Unformatted text preview: University of California, San Diego
Department of Electrical and Computer Engineering ECE109  Spring 2005
Midterm Exam  SOLUTIONS
Wednesday, May 4, 2005 5:00pm 6:20pm
Location: HSS 2250
K. Zeger Name Your UCSD ID Number
Signature INSTRUCTIONS
This exam is open book and open notes. No calculators, laptop computers, or other electronic devices
are allowed.
Write your answers in the spaces provided. Show
all your work. If you need extra space, please use the
back of the previous page. Partial credit will be given
only for substantial progress on a problem. Zero
credit will be given for correct answers that lack adequate explanation of how they were obtained. There
is a maximum total of 30 points on this exam. Simplify your answers as much as possible and leave answers as fractions, not was decimal numbers. GRADING
1. 10 points
2. 10 points
3. 10 points
TOTAL (30 points) ...
1 Problem 1 (10 points)
Al, Bo, and Cy tell the truth according to certain probabilities. The probability that either Al or Cy
lies is and the probability both Al and Cy lies is . The probability that Cy lies but Al tells the truth
is , and the probability Al and Bo lie while Cy tells the truth is . The probability that all three of
them lie is . What is the probability that all but Al are truthful, given that Al and Cy are not both
truthful? ¤ , denote the events that Al, Bo, and Cy are lying, respectively. Then we are §¦¥ &'¤ #%$¥
§¦
¢ #§ ¥ ¨
©
¡ #$¥ © ¨
§ ©
§ ¨ #"!¥
£ ¥ ¨¨
§ ¦ ©
© is a disjoint union, we have § ¥ § ¥ § ¦ ¥ § ¦ ¥ §
)42$"3#%)3#210)(¥ ¢ Therefore, since ¤ § ¦ ©
1X#¥ ¨
§
U)!¥
§
U)!¥ &
E¢ C ¡ C @
W(UV!B£16 § $¥¨ ¨ TP!S¥RE § ¨ $B¥@ ¨ 7#P!¥ R § ¥ ¨
¦ © E § @ © ¦ © § Q ¦ ©
©
H
§
P!¥ I § $¥
¦
&E¢ C ¡ C @
GF!#!DB£A6 5#§ ¥ ¨ 9#$¥ ¨ 8 %¥ ¨ 7#"!¥ ¨ 5 § $¥ ¨
© 6 § © 6 § ¦ © 6 § ©
¦© So, the quantity we are asked to ﬁnd is (using the fact that Note that £ ¡ SOLUTION: Let ,
given: ) was extraneous information that was not needed to solve the problem. 2 & d
e
c C [email protected]
U W A6
E C H
2 6
¦ C @ %QI
C¤ [email protected]
5 ¤I ¤ ¢
QP ¥ ©¡
8 754
6 3 @ E @ E a F 6 ¨¦
c b §8 7 P £© E @ Y FR§8 [email protected] & R§8 754 U £©
¨¦
E
3
FE 6 3
C
C
Y£X FR§8 67534 WV £©
0
U ¨¦
E
C
H
7 Y R§8 [email protected] X WV `©
0U ¨¦
FE 6 3
C
G§8 [email protected]
FE 6
C
¤
TI T
¨¦
£©
H
RB§8 [email protected]
FE 6 93 or ! 3
B R8 754
FE 6 3 C ¤ IE @ ¢
QP5¥¤©¡
6 E £©@
¨¦ C
SB H
7 G§8 754
FE 6 3 C ' 0!¦
1)( iff
8 [email protected]
6 93
C
DB 6
6 A (! ¦ E
W @ 5¥¤£¡
[email protected]¢ ©
¨ & %! ¦ iff (by the
¤ SOLUTION: Note that
quadratic formula). Thus, ¨¦
£©§ 6 $ #"
¦! ©
¨ Let be a random variable whose probability density function is
Find
. for all . Problem 2 (10 points) Problem 3 (10 points)
satisfying all of the following three properties: Find a random variable © ¢ ¤ ©
¨ £X8S2 ¨
¦ 5 2 2 0
¡
©
¨
¤ ©
58D ¨
0 .
. [email protected]¢
5¥¤#¨
7¤ 2 You must precisely specify either the probability mass function
bility density function
if is continuous. if ¦@¤
¡ §@0V¥¤ for all E
W£¨ (iii) (ii) for all real numbers .
¤ (i) is discrete, or the proba [email protected]¢
5¥¤¡ SOLUTION: Property (i) tells us the random variable
is continuous, not discrete. Property (ii)
tells us the positive portion of the pdf of
is conﬁned to the range
. Property (iii) says that
the area under the pdf
from to
should be times the area under the same curve from
to . This can be accomplished by having the pdf increase linearly with (i.e. triangle shaped), as:
for
and
otherwise. This is because
(area of
triangle), so § 8(2
¤ © ¤
¨ ¤
0 § ##£¨ E3¤ § X8D2
@ ¤
5¥¤©¡
[email protected]¢
0
1 & 8'2 ¨
¤ © ¢ ¤ § ©
¨
¡ ©0 ©
¦
[email protected]¢
5¤ #¡ § W£Q@ EFX6 @ E ¨ @ ¤ I
E¨
0
¡ RW`©¦§ T¤ B § $¦
4
£¨ ¤ ¢ and obtain 0 ¤
¤ ¤ . ¤ § 5¥¤©¡
[email protected]¢ ¨
£#0§ so ¡ ©0 ©
¦ We solve for the constant ...
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Full Document
 Spring '08
 KennethZeger
 Probability theory, probability density function, Cumulative distribution function, Probability mass function, Cy

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