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ECE 302 Division 2.
Homework 5 Solutions.
The noarbitrage price and expected payoff of a European call option under the
binomial model.
This problem builds on the option pricing model of Problem 6 from Homework 1 to explore the
concepts of expectation and risk.
The price of one share of XYZ stock at time
t
= 0 is
S
0
= $100. The price may change at integer times
t
= 1
,
2
, . . .
. The prices
S
1
, S
2
, . . .
are random variables. Given that the price at time
t
is
S
t
=
s
t
, the
conditional PMF of
S
t
+1
is:
p
S
t
+1

S
t
(
s
t
+1

s
t
) =
p,
if
s
t
+1
= 1
.
1
s
t
1

p,
if
s
t
+1
= 0
.
9
s
t
0
,
otherwise,
for
t
= 0
,
1
,
2
, . . .
. Here, 0
< p <
1. In other words, at each time, the stock price can be either 10%
larger than the price at the preceding time (with probability
p
), or 10% smaller than the price at the
preceding time (with probability 1

p
). Assume that stock price movements during two consecutive
time periods are independent. In other words, assume that events
{
S
t
+1
> S
t
}
and
{
S
t
> S
t

1
}
are
independent.
A
European call option
on the XYZ stock with strike price $102 expires at time
t
= 2. Recall from
the ±rst lecture that a call option on XYZ with strike $102 gives its holder the right to buy one share
of XYZ from the issuer of the option for $102. The term
European
means that the option can only
be exercised at expiration. (An option that can be exercised at any time during its existence is called
American
.)
We let the prices of the option at times 0, 1, and 2 be denoted by
V
0
,
V
1
, and
V
2
, respectively.
V
1
and
V
2
are random variables. The experimental outcome of
V
2
is determined by the stock price
S
2
. For
example, if
S
2
= 99, the option is worthless at time
t
= 0: it would not make sense for the holder of
the option to exercise it and buy a share of XYZ for $102 if he can buy it for $99 in the market. In
this case,
V
2
= 0.
The riskfree interest rate is
r
= 0
.
05. This means that any amount of cash
C
t
held at time
t
will turn
into
C
t
+1
= 1
.
05
C
t
at time
t
+ 1.
(a) List all values that
S
2
can take with nonzero probabilities. There are altogether three such
values. For each of these values, specify the corresponding option payo²
V
2
.
Solution.
Let’s denote by
s
+
1
the stock price at time
t
= 1 if it is larger than the price at time
t
= 0. Similarly, let us denote by
s

1
the stock price at time
t
= 1 if it is smaller than
t
0
:
s
+
1
=
uS
0
= 1
.
1
·
$100 = $110
s

1
=
dS
0
= 0
.
9
·
$100 = $90
We denote the two corresponding option prices at time
t
= 1 by
v
+
1
and
v

1
. These will be
calculated in Parts (b) and (c). For
t
= 2, there are four possibilities for the stock price which
1
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View Full Documentwe will denote by
s
++
2
,
s
+

2
,
s

+
2
, and
s

2
:
s
++
2
=
u
2
S
0
=
us
+
1
= 1
.
1
·
s
+
1
= 1
.
1
·
$110 = $121
s
+

2
=
duS
0
=
ds
+
1
= 0
.
9
·
s
+
1
= 0
.
9
·
$110 = $99
s

+
2
=
udS
0
=
us

1
= 1
.
1
·
s

1
= 1
.
1
·
$90 = $99
s

2
=
d
2
S
0
=
ds

1
= 0
.
9
·
s

1
= 0
.
9
·
$90 = $81
The corresponding option prices are denoted
v
++
2
,
v
+

2
,
v

+
2
, and
v

2
.
Since
s
+

2
=
s

+
2
, there are only three distinct possibilities for the stock price at
t
= 2: $121,
$99, and $81. When the stock price at
t
= 2 is $99 or $81, the option is worthless. Therefore,
v
+

2
=
v

+
2
=
v

2
= 0
.
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