Germán Rodríguez
Survival Analysis Princeton University

Cox Regression

We continue our analysis of the Gehan data by fitting a proportional hazards model. This is the same dataset used as an example in Cox's original paper: Cox, D.R. (1972) Regression Models and Life tables, (with discussion) Journal of the Royal Statistical Society, 34: 187–220.

The first task is to read and stset the data. I also create a dummy variable for treated.

. infile group weeks relapse using ///
>   http://data.princeton.edu/wws509/datasets/gehan.raw, clear
(42 observations read)

. gen treated = group == 2

. stset weeks, failure(relapse)

     failure event:  relapse != 0 & relapse < .
obs. time interval:  (0, weeks]
 exit on or before:  failure

------------------------------------------------------------------------------
         42  total observations
          0  exclusions
------------------------------------------------------------------------------
         42  observations remaining, representing
         30  failures in single-record/single-failure data
        541  total analysis time at risk and under observation
                                                at risk from t =         0
                                     earliest observed entry t =         0
                                          last observed exit t =        35
> gehan <- read.table("http://data.princeton.edu/wws509/datasets/gehan.dat")

> names(gehan)
[1] "group"   "weeks"   "relapse"

> summarize(gehan, events = sum(relapse), exposure = sum(weeks))
  events exposure
1     30      541

> gehan <- mutate(gehan, treated = as.numeric(group == "treated"))

Here's a run fitting a Cox model with all the defaults

. stcox treated

         failure _d:  relapse
   analysis time _t:  weeks

Iteration 0:   log likelihood =  -93.98505
Iteration 1:   log likelihood = -86.385606
Iteration 2:   log likelihood = -86.379623
Iteration 3:   log likelihood = -86.379622
Refining estimates:
Iteration 0:   log likelihood = -86.379622

Cox regression -- Breslow method for ties

No. of subjects =           42                  Number of obs    =          42
No. of failures =           30
Time at risk    =          541
                                                LR chi2(1)       =       15.21
Log likelihood  =   -86.379622                  Prob > chi2      =      0.0001

------------------------------------------------------------------------------
          _t | Haz. Ratio   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
     treated |   .2210887   .0905501    -3.68   0.000     .0990706    .4933877
------------------------------------------------------------------------------

. di _b[treated]
-1.5091914

Stata reports hazard ratios unless you specify the option nohr.

> library(survival)

> cm <- coxph(Surv(weeks, relapse) ~ treated, data = gehan)

> cm
Call:
coxph(formula = Surv(weeks, relapse) ~ treated, data = gehan)


          coef exp(coef) se(coef)     z       p
treated -1.572     0.208    0.412 -3.81 0.00014

Likelihood ratio test=16.4  on 1 df, p=5.26e-05
n= 42, number of events= 30 

R reports log-relative risks, but also exponentiates the coefficients to obtain hazard ratios. We see that the treatment reduced the risk of relapse by almost 80% at any duration.

Handling Ties

There are various options for handling ties. Cox's original proposal relies on the discrete partial likelihood. A closely-related alternative due to Kalbfleisch and Prentice uses the marginal likelihood of the ranks. Both methods are computationally intensive. A good fast approximation is due to Efron, and a simpler and faster, though somewhat less accurate, method is due to Breslow and Peto. See the notes for details.

In terms of our software, Stata implements all four using the options exactp, exactm, efron and breslow. The default is breslow but I recommend you always use efron. R implements all but the marginal likelihood, using the argument ties with possible values "efron", "breslow", and "exact". The default is "efron". Let us compare them all.

. estimates store breslow

. quietly stcox treated, efron

. estimates store efron

. quietly stcox treated, exactm

. estimates store exactm

. quietly stcox treated, exactp

. estimates store exactp

. estimates table breslow efron exactm exactp

------------------------------------------------------------------
    Variable |  breslow       efron        exactm       exactp    
-------------+----------------------------------------------------
     treated | -1.5091914   -1.5721251   -1.5981915    -1.628244  
------------------------------------------------------------------
> cmb <-  coxph(Surv(weeks, relapse) ~ treated, data = gehan, ties="breslow")

> cmp <-  coxph(Surv(weeks, relapse) ~ treated, data = gehan, ties="exact")

> data.frame(breslow = coef(cmb), efron = coef(cm), exact = coef(cmp))
          breslow     efron     exact
treated -1.509191 -1.572125 -1.628244

As you can see, Efron's approximation is closer to the exact partial likelihood than Breslow's. The marginal likelihood is even closer. Cox reported a log-likelihood of -1.65 in his paper, which he obtained by evaluating the likelihood in a grid of points. The more exact calculations here yield -1.63, so he did pretty well by hand (see page 197 in the paper).

