1 with parameter (1 0) with
the
1
Chapter 5 solutions
1. The joint density function of variables = 1 is given by
X (x) = x
(x) =
=1
(1 ) ( )
Clearly, if is a non-negative constant, the pdf also would be non-negative. In order that
X (x) be a pdf, it should also integrate to 1. Theref
1
Chapter 8 solutions
1. We recall that the set cfw_ = cfw_1 2 , and wish to prove the chain rule
=1
[1 2 ] = [1 ] [2 |1 ] [ |1 2 1 ]
(1)
We can use mathematical induction in this case. By denition of conditional probability, we
know
[1 2 ] = [1 ] [2 |
ECE514 Random Process Fall 2010
HW3 - Due: October 28, 2010
Contact [email protected]
Recommended reading: Section 3.1. Section 3.2 (proofs of Propositions 3.2.2, 3.2.3, and 3.2.4 can be skipped). Section 3.3 (examples 3.3.1 and 3.3.2 could be helpful, wherea
ECE514 Random Process Fall 2010
HW3 - Due: October 28, 2010
Contact [email protected]
Recommended reading: Section 3.1. Section 3.2 (proofs of Propositions 3.2.2, 3.2.3, and 3.2.4 can be skipped). Section 3.3 (examples 3.3.1 and 3.3.2 could be helpful, wherea
Example: Suppose, we have particles emitted in t seconds according to a Poisson law with rate t. Assume also that all particles have energies cfw_ X k , that are distributed according to the Maxwell distribution with mean 3T 2 and variance 3T 2 2. Let S
IID Random Variables Lecture 11 Sequences of IID Random Variables
A great variety of applications require open-ended measurements, which are random in nature Such measurements may not be handled as Ndimensional vectors, but rather as a sequence (i.e., si
Ex:
2 ( x 2 + y 2 + x 2 y 2 + 1) Are x and y independent? First factor out f xy ( x, y ) as
f xy ( x, y ) =
1
Lecture 9
f xy ( x, y ) =
1 [ x (1 + y 2 ) + ( y 2 + 1)]
2 2
=
1
(1 + y 2 ) (1 + x 2 )
1
f x ( x) =
= =
(1 + y
1
1
2
(1 + x 2 )
1 dy ) (1 +
Lecture 8 Multi-dimensional distributions
1.
One may check that the following function are valid jcdfs. That is they satisfy all the properties that a jcdf must posess. Example: (1 e x )( 1 e 2 y ), x 0 , y 0 FXY ( x, y ) = 0 otherwise
y3 ,0 x 1,3 y 5 2
2
A Vector Random Variable
Experiments with more than one measurement
Example: The ftp experiment 2192 bytes received in 0.11 seconds (20 kbytes/s) 26573 bytes received in 0.46 seconds (56 kbytes/s) Defining X=( size, time, speed), we can write X1= (2192,