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Solutions to HW1
1. Problem 3.1 from Stock & Watson.
The central limit theorem suggests that when the sample size (
n
) is large, the distribution of
the sample average
(
Y
)
is approximately
N
(
µ
Y
,σ
2
Y
)
with
σ
2
Y
=
σ
2
Y
n
.
Given a population
µ
Y
=
100
2
Y
=43
.
0
,wehave
:
(a)
n
= 100
2
Y
=
σ
2
Y
n
=
43
100
=0
.
43
,and
Pr(
Y<
101) = Pr(
Y
−
100
√
0
.
43
<
101
−
100
√
0
.
43
)
≈
Φ
(1
.
525) = 0
.
9364
.
(b)
n
=64
2
Y
=
σ
2
Y
n
=
43
64
.
6719
Pr(101
<
103) = Pr(
101
−
100
√
0
.
6719
<
103
−
100
√
0
.
6719
)
≈
Φ
(3
.
6599)
−
Φ
(1
.
2200) = 0
.
9999
−
0
.
8888 = 0
.
1111
(c)
n
= 165
2
Y
=
σ
2
Y
n
=
43
165
.
2606
Pr(
Y>
98)=1
−
Pr(
Y
≤
98)
=1
−
Pr(
Y
−
100
√
0
.
2606
≤
98
−
100
√
0
.
2606
≈
1
−
Φ
(
−
3
.
9178) =
Φ
(3
.
9178) = 1
.
0000
(rounded to four decimal places).
2. Problem 3.2 from Stock & Watson.
Each random draw
Y
i
from the Bernoulli distribution takes a value of either zero or one with
probability
Pr(
Y
i
=1)=
p
and
Pr(
Y
i
=0)=1
−
p
. The mean and the variance of the random
variable
Y
i
are as follows.
E
(
Y
i
)=0
×
Pr(
Y
i
=0)+1
×
Pr(
Y
i
p
var(
Y
i
)=
E
[(
Y
i
−
µ
Y
)
2
]
=(
0
−
p
)
2
×
Pr(
Y
i
=0)+(1
−
p
)
2
×
Pr(
Y
i
=1)
=
p
2
(1
−
p
)+(1
−
p
)
2
p
=
p
(1
−
p
)
(a) The fraction of successes is
b
p
=
#(
successes
)
n
=
#(
Y
i
n
=
P
n
i
=1
Y
i
n
=
Y
1
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View Full Document(b)
E
(
b
p
)=
E
(
P
n
i
=1
Y
i
n
1
n
n
X
i
=1
E
(
Y
i
1
n
n
X
i
=1
p
=
p
(c)
var(
b
p
)=var(
P
n
i
=1
Y
i
n
1
n
2
n
X
i
=1
Var
(
Y
i
1
n
2
n
X
i
=1
p
(1
−
p
p
(1
−
p
)
n
.
The second equality uses the fact that
Y
1
,Y
2
, ......
, Y
n
are iid draws and
Cov
(
Y
i
j
)=0
for
i
6
=
j.
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 Spring '10
 DR.RAHULA

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