9. Estimating Survival Distribution for a PH Model

Objective:

• Estimating the survival distribution for individuals with a certain combination of covariates.

PH model assumption:

• 𝜆 𝑡 𝑧 = 𝜆0(𝑡)exp(𝛽𝑇𝑍)

• Given Z=𝑧∗, it is easily to derive the relationship between S(t|z) and covariates as the following function:

𝑆 𝑡 𝑧∗ = 𝑒− 0𝑡𝜆 𝑢 𝑧∗ 𝑑𝑢 = 𝑒− 0

𝑡𝜆0 𝑢 𝑧

∗ 𝑒𝑥𝑝(𝛽𝑇𝑧∗)𝑑 𝑢 = 𝑒−𝑒𝑥𝑝(𝛽𝑇𝑧∗)𝛬0(𝑡)

This means in order to estimate S(t|𝑍 = 𝑧∗), we only need to estimate 𝛽 and Λ0(𝑡), where 𝛽 can be estimated by MPLE.

Another Goal of the COX model

Estimating 𝜦𝟎(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 = σ𝑥<𝑡𝑑𝑁(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 𝑖𝑡ℎ

individual with covariate 𝑍𝑖 = (𝑍𝑖1 , … , 𝑍𝑖𝑞)𝑇, has hazard

𝜆𝑖 𝑡 = 𝜆0(𝑡)exp(𝛽𝑇𝑍𝑖)

Estimating 𝜦𝟎(𝐭) • 𝑑𝑁𝑖 𝑥 |𝐹 𝑥 ~Bin 𝑌𝑖 x , π𝑖 x ,where π𝑖 x ≈ 𝜆𝑖(𝑥)Δ𝑥

→ 𝐸[𝑑𝑁𝑖 𝑥 |𝐹(𝑥)] = 𝑌𝑖𝜆𝑖(𝑥)Δ𝑥

= 𝜆0(𝑥)exp(𝛽𝑇𝑍𝑖) 𝑌𝑖Δ𝑥

• 𝑑𝑁 𝑥 = σ𝑖=1𝑛 𝑑𝑁𝑖 𝑥

→ 𝐸 𝑑𝑁 𝑥 𝐹 𝑥 = 𝐸[σ𝑖=1𝑛 𝑑𝑁𝑖 𝑥 |𝐹(𝑥)]

= σ𝑖=1𝑛 𝐸[𝑑𝑁𝑖 𝑥 |𝐹(𝑥)]

= σ𝑖=1𝑛 𝜆0(𝑥)exp(𝛽

𝑇𝑍𝑖) 𝑌𝑖Δ𝑥

= 𝜆0(𝑥) Δ𝑥 σ𝑖=1𝑛 exp(𝛽𝑇𝑍𝑖) 𝑌𝑖(𝑥)

• Therefore we estimate 𝜆0(𝑥) Δ𝑥 by using

𝑑𝑁(𝑥)

σ𝑖=1𝑛 exp(𝛽𝑇𝑍𝑖) 𝑌𝑖(𝑥)

Estimating 𝜦𝟎(𝐭)

• Λ0(𝑡) ≈ σ𝑥<𝑡 𝜆0(𝑥) Δ𝑥

→ Λ0 (𝑡) = σ𝑥<𝑡𝑑𝑁(𝑥)

σ𝑖=1𝑛 exp(𝛽𝑇𝑍𝑖) 𝑌𝑖(𝑥)

• Specific, if all the 𝛽′𝑠 were equal to zero, then the

previous formula would reduce to σ𝑥<𝑡𝑑𝑁(𝑥)

𝑌(𝑥), giving

us back the Nelson-Aalen estimator.

Property of 𝜦𝟎(𝐭)• Λ0(𝑡)is approximately unbiased for Λ0(𝑡)

Proof: 𝐸(σ𝑥<𝑡(𝑑𝑁(𝑥)

σ𝑖=1𝑛 exp(𝛽𝑇𝑍𝑖) 𝑌𝑖(𝑥)

))

= σ𝑥<𝑡𝐸[𝑑𝑁(𝑥)

σ𝑖=1𝑛 exp(𝛽𝑇𝑍𝑖) 𝑌𝑖(𝑥)

]

= σ𝑥<𝑡𝐸{𝐸𝑑𝑁 𝑥

σ𝑖=1𝑛 exp(𝛽𝑇𝑍𝑖) 𝑌𝑖 𝑥

𝐹 𝑥 }

since σ𝑖=1𝑛 exp(𝛽𝑇𝑍𝑖) 𝑌𝑖 𝑥 is fixed conditional on F(x)

𝐸𝑑𝑁 𝑥

σ𝑖=1𝑛 exp(𝛽𝑇𝑍𝑖) 𝑌𝑖 𝑥

𝐹 𝑥 =𝐸[𝑑𝑁(𝑥)|𝐹(𝑥)]

