Estimation, Evaluation, and Selection of Actuarial Models

117
Exercise 65 For the Nelson-Åalen estimate,
1.5641 = Ĥ(35) = 2
15 + 3
13 + 2
10 + d 8 + 2
8 − d
(1.5641 − 0.5641)8(8 − d) = d(8 − d)+16
0 = d 2 − 16d +48
d = 4.
The variance is
2
15 2 + 3
13 2 + 2
10 2 + 4 8 2 + 2 4 2 =0.23414.
Exercise 66 The standard deviation is
µ 15
100 2 + 20
65 2 + 13 1/2
40 2 =0.11983.
Exercise 67 In general, for the exponential distribution,
l 0 (θ) = −nθ −1 + n¯xθ −2
l 00 (θ) = nθ −2 − 2n¯xθ −3
E[l 00 (θ)] = nθ −2 − 2nθθ −3 = −nθ −2
Var(ˆθ) = θ 2 /n
where the third line follows from E( ¯X) =E(X) =θ. The estimated variance for Data Set B is
dVar(ˆθ) =1, 424.4 2 /20 = 101, 445.77 and the 95% confidence interval is 1, 424.4±1.96(101, 445.77) 1/2
or 1, 424.4 ± 624.27. Note that in this particular case, Theorem 3.29 gives the exact value of the
variance. That is because
For the gamma distribution,
Var(ˆθ) =Var( ¯X) =Var(X)/n = θ 2 /n.
l(α, θ) = (α − 1)
∂l(α, θ)
∂α
∂l(α, θ)
∂θ
=
=
nX
j=1
nX
ln x j −
j=1
ln x j − nΓ0 (α)
Γ(α)
nX
x j θ −2 − nαθ −1
j=1
nX
x j θ −1 − n ln Γ(α) − nα ln θ
j=1
− n ln θ
∂ 2 l(α, θ)
∂α 2 = −n Γ(α)Γ00 (α) − Γ 0 (α) 2
Γ(α) 2
∂ 2 l(α, θ)
= −nθ −1
∂α∂θ
∂ 2 l(α, θ)
nX
∂θ 2 = −2 x j θ −3 + nαθ −2 = −2n¯xθ −3 + nαθ −2 .
j=1