Estimation, Evaluation, and Selection of Actuarial Models

119
With regard to assumption (ii) of Theorem 3.29,
Z θ
0
Z
∂ 1 θ
∂θ θ dx = −θ −2 dx = − 1
0
θ 6=0.
Exercise 69 From Exercise 67 we have ˆα =0.55616, ˆθ =2, 561.1 and covariance matrix
· ¸
0.021503 −99.0188
.
−99.0188 1, 045, 668
The function to be estimated is g(α, θ) =αθ with partial derivatives of θ and α. The approximated
variance is
£ ¤ · ¸· ¸
0.021503 −99.0188 2, 561.1
2, 561.1 0.55616 = 182, 402.
−99.0188 1, 045, 668 0.55616
The confidence interval is 1, 424.4 ± 1.96 √ 182, 402 or 1, 424.4 ± 837.1.
Exercise 70 The partial derivatives of the mean are
∂e µ+σ2 /2
∂µ
∂e µ+σ2 /2
= e µ+σ2 /2 = 123.017
= σe µ+σ2 /2 =134.458.
∂σ
The estimated variance is then
£ ¤ · ¸· ¸
0.1195 0 123.017
123.017 134.458 = 2887.73.
0 0.0597 134.458
Exercise 71 The first partial derivatives are
∂l(α, β)
∂α
∂l(α, β)
∂β
= −5α − 3β +50
= −3α − 2β +2.
The second partial derivatives are
∂ 2 l(α, β)
∂α 2 = −5
∂ 2 l(α, β)
∂β 2 = −2
∂ 2 l(α, β)
∂α∂β
= −3
and so the information matrix is · 5 3
3 2
¸
.