Estimation, Evaluation, and Selection of Actuarial Models

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1− GAMMADIST(1/x, α, 1/θ,true). The estimates are ˆα =0.83556 and ˆθ =5, 113. The value of
the loglikelihood function is −363.92.
Exercise 29 In each case the likelihood function is f(27)f(82) ···f(243)[1 − F (250)] 13 . The following
table provides the estimates for both the original and censored data sets.
Model Original Censored
exponential ˆθ =1, 424.4 ˆθ =594.14
gamma ˆα =0.55616, ˆθ =2, 561.1 ˆα =1.5183, ˆθ = 295.69
inv. exponential ˆθ =197.72 ˆθ =189.78
inv. gamma ˆα =0.70888, ˆθ = 140.16 ˆα =0.41612, ˆθ =86.290
The censoring tended to disguise the true nature of these numbers and in general had a large impact
on the estimates.
Exercise 30 The calculations are done as in the example, but with θ unknown. The likelihood must
be numerically maximized. For the shifted data the estimates are ˆα =1.4521 and ˆθ =907.98. The
two expected costs are 907.98/0.4521 = 2, 008 and 1, 107.98/0.4521 = 2, 451 for the 200 and 400
deductibles respectively. For the unshifted data the estimates are ˆα =1.4521 and ˆθ = 707.98. The
three expected costs are 707.98/0.4521 = 1, 566, 2,008, and 2, 451 for the 0, 200, and 400 deductibles
respectively. While it is always the case that for the Pareto distribution the two approaches produce
identical answers, that will not be true in general.
Exercise 31 Thesametablecanbeused. Theonlydifference is that observations that were
surrenders are now treated as x-values and deaths are treated as y-values. Observations that ended
at 5.0 continue to be treated as y-values. Once again there is no estimate for a Pareto model. The
gamma parameter estimates are ˆα =1.229 and ˆθ =6.452.
Exercise 32 The contribution to the likelihood for the first five values (number of drivers having
zero through four accidents) is unchanged. However, for the last seven drivers, the contribution is
[Pr(X ≥ 5)] 7 =[1− p(0) − p(1) − p(2) − p(3) − p(4)] 7
and the maximum must be obtained numerically. The estimated values are ˆλ = 0.16313 and
ˆq =0.02039. These answers are similar to those for Example 2.36 because the probability of six or
more accidents is so small.
Exercise 33 There are four cases, with the likelihood function being the product of the probabilities
for those cases raised to a power equal to the number of times each occurred. The following table
provides the probabilities.
Event
Probability
F (1)−F (0.4)
1−F (0.4)
Observed at age 35.4 and died
Observed at age 35.4 and survived 1 − 0.6w
Observed at age 35 and died F (1) = w
Observed at age 35 and survived 1 − w
The likelihood function is
µ 0.6w
L =
1 − 0.4w
and its logarithm is
= w−0.4w
1−0.4w = 1−0.4w
0.6w
1−0.4w = 1−0.4w
1−w
6 µ 1 − w 4
w 8 (1 − w) 12 ∝ w14 (1 − w) 16
1 − 0.4w
(1 − 0.4w) 10
l =14lnw +16ln(1− w) − 10 ln(1 − 0.4w).