Linda Staub & Alexandros Gekenidis
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1 Seminar in Statistics: Survival Analysis Chapter 2 Kaplan-Meier Survival Curves and the Log- Rank Test Linda Staub & Alexandros Gekenidis March 7th, 2011
2 1 Review Outcome variable of interest: time until an event occurs Time = survival time Event = failure Censoring: Don t know survival time exactly True survival time observed survival time Right-censored
3 1 Review T = failure time with distribution F, density f C = censoring time with distribution G, density g Assume that the censoring time C is independent of the variable of interest T X = min (T, C), Δ = 1 *T C+ We observe n i.i.d. copies of (X, Δ)
4 Survivor function S t = Pr (T > t)
5 Alternative (Ordered) Data Layout Risk set: collection of individuals who have survived at least to time t ()
6 2 Kaplan-Meier Curves Example The data: remission times (weeks) for two groups of leukemia patients Group 1 (n=21) treatment Group 2 (n=21) placebo # failed # censored Total 6, 6, 6, 7, 10, 1, 1, 2, 2, 3, 13, 16, 22, 23, 4, 4, 5, 5, 6+, 9+, 10+, 11+, 8, 8, 8, 8, 17+, 19+, 20+, 11, 11, 12, 12, 25+, 32+, 32+, 15, 17, 22, , 25+ Group Group Descriptive statistic: ignoring 's.1, T 8. 6 T denotes censored
7 Table of ordered failure times Group 1 (treatment) t( ) n m q ( ) Group 2 (placebo) t n m q > Group 1 (treatment) 6, 6, 6, 7, 10, 1, 1, 2, 2, 3, 13, 16, 22, 23, 4, 4, 5, 5, 6+, 9+, 10+, 11+, 8, 8, 8, 8, 17+, 19+, 20+, 11, 11, 12, 12, 25+, 32+, 32+, 15, 17, 22, , denotes censored Group 2 (placebo) Remark: no censorship in group 2
8 Computation of KM-curve for group 2 (no censoring) t( ) n m /21 = /21 = /21 = /21 = /21 = /21 = /21 = /21 = /21 = /21 = /21 = /21 =.00 q S t S t = # surviving past t () 21
9 Computation of KM-curve for group 2 (no censoring) t( ) n m /21 = /21 = /21 = /21 = /21 = /21 = /21 = /21 = /21 = /21 = /21 = /21 =.00 q S t S t = # surviving past t () 21
10 Computation of KM-curve for group 2 (no censoring) t( ) n m /21 = /21 = /21 = /21 = /21 = /21 = /21 = /21 = /21 = /21 = /21 = /21 =.00 q S t S t = # surviving past t () 21
11 Computation of KM-curve for group 2 (no censoring) t( ) n m /21 = /21 = /21 = /21 = /21 = /21 = /21 = /21 = /21 = /21 = /21 = /21 =.00 q S t S t = # surviving past t () 21
12 Computation of KM-curve for group 2 (no censoring) t( ) n m /21 = /21 = /21 = /21 = /21 = /21 = /21 = /21 = /21 = /21 = /21 = /21 =.00 q S t S t = # surviving past t () 21
13 KM Curve for Group 2 (Placebo) > time2 <- c(1,1,2,2,3,4,4,5,5,8,8,8,8,11,11,12,12,15,17, 22,23) > status2 <- c(1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1) > fit2 <- survfit(surv(time2, status2) ~ 1) > plot(fit2,conf.int=0, col = 'red', xlab = 'Time (weeks)', ylab = 'Survival Probability') > title(main='km Curve for Group 2 (placebo)')
14 General KM formula Alternative way to calculate the survival probabilities KM formula = product limit formula S t = Pr T > t (i) T t (i) i=1 = S t ( 1) Pr T > t () T t () Proof: blackboard
15 Computation of KM-curve for group 1 (treatment) t( ) n m q S t () /21= /17= /15= /12= /11= /7= /6=.4482 Fraction at t () : Pr T > t () T t () ) failed prior to Not available at t t Censored prior to t
16 Computation of KM-curve for group 1 (treatment) t( ) n m q S t () /21= /17= /15= /12= /11= /7= /6=.4482 Fraction at t () : Pr T > t () T t () ) failed prior to Not available at t t Censored prior to t
