SOLUTIONS TO MIDTERM TEST # 2 – ACTSC 433/833,
WINTER 2012
1. (a) The kernel density estimate for the probability that an individual aged 90 will survive
at least 3 years using the uniform kernel with bandwidth
b
= 0
.
5 is
ˆ
S
(3) =
1
8
(0 +
0 + 0
.
9 + 1 + 2
×
1 + 1 + 1) = 0
.
7375.
(b) The uniform distribution function is
F
(
x
) =
x
8
,
0
≤
x
≤
8. Thus, it is easy to
determine that
D
=
1
.
8
8
< D
8
,
0
.
1
= 0
.
43. Hence, we accept the null hypothesis
and the uniform distribution
U
(0
,
8) is appropriate for the time until death of an
individual aged 90 at a 10% signiﬁcance level.
2. (a) Note that
L
(
θ
) =
θ
3
n
Q
n
i
=1
3
x
4
i
, θ
≤
x
(1)
. Hence, the maximum likelihood estimator
ˆ
θ
n
=
X
(1)
.
(b) We have that
E
(
ˆ
θ
n
) =
R
∞
0
S
X
(1)
(
x
)
dx
=
3
n
3
n

1
θ
→
θ
as
n
→ ∞
. Hence,
ˆ
θ
n
is an
asymptotically unbiased estimator of
θ
.
(c) We have
E
(
ˆ
θ
2
n
) =
R
∞
0
2
xS
X
(1)
(
x
)
dx
=
3
n
3
n

2
θ
2
→
θ
2
. Thus,
V ar
(
ˆ
θ
n
)
→
0 as
n
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This note was uploaded on 04/04/2012 for the course ACTSC 433 taught by Professor Johnny during the Winter '09 term at Waterloo.
 Winter '09
 johnny

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