ECE 302: Probabilistic Methods in Electrical and Computer Engineering
Fall 2014 / Section 1
Instructor: Prof. Stanley H. Chan
Mid Term 1
Solution
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Problem 1. (20 points)
(a) (10 points) Events A1 and A2 are disjoint. It is given that P(A2 ) = 2P(A1 ) and
P(A1 ∪ A2 ) = 1 − 2P(A2 ). Find P(A1 ).
Solution: Let p = P(A1 ). Note that
P(A1 ) + P(A2 ) + P((A1 ∪ A2 )c ) = P(Ω) = 1.
Thus, we have p + 2p + 4p = 1, and so p = 1/7.
(b) (10 points) Events A3 and A4 are independent, and P(A3 ) = P(A4 ) = 3/4. Find
P(A3 ∪ A4 ).
Solution: Independence means P(A3 ∩ A4 ) = P(A3 )P(A4 ). Therefore,
P(A3 ∪ A4 ) = P(A3 ) + P(A4 ) − P(A3 ∩ A4 )
= 3/4 + 3/4 − 9/16 = 15/16.
Problem 2. (40 points)
Let X be a random variable with PMF P(X = k) = pX (k) = c/3k for k = 1, 2, . . ..
(a) (5 points) Determine the value of c.
Solution: Use the fact that
P∞
k=1
pX (k) = 1. This implies
∞
∞
X
c
cX 1
=
3k
3
3k
k=1
k=0
c
1
c
=
= .
3 1 − 1/3
2
So c = 2.
(b) (15 points) Find E[X].
Solution: First, you need to the following results:
∞
X
k=0
1
r =
1−r
k
and
∞
X
krk−1 =
k=1
1
(1 − r)2
and show that
∞
X
krk =
k=1
r
.
(1 − r)2
Then,
E[X] = c
k
∞
1
X
1
3
3
= .
k
=2
3
2
(1 − 13 )2
k=1
(c) (20 points) Suppose Y = |X − 2|. Determine the PMF of Y and E[Y ].
Solution: The PMF of Y is
c
k=0
32 ,
c
c
pY (k) = 31 + 33 ,
k=1
c
,
k
= 2, 3, . . .
3k+2
X
!
∞
1
1
1
1
E[Y ] = c (0)
+ (1)
+ 3 +
k
32
31
3
3k+2
k=2
!
k
∞
1
1
1X
1
=2 0+
+
+
k
3 27
9
3
k=2
!!
k
∞
1
1
1 X
1
1
=2 0+
+
+
k
−
3 27
9
3
3
k=1
1
1
1 3 1
5
=2 0+
+
+
−
= .
3 27
9 4 3
6
2
Problem 3. (40 points)
Consider the following communication channel. A source transmits a string of binary
symbols through a noisy communication channel. Each symbol is 0 or 1 with probability p
and 1 − p, respectively, and is received incorrectly with probability ε0 and ε1 , respectively.
Errors in different symbols transmissions are independent.
1 − ε0
0
0
ε0
ε1
1
1 − ε1
1
Denote S as the source and R as the receiver.
(a) (8 points) What is the probability that a symbol is correctly received?
Solution: Correctly receive means that {R = 1 ∩ S = 1} or {R = 0 ∩ S = 0}. So, by
conditional probability, we have
P(correct) = P(R = 1 | S = 1)P(S = 1) + P(R = 0 | S = 0)P(S = 0)
= (1 − ε1 )(1 − p) + (1 − ε0 )p.
(b) (8 points) Given that a string 1011 is sent, what is the probability that the string is
correctly received?
Solution: The result follows from independence.
P(R = 1011 | S = 1011) = P(R = 1 | S = 1)3 P(R = 0 | S = 0)
= (1 − ε1 )3 (1 − ε0 ).
(c) (8 points) In an effort to improve reliability, each symbol is transmitted three times
and the received string is decoded by majority rule. In other words, a 0 (or 1) is
transmitted as 000 (or 111, respectively), and it is decoded at the receiver as a 0 (or
1) if and only if the received three-symbol string contains at least two 0s (or 1s,
respectively). What is the probability that the symbol is correctly decoded, given
that we send a 0?
Solution: All possible choices of received signal that will be decoded as “0” are 000,
001, 010, 100. Therefore,
P(R = {000, 001, 010, 100} | S = 0) = P(R = 000 | S = 0) + P(R = 001 | S = 0)
+ P(R = 010 | S = 0) + P(R = 100 | S = 0)
= (1 − ε0 )3 + 3(1 − ε0 )2 ε0 .
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(d) (8 points) Suppose that the scheme of part (c) is used. What is the probability that
a 0 was sent given that the string 101 was received?
Solution: This is the Bayes’ theorem and the Law of total probability.
P(R = 101 | S = 0)P(S = 0)
P(R = 101 | S = 0)P(S = 0) + P(R = 101 | S = 1)P(S = 1)
(1 − ε0 )ε20 p
=
.
(1 − ε0 )ε20 p + (1 − ε1 )2 ε1 (1 − p)
P(S = 0 | R = 101) =
(e) (8 points) Suppose the scheme of part (c) is used, and given that a 0 was sent. For
what value of ε0 is there an improvement in the probability of correct decoding?
Solution: We want P(R = {000, 001, 010, 100} | S = 0) > P(R = 0 | S = 0). This
implies that
(1 − ε0 )3 + 3(1 − ε0 )2 ε0 > 1 − ε0
With some calculations we can show that the inequality implies that 0 < ε0 < 12 .
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