HOMEWORK 1
SOLUTION
Northwestern University
Marciano Siniscalchi
Fall 2009
Econ 3310
1.
The Peter Lynch Problem
This was pretty straightforward. For the ﬁrst part, we ﬁrst compute the probability of
k
successes
in
n
= 13 trials, for
k
= 11
,
12
,
13, where the probability of a success is
p
=
1
2
; then we sum these
three numbers to get the probability of at least 11 successes in 13 trials. This is approximately
1
.
123% (i.e. 0
.
01123), a small number.
For the second part, the trick is to note that the probability that at least one fund manager
outperforms Peter Lynch equals 1 minus the probability that no one does. The latter probability
is easy to calculate: the probability that one particular manager does not beat Peter Lynch is
1

0
.
01123, so the number we need is (1

0
.
01123)
500
≈
0
.
003528 (i.e. 0
.
3538%). Therefore, the
probability that at least one manager outperforms Peter Lynch is approximately 1

0
.
003528 =
0
.
9965 = 99
.
65%.
Bottom line: maybe Peter Lynch was a really clever guy, but then again, maybe he was just
lucky—even if all fund managers chose stocks at random over theperiod in question,
someone
just
“had” to outperform Peter Lynch!
2.
Coin Flips!
A) State space
S
=
{
HHHHH,HHHHT,.
..,TTTTT
}
will do. Notice that there are 2
5
= 32
states. Let probability
P
be deﬁned by
P
(
{
HHHHH
}
) =
P
(
{
HHHHT
}
) =
...
=
P
(
{
TTTTT
}
) =
1
32
,
and we are done.
B)
X
=
5 for
s
=
HHHHH
3 for
s
∈ {
HHHHT,HHHTH,HHTHH,HTHHH,THHHH
}
.
.
.

5 for
s
=
TTTTT
C) Probability distribution is a tuple
(

5
,
1
32
;

3
,
5
32
;

1
,
10
32
; 1
,
10
32
; 3
,
5
32
; 5
,
1
32
)
where the ﬁrst
element is the outcome and the second probability. How do you get probabilities? Well you can
write out the whole table and count the number of states for which
X
=
x
holds. For this problem