Proportionality of Hazards

One way to test proportionality of hazards is to introduce interactions with duration. In his original paper Cox tried a linear interaction with time. We will do the same, except that he worked with t - 10 to achieve more orthogonality and we will use t.

Stata's tvc() and texp()

Stata makes it very easy to introduce interactions with time by providing two options

  1. tvc(varlist), to specify the variable(s) we want to interact with time, and

  2. texp(expression) to specify a function of time _t, typically just time, texp(_t), or log-time, texp(log(_t).

Stata will then create a variable equal to the product of the variable specified in tvc() by the time expression specified in texp() and add it to the model.

Let us use this technique to interact treatment and time

. stcox treated, tvc(treated) texp(_t) efron

         failure _d:  relapse
   analysis time _t:  weeks

Iteration 0:   log likelihood =  -93.18427
Iteration 1:   log likelihood =  -85.34729
Iteration 2:   log likelihood = -85.008964
Iteration 3:   log likelihood = -85.008326
Iteration 4:   log likelihood = -85.008326
Refining estimates:
Iteration 0:   log likelihood = -85.008326

Cox regression -- Efron method for ties

No. of subjects =           42                  Number of obs    =          42
No. of failures =           30
Time at risk    =          541
                                                LR chi2(2)       =       16.35
Log likelihood  =   -85.008326                  Prob > chi2      =      0.0003

------------------------------------------------------------------------------
          _t | Haz. Ratio   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
main         |
     treated |   .2057005   .1595878    -2.04   0.042     .0449626    .9410648
-------------+----------------------------------------------------------------
tvc          |
     treated |   1.000865   .0617494     0.01   0.989       .88687    1.129514
------------------------------------------------------------------------------
Note: Variables in tvc equation interacted with _t.

We see that there's no evidence that the treatment effect changes linearly with time. BTW we didn't really have to specify texp(_t) because that's the default.

Another possibility is to allow different treatment effects at early and late durations, say before and after 10 weeks. This is easily done by changing the time expression:

. stcox treated, tvc(treated) texp(_t > 10) efron

         failure _d:  relapse
   analysis time _t:  weeks

Iteration 0:   log likelihood =  -93.18427
Iteration 1:   log likelihood = -84.972656
Iteration 2:   log likelihood =  -84.74237
Iteration 3:   log likelihood = -84.740124
Iteration 4:   log likelihood = -84.740124
Refining estimates:
Iteration 0:   log likelihood = -84.740124

Cox regression -- Efron method for ties

No. of subjects =           42                  Number of obs    =          42
No. of failures =           30
Time at risk    =          541
                                                LR chi2(2)       =       16.89
Log likelihood  =   -84.740124                  Prob > chi2      =      0.0002

------------------------------------------------------------------------------
          _t | Haz. Ratio   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
main         |
     treated |   .2702224   .1426515    -2.48   0.013     .0960215    .7604559
-------------+----------------------------------------------------------------
tvc          |
     treated |   .5475566   .4505425    -0.73   0.464     .1091542    2.746741
------------------------------------------------------------------------------
Note: Variables in tvc equation interacted with _t>10.

The point estimates indicate a 73% reduction in risk in the first ten weeks and an additional 44% reduction after ten weeks, for a total of 85% in the later period. However, the difference in treatment effects between the two periods is not significant.

