Estimation, Evaluation, and Selection of Actuarial Models
Estimation, Evaluation, and Selection of Actuarial Models Estimation, Evaluation, and Selection of Actuarial Models
90 CHAPTER 5. MODELS WITH COVARIATES That is, the outer sum is taken over all uncensored observations less than or equal to t. Thenumerator is the number of observations having an uncensored value equal to y j and the denominator, rather than having the number in the risk set, adds their c-values. As usual, the baseline survival function is estimated as Ŝ0(t) =exp[−Ĥ0(t)]. Example 5.8 For the continuing example (using the original values), estimate the baseline survival function and then estimate the probability that a claim for a 35-year-old wood house will exceed 80% of the house’s value. Compare this to the value obtained from the beta distribution model obtained earlier. Using the estimates obtained earlier, the 10 c-values are as given in the following table. Also included is the jump in the cumulative hazard estimate, followed by the estimate of the cumulative hazard function itself. Values for that function apply from the given y-value up to, but not including, the next y-value. value y c jump Ĥ 0 (y) Ŝ 0 (y) 1 8 8 0.8300 0.8300+···+0.3699 =0.1597 0.1597 0.8524 1 22 22 0.9282 0.9282+···+0.3699 =0.1841 0.3438 0.7091 1 51 51 0.3840 0.3840+···+0.3699 =0.2220 0.5658 0.5679 1 55 55 0.3564 0.3564+···+0.3699 =0.2427 0.8086 0.4455 1 70 70 0.9634 0.9634+···+0.3699 =0.2657 1.0743 0.3415 1 81 81 0.8615 0.8615+···+0.3699 =0.3572 1.4315 0.2390 85 0.3434 90 0.8942 1 93 93 0.3308 0.3308+0.3699 =1.4271 2.8586 0.0574 95 0.3699 For the house as described, c =exp[−0.00373(35) − 0.91994(1)] = 0.34977. The estimated probability is 0.3415 0.34977 =0.68674. From the beta distribution, Ŝ 0 (0.8) = 0.27732 and c = exp[−0.00315(35) − 0.77847(1)] = 0.41118 which gives an estimated probability of 0.27732 0.41118 = 0.59015. ¤ Exercise 91 Suppose the 40 observations in Data Set D2 were from four types of policyholders. Observations 1, 5, ... are from male smokers, observations 2, 6, ... are from male nonsmokers, observations 3, 7, ... are from female smokers, and observations 4, 8, ... are from female nonsmokers. You are to construct a model for the time to surrender and then use the model to estimate the probability of surrendering in the first year for each of the four cases. Construct each of the following three models: 1. Use four different Nelson-Åalen estimates, keeping the four groups separate. 2. Use a proportional hazards model where the baseline distribution has the exponential distribution. 3. Use a proportional hazards model with an empirical estimate of the baseline distribution.
5.2. PROPORTIONAL HAZARDS MODELS 91 Exercise 92 (*) The duration of a strike follows a Cox proportional hazards model in which the baseline distribution has an exponential distribution. The only variable used is the index of industrial production. When the index has a value of 10, the mean duration of a strike is 0.2060 years. When the index has a value of 25, the median duration is 0.0411 years. Determine the probability that a strike will have a duration of more than one year if the index has a value of 5. With regard to variance estimates, recall that the logarithm of the partial likelihood function is l(β) = X j ∗ ln c j ∗ P i∈R(y j ) c i where the sum is taken over all observations that produced an uncensored value. Taking the first partial derivative with respect to β g produces ⎡ ∂ l(β) = X ∂β g j ∗ To simplify this expression, note that ⎣ 1 c j ∗ ∂c j ∗ ∂β g − 1 P i∈R(y j ) c i ∂c j ∗ ∂β g = ∂eβ 1 z j ∗ 1+β 2 z j ∗ 2 +···+β p z j ∗ p ∂β g ∂ ∂β g X i∈R(y j ) = z j ∗ gc j ∗ where z j ∗ g is the value of z g for subject j ∗ . The derivative is ∂ l(β) = X " P i∈R(y z j ∂β ∗ g − j ) z # igc i P g j ∗ i∈R(y j ) c . i The negative second partial derivative is ⎡ ∂2 − l(β) = X ∂β h β g j ∗ ⎢ ⎣ P i∈R(y j ) z igz ih c i P i∈R(y j ) c − i c i ⎤ ⎦ . ³ ⎤ P P i∈R(y j ) z igc i´³ i∈R(y j ) z ihc i´ ³ P i∈R(y j ) c i´2 Using the estimated values these partial derivatives provide an estimate of the information matrix. Example 5.9 Obtain the information matrix and estimated covariance matrix for the continuing example. Then use this to produce a 95% confidence interval for the relative risk of a wood house versus a brick house of the same age. Consider the entry in the outer sum for the observation with z 1 =50and z 2 =1. The risk set contains this observation (with a value of 93 and c =0.330802) and the censored observation with z 1 =20and z 2 =1(with a value of 95 and c =0.369924). For the derivative with respect to β 1 and β 2 the entry is 50(1)(0.330802) + 20(1)(0.369924) [50(0.330802) + 20(0.369924)][1(0.330802) + 1(0.369924)] − 0.330802 + 0.369924 (0.330802 + 0.369924) 2 =0. ⎥ ⎦ .
