## Background

In the previous tutorial I used `stpm2_standsurv`

to obtain standardized survival functions. In this tutorial I show the first of a
number of different measures of the standardized survival function where I obtain centiles of the standardized survival function.

As a reminder a centile of a survival function can be obtained by solving $S(t) = \alpha$ for $t$. For example, for the median survival time we set $\alpha = 0.5$, i.e. the 50th (per)centile. For simple parametric distributions, such as the Weibull, we can solve for $t$ analytically, but for more complex models the centile is obtained through iterative root finding techniques. In `stpm2`

I have used Brent’s root finder when evaluating centiles.

The centile of a standardized survival function is obtained by solving the following equation for t.

$$ E\left(S(t | X=x,Z\right) = \alpha $$

This is done through root finding (using Brent’s root finder) by solving,

$$ \frac{1}{N}\sum_{i=1}^N {S(t | X=x,Z)} - \alpha = 0 $$

Variances can be obtained using M-estimation .

## Example

I use a colon cancer example. I first load and `stset`

the data

```
. //use https://www.pclambert.net/data/colon, clear
. use c:/cansurv/data/colon, clear
(Colon carcinoma, diagnosed 1975-94, follow-up to 1995)
. drop if stage==0
(2,356 observations deleted)
. stset surv_mm, f(status=1,2) scale(12) exit(time 120)
failure event: status == 1 2
obs. time interval: (0, surv_mm]
exit on or before: time 120
t for analysis: time/12
------------------------------------------------------------------------------
13,208 total observations
0 exclusions
------------------------------------------------------------------------------
13,208 observations remaining, representing
8,866 failures in single-record/single-failure data
43,950.667 total analysis time at risk and under observation
at risk from t = 0
earliest observed entry t = 0
last observed exit t = 10
```

I drop those with missing stage information (`stage == 0`

). I am investigating all cause survival (`status=1,2`

).

I fit a model that includes stage, sex and age (using a rectricted cubic splines). I assume proportional hazards, but if I relax this assusmption the syntax for `stpm2_standsurv`

would be identical. Stage is classified as localised, regional and distant and is modelled using two dummy covariates with localised as the reference category.

```
. tab stage, gen(stage)
Clinical |
stage at |
diagnosis | Freq. Percent Cum.
------------+-----------------------------------
Localised | 6,274 47.50 47.50
Regional | 1,787 13.53 61.03
Distant | 5,147 38.97 100.00
------------+-----------------------------------
Total | 13,208 100.00
. gen female = sex==2
. rcsgen age, df(3) gen(agercs) center(60)
Variables agercs1 to agercs3 were created
. stpm2 stage2 stage3 female agercs*, scale(hazard) df(4) nolog eform
Log likelihood = -19665.932 Number of obs = 13,208
------------------------------------------------------------------------------
| exp(b) Std. Err. z P>|z| [95% Conf. Interval]
-------------+----------------------------------------------------------------
xb |
stage2 | 1.758926 .0617187 16.09 0.000 1.642026 1.884149
stage3 | 5.656322 .1390555 70.48 0.000 5.39024 5.935539
female | .8634681 .01891 -6.70 0.000 .8271894 .901338
agercs1 | .9958187 .0053355 -0.78 0.434 .9854161 1.006331
agercs2 | 1.000015 .0000127 1.22 0.223 .9999906 1.00004
agercs3 | .9999654 .0000155 -2.24 0.025 .9999351 .9999957
_rcs1 | 3.484444 .0392205 110.90 0.000 3.408415 3.562169
_rcs2 | 1.1908 .0101285 20.53 0.000 1.171113 1.210817
_rcs3 | .9620673 .0050498 -7.37 0.000 .9522207 .9720158
_rcs4 | 1.015331 .003491 4.43 0.000 1.008512 1.022197
_cons | .1615451 .0046341 -63.55 0.000 .1527132 .1708879
------------------------------------------------------------------------------
Note: Estimates are transformed only in the first equation.
```

The is a clear effect of stage with a hazard ratio of 5.66 for distant stage versus localised stage. Remember that I am modelling all cause survival and one would expect a cause-specific hazard ratio to be higher. The all-cause mortality rate for females is 14% lower than males.