Splitting at Failures

Because only times with observed failures contribute to the partial likelihood we can introduce arbitrary interactions by splitting the data at each failure time. As a sanity check we verify that we obtain the same estimate as before

. gen id = _n  // need an id variable to split

. streset, id(id)
-> stset weeks, id(id) failure(relapse)

                id:  id
     failure event:  relapse != 0 & relapse < .
obs. time interval:  (weeks[_n-1], weeks]
 exit on or before:  failure

------------------------------------------------------------------------------
         42  total observations
          0  exclusions
------------------------------------------------------------------------------
         42  observations remaining, representing
         42  subjects
         30  failures in single-failure-per-subject data
        541  total analysis time at risk and under observation
                                                at risk from t =         0
                                     earliest observed entry t =         0
                                          last observed exit t =        35

. stsplit , at(failures)
(17 failure times)
(384 observations (episodes) created)

. quietly stcox treated, efron

. di _b[treated]
-1.5721251

. estimates store ph
> failure_times <- sort(unique(gehan$weeks[gehan$relapse]))

> gehanx <- survSplit(gehan, cut = failure_times, 
+   event = "relapse", start = "t0", end = "weeks")

> coef(coxph(Surv(t0, weeks, relapse) ~ treated, data=gehanx))
  treated 
-1.572125 

We now introduce a linear interaction with time using our dummy variable for treated. (You could specify the model as group * t0. R will omit t0 because it is implicit in the baseline hazard and complain that the model matrix is singular, but the results will be correct. My approach is a bit cleaner.)

. stcox treated c.treated#c._t, efron

         failure _d:  relapse
   analysis time _t:  weeks
                 id:  id

Iteration 0:   log likelihood =  -93.18427
Iteration 1:   log likelihood =  -85.34729
Iteration 2:   log likelihood = -85.008964
Iteration 3:   log likelihood = -85.008326
Iteration 4:   log likelihood = -85.008326
Refining estimates:
Iteration 0:   log likelihood = -85.008326

Cox regression -- Efron method for ties

No. of subjects =           42                  Number of obs    =         426
No. of failures =           30
Time at risk    =          541
                                                LR chi2(2)       =       16.35
Log likelihood  =   -85.008326                  Prob > chi2      =      0.0003

--------------------------------------------------------------------------------
            _t | Haz. Ratio   Std. Err.      z    P>|z|     [95% Conf. Interval]
---------------+----------------------------------------------------------------
       treated |   .2057005   .1595878    -2.04   0.042     .0449626    .9410648
               |
c.treated#c._t |   1.000865   .0617494     0.01   0.989       .88687    1.129514
--------------------------------------------------------------------------------

. lrtest ph .

Likelihood-ratio test                                 LR chi2(1)  =      0.00
(Assumption: ph nested in .)                          Prob > chi2 =    0.9888

. stjoin // back to normal
(option censored(0) assumed)
(384 obs. eliminated)
> cmx <- coxph(Surv(t0, weeks, relapse) ~ treated + treated:weeks, data=gehanx)

> summary(cmx)
Call:
coxph(formula = Surv(t0, weeks, relapse) ~ treated + treated:weeks, 
    data = gehanx)

  n= 426, number of events= 30 

                    coef  exp(coef)   se(coef)      z Pr(>|z|)  
treated       -1.5813338  0.2057005  0.7758258 -2.038   0.0415 *
treated:weeks  0.0008651  1.0008655  0.0616960  0.014   0.9888  
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

              exp(coef) exp(-coef) lower .95 upper .95
treated          0.2057     4.8614   0.04496    0.9411
treated:weeks    1.0009     0.9991   0.88687    1.1295

Concordance= 0.69  (se = 0.053 )
Rsquare= 0.038   (max possible= 0.354 )
Likelihood ratio test= 16.35  on 2 df,   p=0.0002813
Wald test            = 14.51  on 2 df,   p=0.0007053
Score (logrank) test = 17.69  on 2 df,   p=0.0001443


> c( logLik(cmx), logLik(cm))
[1] -85.00833 -85.00842

We get a Wald test for the interaction term of z = -0.01, and twice a difference in log-likelihoods of 0.001, so clearly there is no evidence of an interaction between treatment and time at risk. Note that these are exactly the same results we got with tvc() and texp().

