13 Solutions for Selected Exercises
245
Since the square bracket is zero, we find the result from (5.22), namely c = N (d 1 ).
Differentiating c once again with respect to S gives us c = ∂∂ c /∂ S = ∂ N (d 1 )/∂ S
for the call option
c = N
(d 1 )
∂d 1
∂ S
=
e
−d
2
1 /2
√
2πσ 2 τ S
.
(13.44)
Exercise 6.2
From (4.45), as a consequence of the call-put parity, we know that p = K e
−r f τ
−
S + c. Differentiating with respect to S, we find p = ∂ p/∂ S = −1 + ∂c/∂ S =
N (d 1 ) − 1, where we used the result from Exercise 6.1. As a second option, we
differentiate p = K e
−r f τ N (−d 2 ) − S N (−d 1 ) and obtain
p =
∂ p
∂ S
= −K e
−r f τ N
(−d 2 )
∂d 2
∂ S
− N (−d 1 ) + S N
(−d 1 )
∂d 1
∂ S
.
(13.45)
Using similar reasoning as in Exercise 6.1 and N (y) + N (−y) = 1, we find the same
result as before: p = N (d 1 ) − 1.
Exercise 6.4
The solution is a straightforward application of (6.9). The numerical solution can be
found in ex6_4.m in the ESM.
Exercise 6.5
Figure 13.6 shows the profit diagram of the butterfly spread. The script ex6_5.m is
available in the ESM.
Exercise 7.1 and 7.2
After setting up the n × 2 matrix, shown in (7.1), and using (7.4) to solve it once
with all error bars σ set to unity. The slope x 1 = a and intercept x 2 = b with their
respective error bars are a = 1.5 ± 0.15 and b = −2.8 ± 0.4, where we use (7.7) to
determine the error bars. In a second fit we take the error bars, given in the table from
the exercise into account, and find ˜
a = 1.7 ± 0.2 and ˜
b = −3.2 ± 0.6. We determine
the R
2 from the predicted measurement values ˆ
y i =
j A i j x j of the first fit and find
R
2
= 0.89.
Exercise 7.3
C(y) is given by C(y) =
2
−1 and, since is diagonal, it equals its transpose:
=
t . Straightforward evaluation then leads to
C(x) = J C(y)J
t
= (A
t
2 A)
−1 A
t
2
2
−1
2 A(A
t
2 A)
−1
= (A
t
2 A)
−1
(A
t
2 A)(A
t
2 A)
−1
= (A
t
2 A)
−1
,
(13.46)
which proves the statement.
245
Since the square bracket is zero, we find the result from (5.22), namely c = N (d 1 ).
Differentiating c once again with respect to S gives us c = ∂∂ c /∂ S = ∂ N (d 1 )/∂ S
for the call option
c = N
(d 1 )
∂d 1
∂ S
=
e
−d
2
1 /2
√
2πσ 2 τ S
.
(13.44)
Exercise 6.2
From (4.45), as a consequence of the call-put parity, we know that p = K e
−r f τ
−
S + c. Differentiating with respect to S, we find p = ∂ p/∂ S = −1 + ∂c/∂ S =
N (d 1 ) − 1, where we used the result from Exercise 6.1. As a second option, we
differentiate p = K e
−r f τ N (−d 2 ) − S N (−d 1 ) and obtain
p =
∂ p
∂ S
= −K e
−r f τ N
(−d 2 )
∂d 2
∂ S
− N (−d 1 ) + S N
(−d 1 )
∂d 1
∂ S
.
(13.45)
Using similar reasoning as in Exercise 6.1 and N (y) + N (−y) = 1, we find the same
result as before: p = N (d 1 ) − 1.
Exercise 6.4
The solution is a straightforward application of (6.9). The numerical solution can be
found in ex6_4.m in the ESM.
Exercise 6.5
Figure 13.6 shows the profit diagram of the butterfly spread. The script ex6_5.m is
available in the ESM.
Exercise 7.1 and 7.2
After setting up the n × 2 matrix, shown in (7.1), and using (7.4) to solve it once
with all error bars σ set to unity. The slope x 1 = a and intercept x 2 = b with their
respective error bars are a = 1.5 ± 0.15 and b = −2.8 ± 0.4, where we use (7.7) to
determine the error bars. In a second fit we take the error bars, given in the table from
the exercise into account, and find ˜
a = 1.7 ± 0.2 and ˜
b = −3.2 ± 0.6. We determine
the R
2 from the predicted measurement values ˆ
y i =
j A i j x j of the first fit and find
R
2
= 0.89.
Exercise 7.3
C(y) is given by C(y) =
2
−1 and, since is diagonal, it equals its transpose:
=
t . Straightforward evaluation then leads to
C(x) = J C(y)J
t
= (A
t
2 A)
−1 A
t
2
2
−1
2 A(A
t
2 A)
−1
= (A
t
2 A)
−1
(A
t
2 A)(A
t
2 A)
−1
= (A
t
2 A)
−1
,
(13.46)
which proves the statement.