Estimation, Evaluation, and Selection of Actuarial Models
Estimation, Evaluation, and Selection of Actuarial Models
Estimation, Evaluation, and Selection of Actuarial Models
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103<br />
t event r s S 9 (t)<br />
0 d (9 times) -<br />
1 x 9 1 8/9<br />
3 x 8 2 2/3<br />
3 x<br />
4 u<br />
5 x 5 1 8/15<br />
6 u<br />
8 x 3 2 8/45<br />
8 x<br />
9 x 1 1 0<br />
The Kaplan-Meier estimate assigns discrete probabilities <strong>of</strong> 8/9−2/3 =2/9, 2/3−8/15 = 2/15,<br />
at times 3 <strong>and</strong> 5 respectively. Adding these two probabilities gives the answer <strong>of</strong> 16/45.<br />
Exercise 19 The mean <strong>of</strong> the data is ˆµ 0 1 = [27 + 82 + 115 + 126 + 155 + 161 + 243 + 13(250)]/20 =<br />
207.95. The expected value <strong>of</strong> a single observation censored at 250 is E(X ∧ 250) = θ(1 − e −250/θ ).<br />
Setting the two equal <strong>and</strong> solving produces ˆθ = 657.26.<br />
Exercise 20 The equations to solve are<br />
Taking logarithms yields<br />
exp(µ + σ 2 /2) = 1, 424.4<br />
exp(2µ +2σ 2 ) = 13, 238, 441.9.<br />
µ + σ 2 /2 = 7.261506<br />
2µ +2σ 2 = 16.398635.<br />
The solution <strong>of</strong> these two equations is ˆµ =6.323695 <strong>and</strong> ˆσ 2 =1.875623 <strong>and</strong> then ˆσ =1.369534.<br />
Exercise 21 The two equations to solve are<br />
0.2 = 1− e −(5/θ)τ<br />
0.8 = 1− e −(12/θ)τ .<br />
Moving the 1 to the left h<strong>and</strong> side <strong>and</strong> taking logarithms produces<br />
0.22314 = (5/θ) τ<br />
1.60944 = (12/θ) τ .<br />
Dividing the second equation by the first equation produces<br />
7.21269 = 2.4 τ .<br />
Taking logarithms again produces<br />
1.97584 = 0.87547τ<br />
<strong>and</strong> so ˆτ =2.25689. Usingthefirst equation,<br />
ˆθ =5/(0.22314 1/2.25689 )=9.71868.
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