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:
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
 Winter '09
 johnny
 Maximum likelihood, Estimation theory, Likelihood function, 7 years, maximum likelihood estimate, 6 years

Click to edit the document details