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13_AS_2_lec_a

# 88889 071429 1 j 1 1 k k 01 02 042857 from the nal

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Unformatted text preview: ˆ S (t ) = S (tmax ) = S (tk ) for t > tmax . 21/42 Refer to Practical Notes 2 and 3 of K&M (p. 99-100) Actuarial Statistics – Module 2: Non-parametric methods Estimating the lifetime distribution: non-parametric approach Kaplan-Meier (product limit) estimator Example 2.1 (ctd) Calculate the Kaplan-Meier estimate of F (t ). j 1 2 3 tj 3 4 11 dj 1 1 2 nj 10 9 7 d ˆ λj = njj 0.1 0.11111 0.28571 ˆ 1 − λj 0.9 0.88889 0.71429 ˆ 1 − j =1 (1 − λk ) k 0.1 0.2 0.42857 From the ﬁnal column, the Kaplan-Meier estimate of F (t ) is 0 for 0 ≤ t < 3 0.1 for 3 ≤ t < 4 ˆ F (t ) = 0.2 for 4 ≤ t < 11 0.42857 for 11 ≤ t ≤ 20 22/42 Actuarial Statistics – Module 2: Non-parametric methods Estimating the lifetime distribution: non-parametric approach Kaplan-Meier (product limit) estimator Variance of Kaplan-Meier estimator ˜ Let F (t ) denote the estimatOR. Greenwood’s formula states (p. 33 of yellow book) ∧ 2 ˜ ˜ Var F (t ) = Var S (t ) ≈ 1 − F (t ) j :tj ≤t dj nj (nj − dj ) Maximum likelihood estimators are asymptotically normally distributed, so we can easily construct conﬁdence intervals, e.g. ∧ S (t ) ± Z1− α 2 where Z1− α is the 1 − 2 distribution. 23/42 α 2 ˜ var S (t ) , percentile from the standard normal Actuarial Statistics – Module 2: Non-parametric methods Estimating the lifetime distribution: non-parametric approach Kaplan-Meier (product limit) estimator Example Consider the following recorded data for the lifetime of a group of 9 rats 2, 4+ , 5, 8, 12, 12+ , 17, 18+ , 18+ , where + denotes the censored. Suppose that you have found the K-M estimate ˆ S (5) = 0.76190. ˆ Derive a 95% conﬁdence for the K-M estimator S (5). 24/42 Actuarial Statistics – Module 2: Non-parametric methods Estimating the lifetime distribution: non-parametric approach Kaplan-Meier (product limit) estimator Solution We have ˜ ˆ var (S (5)) = S (5)2 ( 1 1 + ) 9(9 − 1) 7(7 − 1) = 0.761902 × 0.037698 = 0.02188 so that the conﬁdence interval is ˜ var (S (5)) √ 0.76190 ± 1.96 0.02188 ˆ S (5) ± Z1−0.025 (0.47196, 1.05182) which ﬁnally becomes (0.47196, 1) 25/42 ˆ as S ≤ 1. Actuarial Statistics – Module 2: Non-parametric methods Estimating the lifetime distribution: non-parametric approach Nelson-Aalen estimator 1 Introduction 2 Censoring and truncation Censoring Truncation Likelihood for censored and truncated data 3 Estimating the lifetime distribution: non-parametric approach Preliminaries...
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