Split at 10

As an alternative we could allow different treatment effects before and after 10 weeks. We could use the current dataset, but all we really need to to split at 10, so we'll do just that:

. stsplit dur, at(10)
(21 observations (episodes) created)

. quietly stcox treated, efron // so we can do lrtest

. estimates store ph

. gen after10 = dur == 10

. stcox treated c.treated#c.after10, efron

         failure _d:  relapse
   analysis time _t:  weeks
                 id:  id

Iteration 0:   log likelihood =  -93.18427
Iteration 1:   log likelihood = -84.972656
Iteration 2:   log likelihood =  -84.74237
Iteration 3:   log likelihood = -84.740124
Iteration 4:   log likelihood = -84.740124
Refining estimates:
Iteration 0:   log likelihood = -84.740124

Cox regression -- Efron method for ties

No. of subjects =           42                  Number of obs    =          63
No. of failures =           30
Time at risk    =          541
                                                LR chi2(2)       =       16.89
Log likelihood  =   -84.740124                  Prob > chi2      =      0.0002

-------------------------------------------------------------------------------------
                 _t | Haz. Ratio   Std. Err.      z    P>|z|     [95% Conf. Interval]
--------------------+----------------------------------------------------------------
            treated |   .2702224   .1426515    -2.48   0.013     .0960215    .7604559
                    |
c.treated#c.after10 |   .5475566   .4505425    -0.73   0.464     .1091542    2.746741
-------------------------------------------------------------------------------------

. lrtest ph .

Likelihood-ratio test                                 LR chi2(1)  =      0.54
(Assumption: ph nested in .)                          Prob > chi2 =    0.4638

. drop dur after10

. stjoin  
(option censored(0) assumed)
(21 obs. eliminated)
> gehan10 <- survSplit(gehan, cut = 10,
+   event = "relapse", start = "t0", end = "weeks") %>%
+ mutate(   after10 = as.numeric(t0 == 10), 
+   treated = as.numeric(group == "treated"))

> cm10 <- coxph(Surv(t0, weeks, relapse) ~ treated + treated:after10, 
+   data=gehan10)

> summary(cm10)
Call:
coxph(formula = Surv(t0, weeks, relapse) ~ treated + treated:after10, 
    data = gehan10)

  n= 63, number of events= 30 

                   coef exp(coef) se(coef)      z Pr(>|z|)  
treated         -1.3085    0.2702   0.5279 -2.479   0.0132 *
treated:after10 -0.6023    0.5476   0.8228 -0.732   0.4642  
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

                exp(coef) exp(-coef) lower .95 upper .95
treated            0.2702      3.701   0.09602    0.7605
treated:after10    0.5476      1.826   0.10915    2.7467

Concordance= 0.69  (se = 0.053 )
Rsquare= 0.235   (max possible= 0.948 )
Likelihood ratio test= 16.89  on 2 df,   p=0.0002152
Wald test            = 15.31  on 2 df,   p=0.0004738
Score (logrank) test = 18.91  on 2 df,   p=7.847e-05


> 2*(logLik(cm10) - logLik(cm))     
'log Lik.' 0.5366016 (df=2)

The Wald test now yields z = -0.73 (a chi-squared of 0.53), and the likelihood ratio test concurs with a chi-squared of 0.54 on one d.f. The risk ratio is a bit larger after 10 weeks, but the difference is still not significant. Note that these are exactly the same results we got with tvc() and texp().

Schoenfeld Residuals

Another way to check for proportionality of hazards is to use Schoenfeld residuals (and their scaled counterparts). You can obtain an overall test using the Schoenfeld residuals, or a variable-by-variable test based on the scaled variant. In this case with just one predictor there is only one test, but we'll see later an example with several predictors.

Stata and R offer several possible transformations of time for the test, including a user-specified function, but chose different defaults. In Stata the default is time, but one of the options is km for the Kaplan-Meier estimate of overall survival. In R the default transform is "km" for the K-M estimate, but one of the options is "identity".

. quietly stcox treated, efron

. estat phtest

      Test of proportional-hazards assumption

      Time:  Time
      ----------------------------------------------------------------
                  |                      chi2       df       Prob>chi2
      ------------+---------------------------------------------------
      global test |                      0.00        1         0.9886
      ----------------------------------------------------------------
> zph <- cox.zph(cm , transform="identity")

> zph
            rho    chisq     p
treated 0.00264 0.000205 0.989

The test shows that there is no evidence against the proportional hazards assumption. If there had been, we could get a hint of the nature of the time dependence by plotting the (scaled) residuals against time and using a smoother to glean the trend, if any. In Stata we can do this with the plot() option of the estat phtest command. In R the cox.zph class has a plot() method which uses a spline smoother. I specified df=2 because of the small sample size.

