9. Estimating Survival Distribution for a PH Model

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1 9. Estimating Survival Distribution for a PH Model

2 Objective: Another Goal of the COX model Estimating the survival distribution for individuals with a certain combination of covariates. PH model assumption: λ t z = λ 0 (t)exp(β T Z) Given Z=z, it is easily to derive the relationship between S(t z) and covariates as the following function: S t z = e t 0 λ u z du = e t 0 λ0 u z exp(β T z )du = e exp(βt z )Λ 0 (t) This means in order to estimate S(t Z = z ), we only need to estimate β and Λ 0 (t), where β can be estimated by MPLE.

3 Estimating Λ 0 (t) The same logic of deriving Nelson-Aalen estimate of the cumulative hazard function in one sample problem will be used. Nelson-Aalen estimate for Λ t = σ x<t dn(x) Y(x) In the one-sample problem, all individuals in the sample have the same hazard of failing, implying the same cause-specific hazard. However, in a proportional hazard model, the individuals in the sample do not have hazard of failing at time x but rather have a hazard which depends on their covariate values. That is, for i th individual with covariate Z i = (Z i1,, Z iq ) T, has hazard λ i t = λ 0 (t)exp(β T Z i )

4 Estimating Λ 0 (t) dn i x F x ~Bin Y i x, π i x, where π i x λ i (x)δx E[dN i x F(x)] = Y i λ i (x)δx = λ 0 (x)exp(β T Z i ) Y i Δx dn x = σ i=1 dn i x E dn x F x = E[σ i=1 dn i x F(x)] = σ i=1 E[dN i x F(x)] = σ i=1 λ 0 (x)exp(β T Z i ) Y i Δx = λ 0 (x) Δx σ i=1 exp(β T Z i ) Y i (x) Therefore we estimate λ 0 (x) Δx by using σ n i=1 dn(x) exp(β T Z i ) Y i (x)

5 Estimating Λ 0 (t) Λ 0 (t) σ x<t λ 0 (x) Δx Λ 0 (t) = σ x<t σ n i=1 dn(x) exp(β T Z i ) Y i (x) Specific, if all the β s were equal to zero, then the previous formula would reduce to σ x<t dn(x) Y(x), giving us back the Nelson-Aalen estimator.

6 Property of Λ 0 (t) Λ 0 (t)is approximately unbiased for Λ 0 (t) Proof: E(σ x<t ( = σ x<t E[ σ i=1 dn(x) n exp(β T Z i ) Y i (x) )) σ i=1 dn(x) n exp(β T Z i ) Y i (x) ] = σ x<t E{E σ n i=1 dn x exp(β T Z i ) Y i x F x } since σ n i=1 E σ n i=1 exp(β T Z i ) Y i x is fixed conditional on F(x) dn x exp(β T Z i ) Y i x = σλ 0(x)exp(β T Z i ) Y i Δx σ n i=1 exp(β T Z i ) Y i x = λ 0 x Δx So, E{E x<t σ n i=1 F x = E[dN(x) F(x)] σ i=1 dn x exp(β T Z i ) Y i x n exp(β T Z i ) Y i x = E(σ dn i x F(x)) n exp(β T Z i ) Y i x F x σ i=1 } = λ 0 x Δx Λ 0 (t) x<t

7 Estimate Survival Distribution Estimate the survival distribution for individuals with a certain combination of covariates z 0 (for a randomly sampled subject). λ t Z = λ 0 t exp β T Z λ t Z = z 0 = λ 0 t exp β T z 0

8 Estimate Survival Function S t Z = z 0 = e Λ t Z = z 0 Λ t Z = z 0 = 0 t λ0 t exp β T z 0 du = exp β T z 0 න 0 t λ 0 t du = exp β T z 0 Λ 0 t መS t Z = z 0 = e exp β T z 0 Λ 0 t ( መβ is the MPLE of β)

9 መS t Z = Z 0 Asymptotic Gaussian distribution E መS t Z = Z 0 a S t Z = Z 0 var መS t Z = Z 0 = መS 2 t Z = Z 0 e 2 β T z 0 Q 1 t + Q 2 t; Z 0 Q 1 t = σ ti d i W 2 t i ;β, where d i is the number of deaths at time t i, W t i ; መβ = σ jεr ti e β T z j, R t i = {j Y j t i = 1} Q 2 t; Z 0 = Q 3 t; Z 0 t Var መβ Q 3 t; Z 0 Q 3 t; Z 0 = σ ti t σ ti t W 1 t i ;β W t i ;β W p t i ;β W t i ;β Z 01 Z 0p d i W t i ;β d i W t i ;β, W k t i ; መβ = σ jεr ti Z jk e β T z j Link CL. Confidence intervals for the survival function using Cox's proportional-hazard model with covariates. Biometrics. 1984, 40(3):601-9.

10 Note: Q 2 reflects the uncertainty in the estimation process Q 3 is large when z 0 is far from the average covariate in the risk set Confidence interval of Survival Function መS t Z = Z 0 ± zα 2 var መS t Z = Z 0

11 Example Data on 90 males with larynx cancer Variables Stage of disease (stages 1 to 4) Age at diagnosis of larynx cancer Time of death or on-study time in months Year of diagnosis of larynx cancer Death Indicator (0=alive, 1=dead) See SAS output

12 R Code library(survival); larynx <- read.table(file="data_chap9_larynx.txt", skip=11, col.names=c("stage", "time", "age", "year", "status")); agecat = larynx$age>60; stage.age = as.factor(larynx$stage+(larynx$age>60)*4); larynx = cbind(larynx, agecat, cat=stage.age); larynx = larynx[order(larynx$stage, larynx$agecat), ] larynx.ph = coxph(surv(time,status) ~ stage+agecat, data=larynx, ties='breslow') stage.age2 = larynx[match(levels(larynx$cat), larynx$cat),c("stage", "agecat")]; s <- summary(survfit(larynx.ph, newdata=stage.age2)); cols=rep(c("black","red","green", "blue"),2); ltys = rep(c(1,2), each=4); plot(0,0,type="n", xlab="survival time",ylab="survival probabilities",xlim=c(0,8),ylim=c(0,1)) for (i in 1:8) lines(s$time, s$surv[,i], lty=ltys[i], col=cols[i]); legend("bottomleft", legend=paste("cat ",1:8),lty=ltys, col = cols, cex=0.9,title.adj=0.2);

9 Estimating the Underlying Survival Distribution for a

9 Estimating the Underlying Survival Distribution for a Proportional Hazards Model So far the focus has been on the regression parameters in the proportional hazards model. These parameters describe the