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I am trying to fit a coxph model in R. The study can be described as follows: I have a very large dataset, in counting process form, containing whether or not someone responded to a survey or not. The time variable is the consecutive number of months in which a response was received. The model predicts non-response, i.e. when someone does not respond. There are more than one record per id (itemcode in my dataset). There are no continuous covariates as of now. What is included in the model are seasonal effects--I would like to know how each season increases or lowers the hazard of non-response, relative to fall. The I have already stratified the model. The results are below:

Call:
coxph(formula = Surv(start, cum.goodp, dlq.next) ~ winter + spring + 
summer + strata(sector) + cluster(itemcode), data = nr.sample.split)

n= 651033, number of events= 42508 

       coef exp(coef) se(coef) robust se      z Pr(>|z|)    
winter  0.26850   1.30800  0.01307   0.01283  20.92   <2e-16 ***
spring -0.64040   0.52708  0.01385   0.01342 -47.72   <2e-16 ***
summer  0.29188   1.33894  0.01414   0.01284  22.73   <2e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

   exp(coef) exp(-coef) lower .95 upper .95
winter    1.3080     0.7645    1.2755    1.3413
spring    0.5271     1.8972    0.5134    0.5411
summer    1.3389     0.7469    1.3057    1.3731

Concordance= 0.598  (se = 0.004 )
Rsquare= 0.009   (max possible= 0.636 )
Likelihood ratio test= 5864  on 3 df,   p=0
Wald test            = 4783  on 3 df,   p=0
Score (logrank) test = 5634  on 3 df,   p=0,   Robust = 5015  p=0

 (Note: the likelihood ratio and score tests assume independence of
 observations within a cluster, the Wald and robust score tests do not).

I then estimated a cox.zph function to test the PH assumption, the results of which are below:

           rho   chisq       p
winter -0.1283  691.45 0.00000
spring -0.1151  569.35 0.00000
summer -0.0163    9.36 0.00221
GLOBAL      NA 1096.18 0.00000

Clearly, the PH assumption is not valid for any of the coefficients. Below is a plot of one, summer:

[![Plot of Beta(t) for coefficient "summer"][1]][1]

My question is: since the seasonal dummy variables are static by nature, and their coefficients clearly vary with the time variable, how much does it matter? I get that statistically, it means something, but does the violation of PH assumption invalidate the (intuitively appealing) result that non-response is more likely to happen in the summer and winter? If so, is there a way to handle this so that the PH assumption is not violated? I know about using the tt transform, but I can't seem to figure out the exact form for the function. Any advice, ideas or references would be greatly appreciated.

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  • $\begingroup$ Let me see if I get this straight... your "event" variable here is "nonresponse" which I hope to mean "response", which you have measured time-to-event, e.g. time until responding. And each participant who swings from one season to the next is censored at the end of the season and re-entered into the next season at time 0. My question to you is: what on earth is time zero anyway? No wonder your proportional hazards test is so off. $\endgroup$
    – AdamO
    Apr 5, 2016 at 20:33
  • $\begingroup$ @AdamO...no, the event is non-response as I have said. The event is the incidence of non-response. I am interested in the number of consecutive responses until someone stops responding. Censoring occurs as it normally does. I get similar results when the data is not in counting process form as well. Non-proportional hazards. the seasonal dummies represent the month, e.g. winter is december - feb. Participants don't swing from season to season, time moves cumulatively. $\endgroup$
    – jvalenti
    Apr 5, 2016 at 21:10
  • $\begingroup$ Can you explain any context or background to the problem? Nonresponse to a cancer treatment e.g. means that the endpoint has not been observed yet $\endgroup$
    – AdamO
    Apr 5, 2016 at 21:12
  • $\begingroup$ @AdamO it's not a cancer treatment, it's a survey that is mailed out. they either mail back an answer to the question or they don't. I have data for 60,000+ participants, if you will, measured over sixty months. Time is measured in consecutive responses, i.e. the cumulative sum of uninterrupted responses. Sorry if this wasn't clear. $\endgroup$
    – jvalenti
    Apr 5, 2016 at 21:16
  • $\begingroup$ Then how do you know when they do NOT respond to your survey? The outcome should be response. That makes no sense. $\endgroup$
    – AdamO
    Apr 5, 2016 at 21:18

1 Answer 1

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Since the seasonal dummy variables are static by nature, and their coefficients clearly vary with the time variable, how much does it matter?

