Variance Reduction by Antithetic Variates. A simple and widely used technique for increasing the efficiency and accuracy of Monte Carlo simulations in certain situations with little additional increase in computational complexity is the method of antithetic variates. For each
k
=
1
,
…
,
M
, use the sequence
{
ε
1
(
k
)
,
…
ε
N
−
1
(
k
)
}
in Eq. (4) to simulate a payoff
f
(
S
N
(
k
+
)
)
and also use the sequence
{
−
ε
1
(
k
)
,
…
−
ε
N
−
1
(
k
)
}
in Eq. (4) to simulate an associated payoff
f
(
S
N
(
k
−
)
)
. Thus, the payoffs are simulated in pairs
{
f
(
S
N
(
k
+
)
)
,
f
(
S
N
(
k
−
)
)
}
. A modified Monte Carlo estimate is then computed by replacing each payoff
f
(
S
N
(
k
)
)
in Eq. (6) by the average
[
f
(
S
N
(
k
+
)
)
+
f
(
S
N
(
k
−
)
)
]
2
,
C
^
A
V
(
S
)
=
1
M
∑
k
=
1
M
f
(
S
N
(
k
+
)
)
+
f
(
S
N
(
k
−
)
)
2
(
1
+
r
Δ
t
)
N
. (iii)
Use the parameters specified in Problem 3 to compute several (say,
20
or so) option price estimates using Eq. (6) and an equivalent number of option price estimates using (iii). For each of the two methods, plot a histogram of the estimates and compute the mean and standard deviation of the estimates. Comment on the accuracies of the two methods.
The difference equation (4):
S
n
+
1
(
k
)
=
S
n
(
k
)
+
r
S
n
(
k
)
Δ
t
+
σ
S
n
(
k
)
∈
n
+
1
k
Δ
t
,
S
0
k
=
s
Use the differential equation (4) to generate an ensemble of stock prices
S
N
(
k
)
=
S
(
k
)
(
N
Δ
t
)
,
k
=
1
,
…
,
M
(where
T
=
N
Δ
t
) and then use formula (6) to compare a Monte Carlo estimate of the value of a five-month call option
(
T
=
5
12
years
)
for the following parameter values:
r
=
0.06
,
σ
=
0.2
, and
K
=
$
50
. Find estimates corresponding to current stock prices of
S
(
0
)
=
s
=
$
45
,
$
50
, and
$
55
. Use
N
=
200
time steps for each trajectory and
M
≅
10
,
000
sample trajectories for each Monte Carlo estimate. Check the accuracy of your results by comparing the Monte Carlo approximation with the value computed from the exact Black-Scholes formula
C
(
s
)
=
s
2
erfc
(
−
d
1
2
)
−
K
2
e
r
T
erfc
(
−
d
2
2
)
, (ii)
where
d
1
=
1
σ
T
[
ln
(
s
k
)
+
(
r
+
σ
2
2
)
T
]
,
d
2
=
d
1
−
σ
T
And
erfc
(
x
)
is the complementary error function,
erfc
(
x
)
=
2
π
∫
x
∞
e
−
t
2
d
t
.
The difference equation (4) is given below:
S
n
+
1
(
k
)
=
S
n
(
k
)
+
r
S
n
(
k
)
Δ
t
+
σ
S
n
(
k
)
∈
n
+
1
k
Δ
t
,
S
0
k
=
s