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Solutions to Assignment 5 – ACTSC 433/833, Winter 2012
1. (a) For Line 1:
S
1
(
x
) =
S
0
(
x
) and
f
1
(
x
) =
α
100
α
x
α
+1
, x >
100. For Line 2:
S
2
(
x
) = (
S
0
(
x
))
b
and
f
2
(
x
) =
αb
100
αb
x
αb
+1
, x >
100, where
b
=
e
β
. Thus,
L
(
α,b
) =
f
1
(135)
f
1
(160)
f
1
(240)
f
1
(320)
f
1
(380)
f
2
(115)
f
2
(140)
f
2
(205)
f
2
(245)
f
2
(285)
.
Then, it follows from
∂
ln
L
(
α,b
)
∂α
= 0 and
∂
ln
L
(
α,b
)
∂b
= 0 that ˆ
α
= 1
.
20664 and
ˆ
b
=
1
.
32073. Hence,
ˆ
β
= 0
.
27818.
(b) By
S
1
(
x
) = 0
.
5, we know that the median loss in line 1 is
x
= 100
/
0
.
5
1
/α
. Hence, the
maximum likelihood estimate for the median loss in line 1 is 100
/
0
.
5
1
/
ˆ
α
= 177
.
64 .
(c) The mean loss in line 2 is
μ
2
=
R
∞
0
S
2
(
x
)
dx
= 100 +
100
αb

1
. Thus, ˆ
μ
2
= 100 +
100
ˆ
α
ˆ
b

1
=
268
.
45.
(d) The maximum likelihood estimate for the probability that a loss in line 2 will exceed
200 is
ˆ
S
2
(200) = 0
.
5
ˆ
α
ˆ
b
= 0
.
3313
.
2. (a) For Line 1:
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 Winter '09
 johnny

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