The following question and answer is based on an exchange
that started on Statalist.

Title | Baseline hazard and baseline hazard contribution | |

Author | William Gould, StataCorp |

In Stata’s
stcox model,
I’ve noticed that it is now possible to obtain nonparametric estimates
of the contribution to the baseline hazard (through the
basehc() option in Stata 7 to 10 or through the
postestimation command **predict, basehc** since Stata 11), but it is no
longer possible to get nonparametric estimates of the baseline hazard itself
(which used to be available through the basehazard()
option in Stata 6). After reading Kalbfleisch and Prentice, I’m wondering if
there is some equivocation in the use of the word “baseline”
here. What is the relationship between baseline hazard and baseline hazard
contribution?

Yes, indeed there is some equivocation.

First, what used to be returned by the old (Stata 6)
basehazard() option is exactly what was
returned by the
basehc() option in versions 7–10
and is created now by the postestimation command **predict** with the option **basehc**.

The problem was that what was returned by the old basehazard() option was not (and what is returned by the new basehc() option is not) the baseline hazard; it is the numerator of the baseline hazard, called the hazard contribution by Kalbfleisch and Prentice (2002, p. 115, eq. 3–34). To convert what is returned to a baseline hazard, you could divide it by Delta_t, the time between failures. But don’t do that. I did some simulations and quickly convinced myself that dividing by Delta_t is a poor estimator of the baseline hazard. Results are much better if the estimate is based on the cumulative hazard, using smoothing followed by numerical differentiation techniques.

The command stcurve calculates and plots the smoothed hazard estimate. By default, stcurve plots the estimate at the means of the covariates:

. sysuse cancer, clear(Patient Survival in Drug Trial). stset studytime, failure(died)failure event: died != 0 & died < . obs. time interval: (0, studytime] exit on or before: failure ------------------------------------------------------------------------------ 48 total obs. 0 exclusions ------------------------------------------------------------------------------ 48 obs. remaining, representing 31 failures in single record/single failure data 744 total analysis time at risk, at risk from t = 0 earliest observed entry t = 0 last observed exit t = 39. stcox drug age, nologfailure _d: died analysis time _t: studytime Cox regression -- Breslow method for ties No. of subjects = 48 Number of obs = 48 No. of failures = 31 Time at risk = 744 LR chi2(2) = 36.29 Log likelihood = -81.765061 Prob > chi2 = 0.0000 ------------------------------------------------------------------------------ _t | Haz. Ratio Std. Err. z P>|z| [95% Conf. Interval] -------------+---------------------------------------------------------------- drug | .2153648 .0676904 -4.89 0.000 .1163154 .3987605 age | 1.116351 .0403379 3.05 0.002 1.040025 1.198279 ------------------------------------------------------------------------------. stcurve, hazard

The command stcurve is using kernel density estimation to perform the smoothing we referred to above. We can do this by hand using the baseline hazard contributions and the command kdensity to perform the smoothing:

. sysuse cancer(Patient Survival in Drug Trial). stset studytime, failure(died)failure event: died != 0 & died < . obs. time interval: (0, studytime] exit on or before: failure ------------------------------------------------------------------------------ 48 total obs. 0 exclusions ------------------------------------------------------------------------------ 48 obs. remaining, representing 31 failures in single record/single failure data 744 total analysis time at risk, at risk from t = 0 earliest observed entry t = 0 last observed exit t = 39. stcox drug age, nologfailure _d: died analysis time _t: studytime Cox regression -- Breslow method for ties No. of subjects = 48 Number of obs = 48 No. of failures = 31 Time at risk = 744 LR chi2(2) = 36.29 Log likelihood = -81.765061 Prob > chi2 = 0.0000 ------------------------------------------------------------------------------ _t | Haz. Ratio Std. Err. z P>|z| [95% Conf. Interval] -------------+---------------------------------------------------------------- drug | .2153648 .0676904 -4.89 0.000 .1163154 .3987605 age | 1.116351 .0403379 3.05 0.002 1.040025 1.198279 ------------------------------------------------------------------------------. predict hc0, basehc(17 missing values generated). sum drugVariable | Obs Mean Std. Dev. Min Max -------------+-------------------------------------------------------- drug | 48 1.875 .8410986 1 3. replace drug=r(mean)drug was int now float (48 real changes made). sum ageVariable | Obs Mean Std. Dev. Min Max -------------+-------------------------------------------------------- age | 48 55.875 5.659205 47 67. replace age=r(mean)age was int now float (48 real changes made). predict double xb, xb . gen double hcmean = (1-(1-hc0)^exp(xb))(17 missing values generated). drop if hc0==.(17 observations deleted). sort _t . by _t: keep if _n==1(10 observations deleted). summ _t, meanonly . local tmin = r(min) . local tmax = r(max) . local N = _N . local N1 = `N' + 1 . local obs = `N'+101 . set obs `obs'obs was 21, now 122. gen t0 = `tmin' + (`tmax'-`tmin')*(_n-`N1')/100 /// in `N1'/l(21 missing values generated). gen t1 = t0 if t0>=4.62 & t0<=28.38(48 missing values generated). kdensity _t [iweight=hcmean] if _d, at(t1) generate(hmean) nograph . twoway line hmean t1, ytitle("") /// xtitle("analysis time") /// title("Smoothed hazard estimate")

We can see that stcurve is doing a lot of work for us. First, it obtains the means of the covariates and calculates the hazard contributions at the mean. Next, it creates 101 equally spaced time points at which to calculate the smoothed hazard estimate. Finally, it uses kdensity to do the smoothing.

- Kalbfleisch, J. D., and R. L. Prentice. 2002.
*The Statistical Analysis of Failure Time Data. 2nd ed*. New York: Wiley.