Estimation, Evaluation, and Selection of Actuarial Models

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With additional information
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62Y
Y62
L(θ) = ⎣ f(x j ) ⎦ [1 − F (1000)] 38 = ⎝ θ −1 e −x j/θ⎠ e −38,000/θ
j=1
j=1
= θ −62 e −28,140/θ e −38,000/θ = θ −62 e −66,140/θ
l(θ) = −62 ln θ − 66, 140/θ
l 0 (θ) = −62/θ +66, 140/θ 2 =0
0 = −62θ +66, 140
ˆθ = 66, 140/62 = 1066.77.
Exercise 47 The density function is
f(x) =0.5x −0.5 θ −0.5 e −(x/θ)0.5 .
The likelihood function and subsequent calculations are
Ã
Y10
L(θ) =
θ −0.5 e −x0.5 j
θ −0.5 ∝ θ −5 exp −θ −0.5 P 10
j=1
0.5x −0.5
j
l(θ) = −5lnθ − 488.97θ −0.5
j=1
x 0.5
j
!
= θ −5 exp(−488.97θ −0.5 )
l 0 (θ) = −5θ −1 +244.485θ −1.5 =0
0 = −5θ 0.5 +244.485
and so ˆθ = (244.485/5) 2 = 2391.
Exercise 48 Each observation has a uniquely determined conditional probability. The contribution
to the loglikelihood is given in the following table:
Observation Probability Loglikelihood
Pr(N=1)
1997-1
Pr(N=1)+Pr(N=2) = (1−p)p
(1−p)p+(1−p)p
= 1
2 1+p
−3ln(1+p)
Pr(N=2)
1997-2
Pr(N=1)+Pr(N=2) = (1−p)p 2
(1−p)p+(1−p)p
= p
2 1+p
ln p − ln(1 + p)
Pr(N=0)
1998-0
Pr(N=0)+Pr(N=1) = (1−p)
(1−p)+(1−p)p = 1+p 1 −5ln(1+p)
Pr(N=1)
1998-1
Pr(N=0)+Pr(N=1) = (1−p)p
(1−p)+(1−p)p = p
1+p
2lnp − 2ln(1+p)
Pr(N=0)
1999-0
Pr(N=0) =1 0
The total is l(p) =3lnp − 11 ln(1 + p). Setting the derivative equal to zero gives 0=l 0 (p) =
3p −1 − 11(1 + p) −1 and the solution is ˆp =3/8.
Exercise 49 When three observations are taken without replacement there are only four possible
results. They are 1,3,5; 1,3,9; 1,5,9; and 3,5,9. The four sample means are 9/3, 13/3, 15/3, and
17/3. The expected value (each has probability 1/4) is 54/12 or 4.5, which equals the population
mean. The four sample medians are 3, 3, 5, and 5. The expected value is 4 and so the median is
biased.
Exercise 50
· ¸
E( ¯X) 1
= E
n (X 1 + ···+ X n ) = 1 n [E(X 1)+···+ E(X n )]
= 1 (µ + ···+ µ) =µ.
n