University of Illinois Fall 2006
ECE 413: Solutions to Problem Set 7
1. (a) Yes, Fred and Wilma always alternate in tossing. Note that a game is won by the winner tossing
a Head. The next toss starts a new game, and the loser goes first in the next game. Hence, Fred
and Wilma always alternate in tossing. Alternation holds even for the other game considered on
the exam in which the winner matches the previous toss. The winning toss of a game is considered
to be the first toss of the next game. Since the loser of a game gets to toss the coin immediately
after a winning toss (this toss by the loser of the previous game is the second toss of the next
game), Fred and Wilma still alternate in tossing.
(b) P(Fn+1) = 1
3P(Fn) + 2
3P(Wn) = 1
3P(Fn) + 2
3(1 −P(Fn)) = 2
3−1
3P(Fn).
P(Wn+1) = 2
3P(Fn) + 1
3P(Wn) = 2
3(1 −P(Wn)) + 1
3P(Wn) = 2
3−1
3P(Wn).
(c) Suppose that P(Fn) = a+bαnfor n≥1. Then we have that
a+bαn+1 =2
3−1
3a+bαn⇒bαnα+1
3=2−4a
3for all n≥1.
In order for this to hold for all n > 1, it must be that α=−1
3and a=1
2. Hence,
P(Fn) = 1
2+b−1
3n
and since P(F1) = 2
3, we get that b=−1
2. The calculation for P(Wn) is almost identical, except
that the initial condition P(W1) = 1
3gives a different value + 1
2for b. Hence, we get that
P(Fn) = 1
21−−1
3nand P(Wn) = 1
21 + −1
3n.
Note that P(Fn)> P(Wn) for odd n, and P(Fn)< P (Wn) for even n.
(d) Since 1
3n→0 as n→ ∞, lim
n→∞
P(Fn) = lim
n→∞
P(Wn) = 1
2and the game is asymptotically fair.
2. (a) p0(86) = 90
86(0.9)86(0.1)4,p1(86) = 90
71(0.9)71(0.1)19. The likelihood ratio is p1(86)/p0(86) ≈
0.027 and hence the ML decision is that the connecting flight is late.
(b) Repeat part (a) for the case when the gate agent observes that X= 96. Obviously p0(96) = 0,
while p1(96) = 90
81(0.9)81(0.1)9. The likelihood ratio is p1(86)/p0(86) = ∞and hence the ML
decision is that the connecting flight is on time. In fact, it should be obvious that for X>90, the
ML decision is that the connecting flight is on time.
(c) The MAP decision rule is the same as the ML decision rule.
3. The likelihood matrix and joint probability matrices are as shown below, and the ML and MAP decision
rules are indicated by boldface entries.
Hypothesis Y= 1 Y= 2 Y= 3 P(Hi)Y= 1 Y= 2 Y= 3
H1:X= 1 0.80.1 0.1 0.50.40.05 0.05
H2:X= 2 0.05 0.90.05 0.25 0.0125 0.225 0.0125
H2:X= 3 0.15 0.05 0.80.25 0.0375 0.0125 0.2
Thus, the ML and MAP rules are the same in this case. Other a priori probabilities could make the
MAP rule different from the ML rule . ..
4. (a) The likelihood of a hit is P(H|F) = 2
5or P(H|C) = 1
4, or P(H|S) = 1
6depending on which
hypothesis is true. Since P(H|F) = 2
5is the largest, the maximum-likelihood decision is that it
was a fastball.
