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01 — Article
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Quiz
Quiz: Monte Carlo: Antithetic Variates, Control Variates, Quasi-MC
Module 1 of 4 · Medium
Quick Quiz
1.
A Monte Carlo estimator uses
N
=
10,000
N = 10{,}000
N
=
10
,
000
paths and achieves a standard error of $0.05$. How many paths are needed to reduce the standard error to $0.01$?
$500{,}000$
$50{,}000$
$250{,}000$
$1{,}000{,}000$
2.
Antithetic variates pair samples
Z
i
Z_i
Z
i
and
−
Z
i
-Z_i
−
Z
i
with
Z
i
∼
N
(
0
,
1
)
Z_i\sim\mathcal{N}(0,1)
Z
i
∼
N
(
0
,
1
)
. Why are the paired call payoffs negatively correlated under GBM?
Because
(
S
T
−
K
)
+
(S_T-K)^+
(
S
T
−
K
)
+
is monotone increasing in
Z
Z
Z
, so
Z
Z
Z
and
−
Z
-Z
−
Z
give opposite payoff extremes
Because GBM sample paths are symmetric around their mean level by construction
Because
Z
i
+
(
−
Z
i
)
=
0
Z_i+(-Z_i)=0
Z
i
+
(
−
Z
i
)
=
0
, so the paired inputs cancel and remove all estimator variance
Because
−
Z
i
-Z_i
−
Z
i
follows a different probability distribution from
Z
i
Z_i
Z
i
under the model
3.
With the optimal control coefficient
c
∗
=
C
o
v
(
f
(
X
)
,
Y
)
/
V
a
r
(
Y
)
c^*=\mathrm{Cov}(f(X),Y)/\mathrm{Var}(Y)
c
∗
=
Cov
(
f
(
X
)
,
Y
)
/
Var
(
Y
)
, the variance of the control-adjusted estimator equals:
σ
f
2
(
1
−
ρ
f
,
Y
2
)
\sigma_f^2\,(1-\rho_{f,Y}^2)
σ
f
2
(
1
−
ρ
f
,
Y
2
)
σ
f
2
ρ
f
,
Y
2
\sigma_f^2\,\rho_{f,Y}^2
σ
f
2
ρ
f
,
Y
2
σ
f
2
(
1
−
ρ
f
,
Y
)
\sigma_f^2\,(1-\rho_{f,Y})
σ
f
2
(
1
−
ρ
f
,
Y
)
σ
f
2
/
ρ
f
,
Y
2
\sigma_f^2\,/\,\rho_{f,Y}^2
σ
f
2
/
ρ
f
,
Y
2
4.
The Koksma-Hlawka inequality implies that quasi-Monte Carlo with a Sobol sequence always outperforms standard Monte Carlo for option pricing, regardless of the payoff's smoothness.
True
False
5.
In the Brownian-bridge construction for QMC simulation of a GBM path, Sobol dimension 1 is assigned to
W
T
W_T
W
T
, dimension 2 to
W
T
/
2
W_{T/2}
W
T
/2
, and so on. Why is this preferred to the sequential mapping (dimension
j
j
j
→ time step
j
j
j
)?
Because it makes the simulated path endpoints exact, removing time-discretisation error
Because the best Sobol dimensions map onto the highest-variance components (low effective dimension)
Because the Brownian-bridge path is computationally faster to simulate than the sequential one
Because it removes the need for a Cholesky decomposition of the path covariance matrix
6.
A control variate
Y
=
e
−
r
T
S
T
Y=e^{-rT}S_T
Y
=
e
−
r
T
S
T
with known mean
μ
Y
=
S
0
\mu_Y=S_0
μ
Y
=
S
0
has correlation
ρ
f
,
Y
≈
0.9
\rho_{f,Y}\approx0.9
ρ
f
,
Y
≈
0.9
with an ATM call payoff. What variance-reduction does this achieve?
Residual factor
1
/
0.9
≈
1.11
1/0.9\approx1.11
1/0.9
≈
1.11
(variance increased by 11%)
Residual factor
1
−
0.9
2
=
0.19
1-0.9^2=0.19
1
−
0.
9
2
=
0.19
(an 81% reduction)
Residual factor
0.9
2
=
0.81
0.9^2=0.81
0.
9
2
=
0.81
(no useful reduction)
Residual factor $1-0.9=0.1$ (a 90% reduction)
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