17 Computation of KM-curve for group 1 (treatment) t( ) n m q S t () /21= /17= /15= /12= /11= /7= /6=.4482 Fraction at t () : Pr T > t () T t () ) failed prior to Not available at t t Censored prior to t
18 Computation of KM-curve for group 1 (treatment) t( ) n m q S t () /21= /17= /15= /12= /11= /7= /6=.4482 Fraction at t () : Pr T > t () T t () ) failed prior to = n m n Not available at t t Censored prior to t
19 Computation of KM-curve for group 1 (treatment) t( ) n m q S t () /21= /17= /15= /12= /11= /7= /6=.4482 Fraction at t () : Pr T > t () T t () ) failed prior to Not available at t t Censored prior to t
20 KM-curve for group 1 (treatment) > time1 <- c(6,6,6,7,10,13,16,22,23,6,9,10,11,17,19,20, 25,32,32,34,35) > status1 <- c(1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0) > fit1 <- survfit(surv(time1, status1) ~ 1) > plot(fit1,conf.int=0, col = 'red', xlab = 'Time (weeks)', ylab = 'Survival Probability') > title(main='km Curve for Group 1 (treatment)')
21 KM-estimator = Nonparametric MLE Model T = failure time distr. function F, density f C = censoring time distr. function G, density g Assume that C is independent of T X = min (T, C) Δ = 1 *T C+ We observe n i.i.d. copies of (X, Δ) Derivation of the likelihood for F Claim The density of observing (x, 1) is: f(x)(1 G(x)) The density of observing (x, 0) is: g(x)(1 F(x)) Proof of the Claim: Blackboard Density of observing (x, δ) is: f x 1 G x δ g x 1 F x 1 δ = f x δ 1 F x 1 δ 1 G x δ g x 1 δ
22 The likelihood for F and G of n i.i.d. observations (x 1, δ 1 ),, (x n, δ n ) is: n i=1 f x i δ i 1 F x i 1 δ i 1 G x i δ i g x i 1 δ i T and C independent Ignore part that involves G In order to find the nonparametric maximum likelihood estimator F n, we need to maximize this expression over all possible distribution functions F (with corresponding density f). Optimization problem sup L n (F) F F where F is the class of all distribution functions on R and n L n F = f x i δ i 1 F x i 1 δ i i=1 But: Problem is not well-defined!
23 Solution: Let f be a density w.r.t. counting measure on the observed failure times (instead of a density w.r.t. Lebesgue measure) Replace f(x i ) by F x i = S x i, the ump of the distribution / survival function at x i Parametrizing everything in terms of the survival function S = 1 F: n δ L n F = S x i 1 δ i S x i i=1 i And S satisfies L n S = max S S L n S, where S is the space of all survival functions One can show that the Kaplan-Meier estimator maximizes the likelihood KM-estimator is the NPMLE
24 Comparison of KM Plots for Remission Data > time1 <- c(6,6,6,7,10,13,16,22,23,6,9,10,11,17,19,20,25,32,32,34,35) > status1 <- c(1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0) > time2 <- c(1,1,2,2,3,4,4,5,5,8,8,8,8,11,11,12,12,15,17, 22,23) > status2 <- c(1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1) > fit1 <- survfit(surv(time1, status1) ~ 1) > fit2 <- survfit(surv(time2, status2) ~ 1) > plot(fit1,conf.int=0, col = 'blue', xlab = 'Time (weeks)', ylab = 'Survival Probability') > lines(fit2, col = 'red') > legend(21,1,c('group 1 (treatment)', 'Group 2 (placebo)'), col = c('blue','red'), lty = 1) > title(main='km-curves for Remission Data') Question: Do we have any reason to claim that group 1 (treatment) has better survival prognosis than group 2?