The value that you get is a form of average over time. Unfortunately as you naturally have more cases in time-to-event analyses early on, you can't simply say that the effect is balanced throughout time. From my experience the initial period has a much heavier impact on the estimate than the later and it

I get that statistically, it means something, but does the violation of PH assumption invalidate the (intuitively appealing) result that non-response is more likely to happen in the summer and winter?

Again, it probably doesn't but you can't be sure until you've checked. Something is definitely happening over time and you should at least have a look at the residual plot (plot(cox.zph(...))). It isn't entirely surprising that you have a problem with the PH since seasons are part of the time variable and there will be situations where early summer and late spring are similar.

If so, is there a way to handle this so that the PH assumption is not violated? I know about using the tt transform, but I can't seem to figure out the exact form for the function.

The tt transform is tricky to use with big data. It explodes the matrix and can get a little messy, e.g. if you modify the lung example in the survival package:

library(survival)
coxph(Surv(time, status) ~ ph.ecog + tt(age), data=lung,
      tt=function(x,t,...) {
        print(length(x))
        pspline(x + t/365.25)
      })

It prints 15809 while there are only 228 rows in the original dataset. The principle of the tt() is that it feeds the variables into the transformation function where you are free to use time any way you wish. Note that you can also have different transformation functions depending for each variable:

library(survival)
coxph(Surv(time, status) ~ tt(ph.ecog) + tt(age), data=lung,
      tt=list(
        function(x,t,...) {
          cbind(x, x + t/365.25, (x + t/365.25)^2)
        },
        function(x,t,...) {
          pspline(x + t/365.25)
        }),
      x=T,
      y=T) -> fit
head(fit$x)

Gives:

    tt(ph.ecog)x tt(ph.ecog) tt(ph.ecog) tt(age)1 tt(age)2 tt(age)3 tt(age)4
6              1         3.4        11.7        0        0        0    0.000
3              0         2.4         5.8        0        0        0    0.020
38             1         3.4        11.7        0        0        0    0.000
5              0         2.4         5.8        0        0        0    0.000
6.1            1         3.2        10.4        0        0        0    0.000
3.1            0         2.2         5.0        0        0        0    0.026
    tt(age)5 tt(age)6 tt(age)7 tt(age)8 tt(age)9 tt(age)10 tt(age)11 tt(age)12
6       0.00  0.00000    0.000   0.0052    0.359      0.58     0.053         0
3       0.48  0.48232    0.021   0.0000    0.000      0.00     0.000         0
38      0.00  0.00087    0.266   0.6393    0.094      0.00     0.000         0
5       0.03  0.51933    0.437   0.0136    0.000      0.00     0.000         0
6.1     0.00  0.00000    0.000   0.0078    0.388      0.56     0.044         0
3.1     0.50  0.45457    0.016   0.0000    0.000      0.00     0.000         0

I therefore try to avoid this solution and use the time-split approach that I wrote about here and in my answer to my own question.

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  • $\begingroup$ doesn't stratification allow hazards to be non proportional? The plots show a fairly systematic pattern, for sure, but i read somewhere (see here: statisticalhorizons.com/wp-content/uploads/2012/01/…) on p. 421, section 15, that stratification is a way around the PH problem. Is this true in your experience? $\endgroup$
    – jvalenti
    Apr 5, 2016 at 21:09
  • $\begingroup$ @jvalenti: Stratification allows a confounder to be non-proportional. It splits the calculations into the strata and then recombines them at the end. This takes care of the non-PH but you wont get an estimate for that confounder, hence you can't use that for dealing with non-PH for your primary outcome. A practical problem that I've found with stratification is that you sometimes have a strata that doesn't allow estimation for one of the model variables due to too few observation within that strata - this causes the entire approach to fail. $\endgroup$
    – Max Gordon
    Apr 6, 2016 at 4:04

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