## Cox Model for Recidivism Data

We return to the recidivism data that we analyzed using parametric models. Let us start by reading the data and setting them up for survival analysis

. use http://www.stata.com/data/jwooldridge/eacsap/recid, clear . gen fail = 1 - cens . stset durat, failure(fail) failure event: fail != 0 & fail < . obs. time interval: (0, durat] exit on or before: failure ------------------------------------------------------------------------------ 1445 total observations 0 exclusions ------------------------------------------------------------------------------ 1445 observations remaining, representing 552 failures in single-record/single-failure data 80013 total analysis time at risk and under observation at risk from t = 0 earliest observed entry t = 0 last observed exit t = 81

> library(foreign) > recid <- read.dta("http://www.stata.com/data/jwooldridge/eacsap/recid.dta") > recid$fail = 1 - recid$cens

### A Proportional Hazards Model

We fit a proportional hazards model using the efron method for handling ties:

. local model workprg priors tserved felon drugs black married educ age . stcox `model', efron failure _d: fail analysis time _t: durat Iteration 0: log likelihood = -3891.7741 Iteration 1: log likelihood = -3841.8477 Iteration 2: log likelihood = -3822.6663 Iteration 3: log likelihood = -3820.9451 Iteration 4: log likelihood = -3820.9142 Iteration 5: log likelihood = -3820.9142 Refining estimates: Iteration 0: log likelihood = -3820.9142 Cox regression -- Efron method for ties No. of subjects = 1,445 Number of obs = 1,445 No. of failures = 552 Time at risk = 80013 LR chi2(9) = 141.72 Log likelihood = -3820.9142 Prob > chi2 = 0.0000 ------------------------------------------------------------------------------ _t | Haz. Ratio Std. Err. z P>|z| [95% Conf. Interval] -------------+---------------------------------------------------------------- workprg | 1.054341 .0952997 0.59 0.558 .8831665 1.258691 priors | 1.100222 .014745 7.13 0.000 1.071699 1.129505 tserved | 1.012483 .0016826 7.47 0.000 1.009191 1.015786 felon | .7361351 .077207 -2.92 0.003 .5993523 .904134 drugs | 1.30457 .1278366 2.71 0.007 1.076607 1.580803 black | 1.485848 .1301791 4.52 0.000 1.251406 1.76421 married | .8741874 .0953226 -1.23 0.218 .7059735 1.082482 educ | .9804111 .0190581 -1.02 0.309 .9437606 1.018485 age | .9966611 .0005162 -6.46 0.000 .9956499 .9976733 ------------------------------------------------------------------------------ . estimates store efron // for later comparisons

> library(survival) > mf <- Surv(durat, fail) ~ + workprg + priors + tserved + felon + drugs + black + married + educ + age > cm <- coxph(mf, data = recid) > cm Call: coxph(formula = mf, data = recid) coef exp(coef) se(coef) z p workprg 0.052916 1.054341 0.090388 0.59 0.5583 priors 0.095512 1.100222 0.013402 7.13 1.0e-12 tserved 0.012406 1.012483 0.001662 7.47 8.3e-14 felon -0.306342 0.736135 0.104882 -2.92 0.0035 drugs 0.265874 1.304570 0.097991 2.71 0.0067 black 0.395985 1.485848 0.087613 4.52 6.2e-06 married -0.134460 0.874187 0.109041 -1.23 0.2175 educ -0.019783 0.980411 0.019439 -1.02 0.3088 age -0.003344 0.996661 0.000518 -6.46 1.1e-10 Likelihood ratio test=142 on 9 df, p=0 n= 1445, number of events= 552

The estimated coefficients are similar to those obtained using a Weibull model. Here's a side by side comparison. Recall that R uses the AFT metric, so we need to convert to PH to compare.