σ𝑖=1𝑛 exp(𝛽𝑇𝑍𝑖) 𝑌𝑖 𝑥

=𝐸(σ𝑑𝑁𝑖 𝑥 |𝐹(𝑥))

σ𝑖=1𝑛 exp(𝛽𝑇𝑍𝑖) 𝑌𝑖 𝑥

=σ𝜆0(𝑥)exp(𝛽

𝑇𝑍𝑖) 𝑌𝑖Δ𝑥

σ𝑖=1𝑛 exp(𝛽𝑇𝑍𝑖) 𝑌𝑖 𝑥

= 𝜆0 𝑥 Δ𝑥

𝑆𝑜, 𝑥<𝑡

𝐸{𝐸𝑑𝑁 𝑥

σ𝑖=1𝑛 exp(𝛽𝑇𝑍𝑖) 𝑌𝑖 𝑥

𝐹 𝑥 } =𝑥<𝑡

𝜆0 𝑥 Δ𝑥 ≈ Λ0 (𝑡)

Estimate Survival Distribution

• Estimate the survival distribution forindividuals with a certain combination ofcovariates 𝑧0 (for a randomly sampled subject).

• 𝜆 𝑡 𝑍 = 𝜆0 𝑡 𝑒𝑥𝑝 𝛽𝑇𝑍

• 𝜆 𝑡 𝑍 = 𝑧0 = 𝜆0 𝑡 𝑒𝑥𝑝 𝛽𝑇𝑧0

Estimate Survival Function

• 𝑆 𝑡 𝑍 = 𝑧0 = 𝑒−Λ 𝑡 𝑍 = 𝑧0

• Λ 𝑡 𝑍 = 𝑧0 = 0𝑡𝜆0 𝑡 𝑒𝑥𝑝 𝛽𝑇𝑧0 𝑑𝑢

= 𝑒𝑥𝑝 𝛽𝑇𝑧0 න0

𝑡

𝜆0 𝑡 𝑑𝑢 = 𝑒𝑥𝑝 𝛽𝑇𝑧0 Λ0 𝑡

• መ𝑆 𝑡 𝑍 = 𝑧0 = 𝑒−𝑒𝑥𝑝𝛽𝑇𝑧0 Λ0 𝑡

( መ𝛽 is the MPLE of 𝛽)

መ𝑆 𝑡 𝑍 = 𝑍0• Asymptotic Gaussian distribution

• 𝐸 መ𝑆 𝑡 𝑍 = 𝑍0 →𝑎𝑆 𝑡 𝑍 = 𝑍0

• ෞ𝑣𝑎𝑟 መ𝑆 𝑡 𝑍 = 𝑍0 = መ𝑆2 𝑡 𝑍 = 𝑍0 𝑒2𝛽𝑇𝑧0 𝑄1 𝑡 + 𝑄2 𝑡; 𝑍0

• 𝑄1 𝑡 = σ𝑡𝑖

𝑑𝑖

𝑊2 𝑡𝑖;𝛽, where 𝑑𝑖 is the number of deaths at time 𝑡𝑖 ,

𝑊 𝑡𝑖; መ𝛽 = σ𝑗𝜖𝑅 𝑡𝑖𝑒𝛽𝑇𝑧𝑗, 𝑅 𝑡𝑖 = {𝑗|𝑌𝑗 𝑡𝑖 = 1}

• 𝑄2 𝑡; 𝑍0 = 𝑄3 𝑡; 𝑍0𝑡 𝑉𝑎𝑟 መ𝛽 𝑄3 𝑡; 𝑍0

𝑄3 𝑡; 𝑍0 =

σ𝑡𝑖≤𝑡𝑊 1 𝑡𝑖;𝛽

𝑊 𝑡𝑖;𝛽− 𝑍01

𝑑𝑖

𝑊 𝑡𝑖;𝛽

⋮

σ𝑡𝑖≤𝑡𝑊 𝑝 𝑡𝑖;𝛽

𝑊 𝑡𝑖;𝛽− 𝑍0𝑝

𝑑𝑖

𝑊 𝑡𝑖;𝛽

, 𝑊 𝑘 𝑡𝑖; መ𝛽 = σ𝑗𝜖𝑅 𝑡𝑖𝑍𝑗𝑘𝑒

𝛽𝑇𝑧𝑗

Link CL. Confidence intervals for the survival function using Cox's proportional-hazard model with covariates. Biometrics. 1984, 40(3):601-9.

Note:

• 𝑄2 reflects the uncertainty in the estimation process

• 𝑄3 is large when 𝑧0 is far from the average covariate in the risk set

• Confidence interval of Survival Function

መ𝑆 𝑡 𝑍 = 𝑍0 ± 𝑧𝛼2

ෞ𝑣𝑎𝑟 መ𝑆 𝑡 𝑍 = 𝑍0

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

R Codelibrary(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);