2
n
1
i
i
)
x
(x
n
1
−
∑
=
=
2
n
1
i
2
i
x
x
n
1
−
∑
=
2
1
)
(
x
x
n
i
i
−
∑
=
=
2
1
2
x
x
n
i
i
−
∑
=
18
.
To estimate E[X], X1,……,X16 have been simulated with the following values resulting,
10,11,10.5,11.5,14,8,13,6,15,10,11.5,10.5,12,8,16,5. Based on these data, if we want the
standard deviation of the estimator of E[X] to be less than 0.1, roughly how many
additional simulation runs will be needed?
(10 marks)
Solution
X1,….,X16 have the resulting values 10, 11, 10.5, 11.5, 14, 8, 13, 6, 15, 10, 11.5, 10.5,
12, 8, 16, 5 respectively.
Thus, n = 16
∑
=
=
∴
n
i
i
n
x
X
1
=
∑
=
16
1
16
i
i
x
=
16
172
= 10.75
We have,
S
2
=
1
)
(
2
1
−
−
∑
=
n
x
x
n
i
i
Thus,
2
16
1
)
(
x
x
i
i
−
∑
=
= 136.4375
095833
.
9
1
16
4375
.
136
2
=
−
=
∴
S
The standard deviation of the estimator of E[X} to be less than 0.1.
1
.
0
015929873
.
3
=
k
15929873
.
30
=
k
k
=
909.5653
k
≈
910 runs
19. Suppose we want to use simulation to estimate
1
0
u
x
E e
e dx
θ
⎡
⎤
=
=
⎣
⎦
∫
. Using antithetic
variables, discuss what kind of improvement is possible.
(10 marks)
Solution
θ
= E[e
U
] =
∫
1
0
x
dx
e
∴
θ
= e-1