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90 CHAPTER 5. MODELS WITH COVARIATES<br />
That is, the outer sum is taken over all uncensored observations less than or equal to t. Thenumerator<br />
is the number <strong>of</strong> observations having an uncensored value equal to y j <strong>and</strong> the denominator,<br />
rather than having the number in the risk set, adds their c-values. As usual, the baseline survival<br />
function is estimated as Ŝ0(t) =exp[−Ĥ0(t)].<br />
Example 5.8 For the continuing example (using the original values), estimate the baseline survival<br />
function <strong>and</strong> then estimate the probability that a claim for a 35-year-old wood house will exceed 80%<br />
<strong>of</strong> the house’s value. Compare this to the value obtained from the beta distribution model obtained<br />
earlier.<br />
Using the estimates obtained earlier, the 10 c-values are as given in the following table. Also<br />
included is the jump in the cumulative hazard estimate, followed by the estimate <strong>of</strong> the cumulative<br />
hazard function itself. Values for that function apply from the given y-value up to, but not including,<br />
the next y-value.<br />
value y c jump Ĥ 0 (y) Ŝ 0 (y)<br />
1<br />
8 8 0.8300<br />
0.8300+···+0.3699<br />
=0.1597 0.1597 0.8524<br />
1<br />
22 22 0.9282<br />
0.9282+···+0.3699<br />
=0.1841 0.3438 0.7091<br />
1<br />
51 51 0.3840<br />
0.3840+···+0.3699<br />
=0.2220 0.5658 0.5679<br />
1<br />
55 55 0.3564<br />
0.3564+···+0.3699<br />
=0.2427 0.8086 0.4455<br />
1<br />
70 70 0.9634<br />
0.9634+···+0.3699<br />
=0.2657 1.0743 0.3415<br />
1<br />
81 81 0.8615<br />
0.8615+···+0.3699<br />
=0.3572 1.4315 0.2390<br />
85 0.3434<br />
90 0.8942<br />
1<br />
93 93 0.3308<br />
0.3308+0.3699<br />
=1.4271 2.8586 0.0574<br />
95 0.3699<br />
For the house as described, c =exp[−0.00373(35) − 0.91994(1)] = 0.34977. The estimated<br />
probability is 0.3415 0.34977 =0.68674. From the beta distribution, Ŝ 0 (0.8) = 0.27732 <strong>and</strong> c =<br />
exp[−0.00315(35) − 0.77847(1)] = 0.41118 which gives an estimated probability <strong>of</strong> 0.27732 0.41118 =<br />
0.59015. ¤<br />
Exercise 91 Suppose the 40 observations in Data Set D2 were from four types <strong>of</strong> policyholders.<br />
Observations 1, 5, ... are from male smokers, observations 2, 6, ... are from male nonsmokers,<br />
observations 3, 7, ... are from female smokers, <strong>and</strong> observations 4, 8, ... are from female nonsmokers.<br />
You are to construct a model for the time to surrender <strong>and</strong> then use the model to estimate<br />
the probability <strong>of</strong> surrendering in the first year for each <strong>of</strong> the four cases. Construct each <strong>of</strong> the<br />
following three models:<br />
1. Use four different Nelson-Åalen estimates, keeping the four groups separate.<br />
2. Use a proportional hazards model where the baseline distribution has the exponential distribution.<br />
3. Use a proportional hazards model with an empirical estimate <strong>of</strong> the baseline distribution.
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