Estimation, Evaluation, and Selection of Actuarial Models

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52 CHAPTER 3. SAMPLING PROPERTIES OF ESTIMATORS Exercise 58 Repeat the previous example, using time to surrender as the variable. Interpret 2 q 3 as the probability of surrendering before the five years expire. The previous example indicated that the usual method of constructing a confidence interval can lead to an unacceptable result. An alternative approach can be constructed as follows. Let Y =ln[− ln S n (t)]. Using the delta method (see Theorem 3.33 in the next Subsection), the variance of Y can be approximated as follows. The function of interest is g(x) = ln(− ln x). Its derivative is g 0 (x) = 1 be approximated by − ln x −1 x =(x ln x)−1 . According to the delta method, the variance of Y can {g 0 [E(S n (t))]} 2 Var[S n (t)] = Var[S n(t)] [S n (t)lnS n (t)] 2 wherewehaveusedthefactthatS n (t) is an unbiased estimator of S(t). Then, an estimated 95% confidence interval for θ =ln[− ln S(t)] is q dVar[S n (t)] ln[− ln S n (t)] ± 1.96 S n (t)lnS n (t) . Because S(t) =exp(−e θ ), putting each endpoint through this formula will provide a confidence interval for S(t). Fortheupperlimitwehave(whereˆv = Var[S d n (t)]) n exp −e ln[− ln S n(t)]+1.96 √ˆv/[S o n (t)lnS n (t)] = exp n[ln o S n (t)]e 1.96√ˆv/[S n (t)lnS n (t)] " = S n (t) U 1.96 √ˆv # , U =exp . S n (t)lnS n (t) Similarly, the lower limit is S n (t) 1/U . This interval will always be inside the range zero to one and is referred to as the log-transformed confidence interval. Example 3.27 Obtain the log-transformed confidence interval for S(3) as in Example 3.26. We have " 1.96 √ # 0.0034671 U =exp =0.32142. 0.8923 ln(0.8923) The lower limit of the interval is 0.8923 1/0.32142 =0.70150 and the upper limit is 0.8923 0.32142 = 0.96404. ¤ Exercise 59 Obtain the log-transformed confidence interval for S(3) in Exercise 58. Similar results are available for the Nelson-Åalen estimator. An intuitive derivation of a variance estimate proceeds as follows. As in the derivation for the Kaplan-Meier estimator, all results are obtained assuming the risk set numbers are known, not random. The number of deaths at death time t i has approximately a Poisson distribution 3 with parameter r i h(t i ) and so its variance is 3 A binomial assumption (as was used for the Kaplan-Meier derivation) could also have been made. Similarly, a Poisson assumption could have been used for the Kaplan-Meier derivation. The formulas in this Note are the ones most commonly used.
3.3. VARIANCE AND CONFIDENCE INTERVALS 53 r i h(t i ), which can be approximated by r i (s i /r i )=s i . Then (also assuming independence), Ã jX ! dVar[Ĥ(y j)] = Var d s i jX dVar(s i ) . jX s i = r i=1 i r 2 = i=1 i r 2 . i=1 i The linear confidence interval is simply Ĥ(t) ± z α/2 q d Var[Ĥ(y j)]. A log-transformed interval similar to the one developed for the survival function 4 is ⎡ q Ĥ(t)U, whereU =exp⎣± z α/2 dVar[Ĥ(y ⎤ j)] ⎦ . Ĥ(t) Example 3.28 Construct an approximate 95% confidence interval for H(3) by each formula using all 40 observations in Data Set D2. The point estimate is Ĥ(3) = 30 1 + 26 2 The linear confidence interval is =0.11026. The estimated variance is 1 30 2 + 2 26 2 =0.0040697. 0.11026 ± 1.96 √ 0.0040697 = 0.11026 ± 0.12504 for an interval of (−0.01478, 0.23530). For the log-transformed interval, " # U =exp ± 1.96(0.0040697)1/2 =exp(±1.13402) = 0.32174 to 3.10813. 0.11026 The interval is from 0.11026(0.32174) = 0.03548 to 0.11026(3.10813) = 0.34270. ¤ Exercise 60 Construct 95% confidence intervals for H(3) by each formula using all 40 observations in Data Set D2 with surrender being the variable of interest. Exercise 61 (*) Ten individuals were observed from birth. All were observed until death. The following table gives the death ages: Age 2 3 5 7 10 12 13 14 Number of Deaths 1 1 1 2 1 2 1 1 Let V 1 denote the estimated conditional variance of 3ˆq 7 , if calculated without any distribution assumption. Let V 2 denote the conditional variance of 3ˆq 7 , if calculated knowing that the survival function is S(t) =1− t/15. Determine V 1 − V 2 . Exercise 62 (*)Fortheintervalfrom0to1year,theexposure(r) is15andthenumberofdeaths (s) is3.Fortheintervalfrom1to2yearstheexposureis80andthenumberofdeathsis24.For 2to3yearsthevaluesare25and5;for3to4yearstheyare60and6;andfor4to5yearsthey are 10 and 3. Determine Greenwood’s approximation to the variance of Ŝ(4). 4 The derivation of this interval uses the transformation Y =lnĤ(t).
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