I will now predict the standardized survival functions, one where I force all subjects to be male and one where I force everyone to be female.

```
. range tt 0 10 100
(13,108 missing values generated)
. stpm2_standsurv, at1(female 0) at2(female 1) timevar(tt) atvar(ms_male ms_female) ci
. twoway (line ms_male ms_female tt, sort) ///
> , yline(0.5, lpattern(dash) lcolor(black)) ///
> yline(0.5, lpattern(dash) lcolor(black)) ///
> xtitle("Years since diagnosis") ///
> ytitle("S(t)", angle(h)) ///
> ylabel(0(0.2)1, format(%3.1f) angle(h)) ///
> legend(order(1 "Male" 2 "Female") ring(0) pos(1) cols(1))
```

The graph of the two standardised survival functions can be seen below.

As expected (given the hazard ratio) females have better survival than males. I have added a horizontal reference line at $S(t)=0.5$. Where this line crosses the survival curves gives the median survival time. Reading from the graph, this is just under 2 years for the males and just under 2.5 years for females. Using the `centile`

option of `stpm2_standsurv`

will estimate these values more accurately with 95% confidenec intervals. We are also interested in contrasts of the centiles, so use of the `contrast`

option will calculate either a difference or ratio of the median survival times with a 95% confidence interval.

```
. stpm2_standsurv, at1(female 0) at2(female 1) timevar(tt) centile(50) ///
> atvar(med_male med_female) contrast(difference) ci
. list med_male* in 1, ab(15)
+-----------------------------------------+
| med_male med_male_lci med_male_uci |
|-----------------------------------------|
1. | 1.9801987 1.8728516 2.0936987 |
+-----------------------------------------+
. list med_female* in 1, ab(15)
+----------------------------------------------+
| med_female med_female_lci med_female_uci |
|----------------------------------------------|
1. | 2.4249751 2.3000843 2.5566472 |
+----------------------------------------------+
. list _contrast* in 1, ab(18)
+----------------------------------------------------+
| _contrast2_1 _contrast2_1_lci _contrast2_1_uci |
|----------------------------------------------------|
1. | .44477636 .30074772 .588805 |
+----------------------------------------------------+
```

The median survival time is 1.98 years for males with a 95% CI (1.87 to 2.09). The median for females is 2.42 years (95% CI, 2.30 to 2.56). As I used the `contrast`

option I also get the difference in the median of the standardised survival curves with a 95% CI. Thus the time at which 50% of females have died is 0.44 years more than the time at which 50% of males have died, 95% CI (0.30 to 0.59).

It is possible to predict for multiple centiles by passing a *numlist* to the `centiles`

option. For example, the code below calculates centiles between 10 and 60 at 10 unit intervals.

```
. stpm2_standsurv, at1(female 0) at2(female 1) timevar(tt) centile(10(10)60) ///
> atvar(cen_males cen_females) contrast(difference) ci ///
> centvar(centiles) contrastvar(cendiff)
. list centile cen_males cen_females cendiff in 1/6, sep(0) noobs
+----------------------------------------------+
| centiles cen_males cen_fem~s cendiff |
|----------------------------------------------|
| 10 .14919254 .16940399 .02021145 |
| 20 .32365079 .38459929 .0609485 |
| 30 .63821705 .78393263 .14571558 |
| 40 1.1648032 1.4192444 .25444124 |
| 50 1.9801987 2.4249751 .44477636 |
| 60 3.4239223 4.277573 .85365069 |
+----------------------------------------------+
```

We can then plot the difference in these various centiles.

```
. twoway (rarea cendiff_lci cendiff_uci centile, sort color(red%30)) ///
> (line cendiff centile, color(black)) ///
> , xtitle(centile) xlabel(,format(%3.0f)) ///
> ytitle("Difference in centile") ylabel(0(0.2)1.2,format(%3.1f) angle(h)) ///
> legend(off)
```

There are probably more innovative ways of presenting such data.

## Acknowledgement

I would like to acknowledge David Druker of StataCorp who I discussed these ideas with at two Nordic Stata User group meetings. David has written a command that estimates centiles of standardized distributions using a two parameter gamma distribution which is available here.

## References

Stefanski, L. & Boos, D. The Calculus of M-Estimation. *The American Statistician* 2002;**56**:29-38