. quietly streg `model', distribution(weibull) . estimates store weibull . estimates table efron weibull, equation(1:1) // match equations ---------------------------------------- Variable | efron weibull -------------+-------------------------- #1 | workprg | .05291557 .05775252 priors | .0955123 .09672029 tserved | .01240588 .01285382 felon | -.30634168 -.3201125 drugs | .26587357 .27186642 black | .39598547 .41429122 married | -.13446046 -.12972325 educ | -.01978331 -.02192692 age | -.0033445 -.00345801 _cons | -3.3512677 -------------+-------------------------- ln_p | _cons | -.22080169 ----------------------------------------

> wm <- survreg(mf, dist = "weibull", data = recid) > cbind( coef(cm), -coef(wm)[-1]/wm$scale ) [,1] [,2] workprg 0.052915570 0.057752517 priors 0.095512305 0.096720295 tserved 0.012405884 0.012853823 felon -0.306341680 -0.320112498 drugs 0.265873573 0.271866422 black 0.395985470 0.414291221 married -0.134460460 -0.129723255 educ -0.019783310 -0.021926923 age -0.003344497 -0.003458007

For example the Cox coefficient for blacks indicates that African Americans face a 48.6% higher risk of recidivism at any duration since release from prison than non-blacks with the same observed characteristics. The Weibull analysis yielded an estimate of 51.3% higher hazard.

### The Treatment of Ties

Let us compare all available methods of handling ties. If you run this code expect the exact calculation to take substantially longer than the others.

. quietly stcox `model', breslow // default . estimates store breslow . quietly stcox `model', exactm . estimates store exactm . quietly stcox `model', exactp . estimates store exactp . estimates table breslow efron exactm exactp ------------------------------------------------------------------ Variable | breslow efron exactm exactp -------------+---------------------------------------------------- workprg | .05284355 .05291557 .05291409 .05315276 priors | .09503255 .0955123 .09551509 .09645497 tserved | .01229776 .01240588 .01240713 .01254959 felon | -.30525144 -.30634168 -.3063558 -.30993357 drugs | .26393017 .26587357 .26588358 .26686928 black | .39366079 .39598547 .39600249 .398402 married | -.13422414 -.13446046 -.13446367 -.13577505 educ | -.01971757 -.01978331 -.01978529 -.02012032 age | -.0033222 -.0033445 -.00334456 -.00335309 ------------------------------------------------------------------ . estimates restore efron // preferred (results efron are active now)

> cm_b <- coxph(mf, data=recid, ties="breslow") > cm_e <- coxph(mf, data=recid, ties="exact") > data.frame(breslow = coef(cm_b), efron = coef(cm), exactp = coef(cm_e)) breslow efron exactp workprg 0.052843554 0.052915570 0.053152759 priors 0.095032546 0.095512305 0.096454966 tserved 0.012297764 0.012405884 0.012549587 felon -0.305251441 -0.306341680 -0.309933573 drugs 0.263930169 0.265873573 0.266869280 black 0.393660789 0.395985470 0.398402001 married -0.134224141 -0.134460460 -0.135775053 educ -0.019717572 -0.019783310 -0.020120315 age -0.003322199 -0.003344497 -0.003353089

As is often the case, the Efron method comes closer to the exact partial likelihood estimate with substantial;y less computational effort, although in this application all methods yield very similar results.

### Baseline Hazard

After fitting a Cox model we can obtain estimates of the baseline survival or cumulative hazard using extensions of the Kaplan-Meier and Nelson-Aalen estimators.

Estimates of the hazard itself, which may be obtained by differencing the estimated cumulative hazard or negative the estimated survival, are usually too "spiky" to be useful, but can be smoothed to glean the general shape.

Stata makes these calculations extremely easy via the `stcurve`

command.
By default the command computes the baseline hazard or survival setting all
covariates to their means, not zero. You can request other values via the
options `at(varname=value)`

. Below I plot the hazards for blacks and others
with all other variables set to their means. To make sure you understand
exactly what Stata is doing under the hood I also do this "by hand".

In R we can obtain the survival via `survfit()`, which accepts a Cox model as argument. You are encouraged to always call this function with a new data frame that specifies exactly what you want to calculate. The default is the mean of all covariates used in the Cox fit. Below I calculate the means explicitly and construct a data frame with two rows, representing blacks and others with all other variables set to their means. I then compute the survival, and difference the negative log to obtain the hazard.