. estat phtest, plot(treated)

. graph export phplot.png, width(500) replace
(file phplot.png written in PNG format)

Residual Plot

> plot(zph, df=2)   

Residual Plot The residuals show no time trend at all, showing that the treatment hazard ratio is fairly constant over time. (We will confirm this result below with a plot of cumulative hazards that provides more direct evidence.)

Baseline Survival

The emphasis in the Cox model is on hazard ratios, but one can calculate a Kaplan-Meier or a Nelson-Aalen estimate of the baseline survival, as shown in the notes. The baseline is defined as the case where all covariate values are zero, and this may not make sense in your data. A popular alternative is to estimate the "baseline" at average values of all covariates. In our case a much better approach is to estimate and plot the estimated survival functions for the two groups. Stata makes this very easy via the stcurve command

. stcurve, surv at(treated=0) at(treated=1)

. graph export coxsurv.png, width(500) replace
(file coxsurv.png written in PNG format)

It is instructive to compute these "by hand" and compare them with separate Kaplan-Meier estimates for each group, which I will plot using different symbols for treated and controls. The plots connect the point estimates using a step function.

. predict S0, basesurv            // control (not mean!)

. gen S1 = S0^exp(_b[treated])    // treated

. sts gen KM = s, by(treated)     // two Kaplan-Meiers

. twoway (scatter S0  _t, c(J) ms(none) sort)  /// baseline
>   (scatter S1 _t , c(J) ms(none) sort)     /// treated
>   (scatter KM _t if treated, msymbol(circle_hollow)) /// KM treated
>   (scatter KM _t if !treated, msymbol(X)) /// KM base
>   , legend(off) ///
> title(Kaplan-Meier and Proportional Hazards Estimates)   

. graph export coxkm.png, width(500) replace      
(file coxkm.png written in PNG format)

Kaplan-Meier and Cox

> sf <- survfit(cm, newdata=list(treated=c(1,0)))

> km <- survfit(Surv(weeks, relapse) ~ treated, data=gehan)

> dsf <- data.frame(time = rep(c(0,sf$time), 2),
+   survival = c(1, sf$surv[,1], 1, sf$surv[,2]), 
+   group = factor(rep(c("treated","control"),
+       rep(length(sf$time) + 1,2))))

> dkm <- data.frame(time = km$time, 
+   survival = km$surv,
+   group = factor(rep(c("treated","control"),
+       km$strata)))    

> ggplot(dsf, aes(time, survival, color = group)) + geom_step() +
+   geom_point(data = dkm, aes(time, survival, shape=group),color="black") +
+   scale_shape_manual(values = c(1, 4)) 

> ggsave("coxkmr.png", width=500/72, height=400/72, dpi=72)     

Kaplan-Meier and Cox The figure looks just like Figure 1 in Cox's paper. You can get pretty much the same thing using Stata's stcoxkm, by(treated), although you will have to tweak the default options. If the purpose of the graph is to check the proportional hazards assumption a much better alternative is to plot the log-log transformation of the survival function, namely -log(-log(S(t)), against log(t) for each group. Under the proportional hazards assumption the resulting curves should be parallel. This plot is useful because the eye is much better at judging whether curves are parallel than whether they are proportional. Stata makes this easy with the stphplot command

. stphplot, by(treated) legend(off) title(Plot of -log(-log(S(t))))

         failure _d:  relapse
   analysis time _t:  weeks
                 id:  id

. graph export coxphplot.png, width(500) replace
(file coxphplot.png written in PNG format)

Log-log plot

> dkm <- mutate(dkm,  lls = -log(-log(survival)))

> ggplot(dkm, aes(log(time), lls, color=group)) + geom_point() +
+   geom_line() +   ylab("-log(-log(S(t)))")

> ggsave("coxphplotr.png", width=500/72, height=400/72, dpi=72)

Log-log plot

The two lines look quite parallel indeed.