EECS 126 Midterm 2 - Probability and Statistics, Exams of Probability and Statistics

The spring 2004 midterm exam for the university of california, berkeley's eecs 126 course on probability and statistics. The exam includes multiple-choice and problem-solving questions related to exponential random variables, cumulative distribution functions, and wireless communication systems. Students are required to write their solutions in the blue book and the exam is closed-book, no-notes, and no-calculators.

Typology: Exams

2012/2013

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UNIVERSITY OF CALIFORNIA
College of Engineering
Department of Electrical Engineering and
Computer Sciences
Professor Tse Spring 2004
EECS 126 — Midterm #2
Closed book, no notes, no calculators.
Please write ALL of your solutions in the blue book. The question sheet itself will be collected at
the end of the exam.
Please show your steps clearly. True/false answer without explanation gets no mark.
BE SURE TO WRITE YOUR NAME ON YOUR BLUE BOOK!!!!
Formula for pdf of rv:
Question Points
possible Yo ur
points
116
220
316
4 – Part I 17
4 – Part II 31
Total 100
N01
,()
f
x
() 1
2π
---------- e
x2
2
--------
=
pf3
pf4
pf5

Partial preview of the text

Download EECS 126 Midterm 2 - Probability and Statistics and more Exams Probability and Statistics in PDF only on Docsity!

UNIVERSITY OF CALIFORNIA College of Engineering Department of Electrical Engineering and Computer Sciences

Professor Tse Spring 2004

EECS 126 — Midterm

• Closed book, no notes, no calculators.

• Please write ALL of your solutions in the blue book. The question sheet itself will be collected at

the end of the exam.

• Please show your steps clearly. True/false answer without explanation gets no mark.

• BE SURE TO WRITE YOUR NAME ON YOUR BLUE BOOK!!!!

Formula for pdf of rv:

Question

Points

possible

Your

points

4 – Part I 17

4 – Part II 31

Total 100

N ( 0 1, ) f ( )x 1 2 π

----------e

  • x^2 =^ --------^2

Problem 1 (16 points)

[7 pts.] a. is an exponential rv with parameter :

[3 pts.] (i) Compute.

[4 pts.] (ii) Sketch the pdf for 2 different values of on the same plot.

[9 pts.] b. has a pdf as follows:

[5 pts.] (i) Sketch the pdf of , where , on the same as plot as pdf of.

Consider all possible cases.

[4 pts.] (ii) Find the mean and variance of.

X μ

f (^) X ( )x = ae – μ^ x x ≥ 0

a

μ 1 , μ 2 ,μ 1 >μ 2

Y

y a +b 2

a ------------ b

Z = cY +d d > 0 Y

Z

Problem 3 (16 points)

The cumulative distribution function of a rv looks as follows:

[2] a) Is this a continuous or a discrete rv? Or neither? Explain.

[4] b) Can you determine the values of and from the given information? If so, find their

values.

[5] c) Suppose you want to generate this rv on MATLAB but you only have access to rand(·),

which generates a uniform rv in [0,1]. Explain how you would simulate.

[5] d) Suppose now you don’t even have access to rand(·) but you do have access to the routine

Bern( ), which generates a Bernoulli random variable with parameter. Explain how

you would use this routine to generate.

X

(Note: Not to scale)

x^2

10

b

a

F (^) X ( )x

x

a b

X

p p X

Problem 4 (Part I – 17 points; Part II – 31 points)

Consider a wireless communication system with a single transmit antenna and two

receive antennas. The channel is described by

where is the transmitted symbol; are the received symbols in antenna and ,

respectively; and are independent noise at the two antennas independent of.

and are the channel gains.

Part I – In this part you can assume that and are determinative and known numbers.

[6 pts.] a. Suppose you transmit , equiprobable. Find the MAP detection rule of based on

alone and the error probability in terms of the signal-to-noise ratio.

[8 pts.] b. Repeat part (a), but now base the detection of based on both and.

[3 pts.] c. In a wireless environment, sometimes or can be very small due to channel fading.

Based on your answers to part (a) and (b), what do you think is the advantage of having two

receive antennas?

X

h (^) a

h (^) b

Y (^) a

Y (^) b

Y (^) a = h (^) a X +W (^) a Y (^) b = h (^) b X +W (^) b

X Y (^) a , Y (^) b a b W (^) a , W (^) b N ( 0 ,σ^2 ) X h (^) a h (^) b

h (^) a h (^) b

X = ±a X Y (^) a

X Y (^) a Y (^) b

h (^) a h (^) b