. stcurve, at(black=1) at(black=0) hazard . // by hand: . preserve . quietly sum black . scalar mb = r(mean) . predict H, basech . keep if fail (893 observations deleted) . bysort _t: keep if _n == 1 (478 observations deleted) . gen h= H[_n + 1] - H (1 missing value generated) . gen rr0 = exp((1 - mb) * _b[black]) . gen rr1 = exp( - mb * _b[black]) . gen h0 = h * rr0 (1 missing value generated) . gen h1 = h * rr1 (1 missing value generated) . line h0 h1 _t, col(red blue) legend(off) . lowess h0 _t, gen(s0) bw(0.5) nograph . lowess h1 _t, gen(s1) bw(0.5) nograph . line h0 h1 s0 s1 _t, col(red blue red blue) legend(off) aspect(1) . restore . graph export recidhaz.png, width(500) replace (file recidhaz.png written in PNG format)

> means <- summarize_each(recid, funs(mean)) > nd <- rbind(means, means) %>% mutate(black = 0:1) > sf <- survfit(cm, newdata = nd) # type will match estimate > nr <- length(sf$time) > ndh <- data.frame( + ethn = factor(rep(c("black","other"), rep(nr - 1, 2))), + time = (sf$time[-1] + sf$time[-nr])/2, + hazard = c(diff(-log(sf$surv[,2])), diff(-log(sf$surv[,1]))) + ) > ggplot(ndh, aes(time, hazard, color = ethn)) + geom_line() + + geom_smooth(span = 0.5, se = FALSE) > ggsave("recidhazr.png", width = 500/72, height = 400/72, dpi = 72)

The estimates suggests that the hazard raises a bit in the first few weeks after release and then declines with duration. This result is consistent with the observation that a log-normal model fits better than a Weibull.

### Schoenfeld Residuals

We now check the proportional hazards assumption using scaled Schoenfeld
residuals. Recall that our software uses different defaults so results
will differ. Stata computes the test using the original time scale.
R computes it using the overall survival function as the time scale.
We specify `transform = "identity"`

to obtain exactly the same results.

. estat phtest, detail Test of proportional-hazards assumption Time: Time ---------------------------------------------------------------- | rho chi2 df Prob>chi2 ------------+--------------------------------------------------- workprg | 0.04790 1.27 1 0.2595 priors | -0.08288 3.06 1 0.0802 tserved | -0.09808 3.59 1 0.0580 felon | 0.00891 0.04 1 0.8423 drugs | -0.00486 0.01 1 0.9090 black | 0.04834 1.27 1 0.2591 married | 0.04923 1.38 1 0.2399 educ | -0.06091 1.83 1 0.1766 age | 0.04229 1.31 1 0.2531 ------------+--------------------------------------------------- global test | 12.76 9 0.1740 ----------------------------------------------------------------

> zph <- cox.zph(cm, transform = "identity"); zph rho chisq p workprg 0.04790 1.2712 0.2595 priors -0.08288 3.0601 0.0802 tserved -0.09808 3.5937 0.0580 felon 0.00891 0.0396 0.8423 drugs -0.00486 0.0131 0.9090 black 0.04834 1.2736 0.2591 married 0.04923 1.3814 0.2399 educ -0.06091 1.8256 0.1766 age 0.04229 1.3063 0.2531 GLOBAL NA 12.7561 0.1740

The overall chi-squared statistic of 12.76 on 9 d.f. indicates no
significant departure from the proportional hazards assumption. The only
variable that might deserve further scrutiny is time served, which
has the largest chi-squared statistic, although it doesn't reach the
conventional five percent level. Just out of curiosity we can plot
the scaled Schoenfeld residuals against time. To do this
we use the `plot(varname)`

option of `stphtest`

.
we use the generic `plot()`

function. By default this plots all
covariates you can use an index to specify the target of interest.

. stphtest, plot(tserved) . graph export recidpht.png, width(500) replace (note: file recidpht.png not found) (file recidpht.png written in PNG format)

> plot(zph[3])

We see no evidence of a trend in the effect of time served, so we have no evidence against the proportionality assumption.

More detailed exploration of this issue can be pursued by introducing interactions with duration, as we demonstrated using Cox's example.