Estimation, Evaluation, and Selection of Actuarial Models

5.2. PROPORTIONAL HAZARDS MODELS 89
value y c contribution to L
c
8 8 c 1 =exp(50β 1 )
1
c 1 +···+c 10
c
22 22 c 2 =exp(20β 1 )
2
c 2 +···+c 10
c
51 51 c 3 =exp(10β 1 + β 2 )
3
c 3 +···+c 10
c
55 55 c 4 =exp(30β 1 + β 2 )
4
c 4 +···+c 10
c
70 70 c 5 =exp(10β 1 )
5
c 5 +···+c 10
c
81 81 c 6 =exp(40β 1 )
6
c 6 +···+c 10
85 c 7 =exp(40β 1 + β 2 )
90 c 8 =exp(30β 1 )
93 93 c 9 =exp(50β 1 + β 2 )
95 c 10 =exp(20β 1 + β 2 )
c 9
c 9 +c 10
The product is maximized when ˆβ 1 = −0.00373 and ˆβ 2 = −0.91994 and the logarithm of the
partial likelihood is −11.9889. When β 1 is forced to be zero, the maximum is at ˆβ 2 = −0.93708
and the logarithm of the partial likelihood is −11.9968. There is no evidence in this sample that
age of the house has an impact when using this model. ¤
Three issues remain. One is to estimate the baseline hazard rate function, one is to deal with
thecasewheretherearemultipleobservationsatthesamevalue,andthefinal one is to estimate
the variances of estimators. For the second problem, there are a number of approaches in the
literature. The question raised earlier could be rephrased as “Given that it is known there are s j
uncensored observations of y j , what is the probability that it was the s j persons who actually had
that value? And do this conditioned on equalling or exceeding that value.” A direct interpretation
of this statement would have the numerator reflect the probability of the s j observations that were
observed. The denominator would be based on all subsets of R(y j ) with s j members. This is a lot
of work. A simplified version due to Breslow treats each of the s j observations separately, but for
the denominator, uses the same risk set for all of them. The effect is to require no change from the
algorithm introduced above.
Example 5.7 In the previous example, suppose that the observation of 81 had actually been 70.
Give the contribution to the partial likelihood function for these two observations.
Using the notation from that example, the contribution for the first observation of 70 would
c
still be 5
c
c 5 +···+c 10
. However, the second observation of 70 would now contribute 6
c 5 +···+c 10
.Notethat
the numerator has not changed (it is still c 6 ); however, the denominator reflectsthefactthatthere
are six observations in R(70). ¤
With regard to estimating the hazard rate function, we firstnotethatthecumulativehazard
rate function is
Z t
Z t
H(t|z) = h(u|z)du = h 0 (u)cdu = H 0 (t)c.
To employ an analog of the Nelson-Åalen estimate, we use
0
Ĥ 0 (t) = X y j ≤t
0
s j
P
i∈R(y j ) c .
i