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Chapter 10
Binomial Option Pricing
±
Question 10.1
1.
Since
25
u
C
=
and
0
d
C
=
we have
25
50
0.50.
Δ=
=
To solve the bond amount, one could use
Equation (10.2); however, once we know the option’s
,
Δ
finding the replicating bond position is
a simple algebra exercise (i.e., no memorization!). We will replicate the upvalue of the call:
04
04
5 130
25
40
38 4316
eB
B
e
.−
.
.×
+
=
⇒
=−
.
The value of the option is the value of the replicating portfolio:
0
5 100
38 4316
11 5684
C
=. ×
−
.
=
.
2.
Since
0
u
P
=
and
25
d
P
we have
25
50
0 50.
−
=− .
We will replicate the upvalue of the put:
04
04
5 130
0
65
62 4513
B
e
.
−.
×
+
=
⇒
==
.
The value of the option is the value of the replicating portfolio:
0
5 100
62 4513 12 4513
P
=−. ×
+
.
=
.
±
Question 10.2
1.
We will use riskneutral pricing for this problem. Our riskneutral probability of an up movement is
()
0
4
*
8
0 4816
13 8
rh
ed
e
p
ud
δ
−.
−−
.
=
.
−
.
The noarbitrage value of the European call is:
**
0
4
0
[
(1
)]
35 0 4816
16 1958
rh
CeC
pC
p
e
.
=+
−
=
×
×
.
=
.
For the replicating strategy, the option’s delta is
35 50
0 70.
/
= .
We can use the option value
result to solve for the bond amount (once again, no memorization required). The option value is the
total cost of replication. The cost of 0.7 shares is $70. Since the option value is $16.1958, we must
borrow $70
$16.1958
$53.8042,
−=
i.e.,
$53.8042.
B
2.
If we observe a price of $17, then the option price is too high relative to its theoretical value. We sell
the option for $17 and synthetically create a call option for $16.196. In order to do so, we buy 0.7 units
of the share and borrow $53.804. These transactions yield no risk and a profit of $0.804.
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View Full Document Chapter 10
Binomial Option Pricing
113
3.
If we observe a price of $15.50, then the option price is too low relative to its theoretical value.
We buy the option and synthetically create a short position in the option (we’ll receive $16.1958).
In order to do so, we short 0.7 shares and lend $53.8042. These transactions yield no risk and a
profit of $0.696.
±
Question 10.3
1.
We will use riskneutral pricing for this problem. Our riskneutral probability of an up movement is
()
0
4
*
8
0 4816
13 8
rh
ed
e
p
ud
δ
−.
−−
.
==
=
.
−
.
The noarbitrage value of the European put is:
**
0
4
0
[
(1
)]
15 0 5184
7 4708
rh
PeP
pP p e
.
=
+−=×
×
.
=
.
For the replicating strategy, the option’s delta is
15 50
0 30.
Δ=−
/
=− .
We can use the option value
result to solve for the bond amount (once again, no memorization required). The option value is the
total cost of replication. By shorting 0.3 shares, we receive $30. Since the option value is $7.4708,
we must lend $37.4708, i.e.,
B
=
$37.4708.
2.
If we observe a price of $8, then the option price is too high relative to its theoretical value. We sell
the option for $8 and synthetically create a long put option for $7.471. In order to do so, we short
0.3 units of the share and lend $37.4708. These transactions yield no risk and a profit of $0.5292.
3.
If we observe a price of $6, then the put option price is too low relative to its theoretical value. We
buy the option and synthetically create a short position in the option (we’ll receive $7.4708). In order
to do so, we buy 0.3 units of the share and borrow $37.4708. These transactions yield no risk and a
profit of $1.4708.
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This note was uploaded on 01/16/2011 for the course FIN 512 taught by Professor Staff during the Fall '08 term at University of Illinois, Urbana Champaign.
 Fall '08
 Staff
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