Statistics 262: Intermediate Biostatistics Non-parametric Survival Analysis

Transcription

1 Statistics 262: Intermediate Biostatistics Non-parametric Survival Analysis Jonathan Taylor & Kristin Cobb Statistics 262: Intermediate Biostatistics p.1/??

2 Overview of today s class Kaplan-Meier Curve Nelson-Aalen hazard estimator Comparing populations: log-rank test Statistics 262: Intermediate Biostatistics p.2/??

3 Kaplan-Meier curve Provides a natural estimate of the survivor function S(t) for a population. Based on (T i, δ i ) 1 i n for non-informative right-censored data. Recall that, for right-censored data we either observe the failure or know that it is in interval [T i, + ). Statistics 262: Intermediate Biostatistics p.3/??

4 Computing the KM curve Based on fact that, for any collection of times (0 = s 0 < s 1 < < s k = t) t and any random time T P (T > t) = k j=1 P (T > s k T > s k 1 ). Taking the points to be the points where the distribution of T jumps P (T > t) = t l t P (T > t l T t l ). Statistics 262: Intermediate Biostatistics p.4/??

5 Formula for the KM curve Natural estimate of conditional probability is 1 d l n l. Ŝ(t) = t l t 1 d l n l. Statistics 262: Intermediate Biostatistics p.5/??

6 Properties of KM curve It is the so-called NPMLE (non-parametric maximum likelihood estimator). L(F (t i, δ i ), 1 i n) n = (1 F (t i )) δ i (F (t i ) F (t i ))1 δ i. i=1 Using delta-rule V ar(ŝ(t)) = Ŝ(t)2 tl t d l n l (n l d l ). Statistics 262: Intermediate Biostatistics p.6/??

7 Delta rule Based on Taylor series approximation f( θ) f(θ) + f(θ) ( θ θ). Given an estimate of some parameter θ V ar(f( θ) ( f( θ) 2 ) Var( θ). Statistics 262: Intermediate Biostatistics p.7/??

8 Confidence interval for Ŝ(t) Estimate of variance above gives (1 α) 100% CI Ŝ(t) ± z 1 α/2 V ar(ŝ(t)). Also can compute CI for log( log(s(t)) log( log(ŝ(t)))±z 1 α/2 V ar(log( log(ŝ(t)))) Statistics 262: Intermediate Biostatistics p.8/??

9 Variance of log( log(ŝ(t))) V ar = CI for S(t): (log( log(ŝ(t))) ) 1 (log(ŝ(t)))2 t l t d l n l (n l d l ). (exp( exp(c u )), exp( exp(c l ))) where c l and c u are the terms in the ± on previous page. Statistics 262: Intermediate Biostatistics p.9/??

10 Relation to hazard rates For a discrete random variable, the quantity P (T > t T t) = 1 λ(t) is a discrete analog of the hazard rate because lim dt 0 P (t T < t + dt T t) = λ(t). For continuous random variables with hazard rate h ( t ) S(t) = exp h(t) dt. 0 Statistics 262: Intermediate Biostatistics p.10/??

11 Another way to specify a discrete probability Statistics 262: Intermediate Biostatistics p.11/?? distribution. Discrete hazard rates Replacing sum with integral ( S(t) = exp ) λ(t) t l t is a survivor function (not the same as the KM curve). Define cumulative hazard Λ(t) = t l t λ(t l ).

12 Nelson-Aalen estimate of hazard Leads to Λ(t) = tl t d l n l. V ar ( Λ(t) ) = t l t d l n 2 l. Statistics 262: Intermediate Biostatistics p.12/??

13 Comparing two groups: log-rank tes Suppose we want to compare survival experience between two groups. We observe (T 1i, δ 1i, 1 i n 1 ) and (T 2i, δ 2i, 1 i n 2 ). For any fixed t we might test H 0 : S 1 (t) = S 2 (t). We could test this by a standard χ 2 test. Can also test whole curve simultaneously H 0 : S 1 (t) = S 2 (t), t 0. Statistics 262: Intermediate Biostatistics p.13/??

14 Mantel-Haenszen procedure For a 2 2 K contingency table, the Mantel-Haenszen is a test of whether π 1jk = π 2jk for all 1 k K. In our setting first index is group, second is alive or dead and the third (confounder) index is time (all observed failure times). Plugging into Mantel-Haenszen formula ( K ( )) 2 X 2 i=1 d 1i n 1id i n i =. n i=1 n 1i n 2i d i (n i d i ) n 2 i (n i 1) Statistics 262: Intermediate Biostatistics p.14/??

15 Different weight functions Depending on what parts of the survival experience we want to have power for, we can look at weighted Mantel-Haenszen formula ( K ( )) 2 i=1 w i d 1i n 1id i n i X 2 = n i=1 w2 i n 1i n 2i d i (n i d i ) n 2 i (n i 1) Generalized Wilcoxon: w i = n i. Tarone and Ware: w i = n i.. Statistics 262: Intermediate Biostatistics p.15/??

16 More weight functions Peto-Prentice test: using a modified survival estimate S(t) = tj t n j + 1 d j n j + 1 and weight w i = S(t i 1 ) n i /(n i + 1). Fleming-Harrington: w i = Ŝ(t i) ρ for some ρ. With ρ = 1 should be close to Peto-Prentice test. Fleming-Harrington & Peto-Prentice limiting distributions depend less on censoring Statistics 262: Intermediate Biostatistics p.16/?? mechanism than Generalized Wilcoxon.

17 Problems with Mantel-Haenszen Contingency tables are not really independent. Asymptotically, distribution of (square root of) numerator is Gaussian, denominator is consistent estimate of its variance. Proof based on counting processes and stochastic integrals. Statistics 262: Intermediate Biostatistics p.17/??

18 Another look at log-rank test If survival curves are the same, then for all t 0. Λ 1 (t) = Λ 2 (t) We can estimate Λ i (t) with H i (t), the Nelson-Aalen estimates of hazard. As H i (t) are sums, it is reasonable to believe that they are asymptotically Gaussian. Statistics 262: Intermediate Biostatistics p.18/??

19 Mini intro to counting processes Any linear combinations of the H i (t) s will also be Gaussian. Can define integral with respect to N i (t) the counting process of the failure times in i-th group. Specifically N i (t) = # number of failures in group i (possibly unobserv Y i (t) = # number of subjects in i-th group at risk at time t Then, H i (t) = t 0 1 Y i (s) dn i(s). Statistics 262: Intermediate Biostatistics p.19/??

20 Log-rank test Log rank statistic can be expressed as W = 0 g 1 (s)dn 1 (s) + 0 g 2 (s)dn 2 (s). where weights are chosen so that the expected value is 0 if H 0 is true. Counting processes tell us how to estimate its variance. It says that the contribution from each failure time are uncorrelated (martingale property). Statistics 262: Intermediate Biostatistics p.20/??

21 Log-rank test Under H 0 : S 1 (t) = S 2 (t), t 0 X 2 = W 2 / V ar(w ) χ 2 1. Reject H 0 if X 2 χ 2 1 α,1. Statistics 262: Intermediate Biostatistics p.21/??

22 More than two populations If we have l groups, we can test H 0 : S 1 (t) =..., S l (t), t 0. Test is again Mantel-Haenszen test with observed failure time as confounder. Equivalent to X 2 = (W 1,..., W l 1 )Σ 1 (W 1,..., W l 1 ) T χ 2 l 1 where W j = l m=0 0 g mj (s)dn m (s). Statistics 262: Intermediate Biostatistics p.22/??