25 3 The Log-Rank Test We look at 2 groups extensions to several groups possible When are two KM curves statistically equivalent? testing procedure compares the two curves we don t have evidence to indicate that the true survival curves are different Nullhypothesis H0 : no difference between (true) survival curves Goal: To find an expression (depending on the data) from which we know the distribution (or at least approximately) under the nullhypothesis
26 Derivation of test statistic Remission data: n=42 t # failures m 1 m 2 # in risk set n 1 n Expected cell counts: n 1 e1 m1 m 2 n1 n 2 Proportion in risk set n 2 e2 m1 m 2 n1 n 2 # of failures over both groups
27 O O O 1 2 i E E E 1 2 i # failure times m i ei O2 E2 Log-rank statistic = Var O E 2 2 Remark: We could also work with O1 E 1 and would get the same statistic! Why?
28 Distribution of log-rank statistic H 0 : no difference between survival curves Log-rank statistic for two groups = O Var 2 2 E2 2 1 O2 E2 Idea of the Proof: If X is standard normal disitributed then X 2 has a χ 2 distribution with 1 df (assuming X to be one-dim) Set X = O 2 E 2 Var O 2 E 2 Then X is standardized and appr. normal distributed for large samples Hence X 2, which is exactly our statistic, has appr. a χ 2 distribution.
29 Log-Rank Test for Remission data R-code > time <- c(6,6,6,7,10,13,16,22,23,6,9,10,11,17,19,20,25,32,32,34,35,1,1,2,2,3,4,4,5,5,8,8,8,8,11,11, 12,12,15,17,22,23) > status <- c(1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1) > treatment <- c(1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2) > fit <- survdiff(surv(time, status) ~ treatment) Result p-value is the probability of obtaining a test statistic at least as extreme as the one that was actually observed! > fit Call: survdiff(formula = Surv(time, status) ~ treatment) N Observed Expected (O-E)^2/E (O-E)^2/V treatment= treatment= Chisq = 16.8 on 1 degrees of freedom, p = 4.17e-05 What does this tell us?
30 The Log-Rank Test for Several Groups H 0 : All survival curves are the same Log-rank statistics for > 2 groups involves variances and covariances of O i E i G ( 2) groups: log-rank statistic ~χ 2 with G 1 df
31 Remarks Alternatives to the Log-Rank Test Wilcoxen Tarone-Ware Peto Flemington-Harrington Variations of the log rank test, derived by applying different weights at the th failure time Weighting the Test statistic: Var w( t )( mi ei ) w( t )( mi ei ) 2 Weight at th failure time
32 Remarks Choosing a Test Results of different weightings usually lead to similar conclusions The best choice is test with most power There may be a clinical reason to choose a particular weighting Choice of weighting should be a priori! Not fish for a desired p-value!
33 Stratified log rank test Variation of log rank test Allows controlling for additional ( stratified ) variable Split data into stratas, depending on value of stratified variable Calculate O E scores within strata Sum O E across strata
34 Stratified log rank test - Example Remission data Stratified variable: 3-level variable (LWBC3) indicating low, medium, or high log white blood cell count (coded 1, 2, and 3, respectively) Treated Group: rx = 0 Placebo Group: rx = 1 Recall: Non-stratified test χ 2 -value of and corresponding p-value rounded to
35 Stratified Log-Rank Test for Remission data R-code > data <- read.table(" > lwbc3 <- c(1,1,1,2,1,2,2,1,1,1,3,2,2,2,2,2,3,3,2,3,3,1,2,2,1,1,3,3,1,3,3,2,3,3,3,3,2,3,3,3,2,3) > fit <- survdiff(surv(data$v1,data$v2)~data$v5+strata(lwbc3)) Result > fit Call: survdiff(formula = Surv(data$V1, data$v2) ~ data$v5 + strata(lwbc3)) N Observed Expected (O-E)^2/E (O-E)^2/V data$v5= data$v5= Chisq = 10.1 on 1 degrees of freedom, p =
36 Stratified vs. unstratified approach
37 Stratified vs. unstratified approach Limitation: Sample size may be small within strata
38 Stratified vs. unstratified approach Limitation: Sample size may be small within strata In next chapter: controlling for other explanatory variables!
39 References KLEINBAUM, D.G. and KLEIN, M. (2005). Survival Analysis. A self-learning text. Springer. MAATHUIS, M. (2007). Survival analysis for interval censored data. Part I.
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