Thank you Terry for the explanation!
John
________________________________
From: "Therneau, Terry M., Ph.D." <thern...@mayo.edu>
Sent: Monday, June 30, 2014 6:04 AM
Subject: Re: standard error of survfit.coxph()
1. The computations "behind the scenes" produce the variance of the cumulative
hazard.
This is true for both an ordinary Kaplan-Meier and a Cox model.
Transformations to other
scales are done using simple Taylor series.
H = cumulative hazard = log(S); S=survival
var(H) = var(log(S)) = the starting point
S = exp(log(S)), so var(S) is approx [deriv of exp(x)]^2 * var(log(S)) =
S^2 var(H)
var(log(log(S)) is approx (1/S^2) var(H)
2. At the time it was written, summary.survfit was used only for printing out
the survival
curve at selected times, and the audience for the printout wanted std(S).
True, that was
20 years ago, but I don't recall anyone ever asking for summary to do anything
else. Your
request is not a bad idea.
Note however that the primary impact of using log(S) or S or log(log(S))
scale is is on
the confidence intervals, and they do appear per request in the summary output.
Terry T.
On 06/28/2014 05:00 AM, r-help-requ...@r-project.org wrote:
> Message: 9
> Date: Fri, 27 Jun 2014 12:39:29 -0700
> To:"r-help@r-project.org" <r-help@r-project.org>
> Subject: [R] standard error of survfit.coxph()
> Message-ID:
> <1403897969.91269.yahoomail...@web122906.mail.ne1.yahoo.com>
> Content-Type: text/plain
>
> Hi, can anyone help me to understand the standard errors printed in the
> output of survfit.coxph()?
>
> time<-sample(1:15,100,replace=T)
>
> status<-as.numeric(runif(100,0,1)<0.2)
> x<-rnorm(100,10,2)
>
> fit<-coxph(Surv(time,status)~x)
> ??? ### method 1
>
> survfit(fit, newdata=data.frame(time=time,status=status,x=x)[1:5,],
> conf.type='log')$std.err
>
> ???????????? [,1]??????? [,2]??????? [,3]??????? [,4]?????? [,5]
> ?[1,] 0.000000000 0.000000000 0.000000000 0.000000000 0.00000000
> ?[2,] 0.008627644 0.008567253 0.008773699 0.009354788 0.01481819
> ?[3,] 0.008627644 0.008567253 0.008773699 0.009354788 0.01481819
> ?[4,] 0.013800603 0.013767977 0.013889971 0.014379928 0.02353371
> ?[5,] 0.013800603 0.013767977 0.013889971 0.014379928 0.02353371
> ?[6,] 0.013800603 0.013767977 0.013889971 0.014379928 0.02353371
> ?[7,] 0.030226811 0.030423883 0.029806263 0.028918817 0.05191161
> ?[8,] 0.030226811 0.030423883 0.029806263 0.028918817 0.05191161
> ?[9,] 0.036852571 0.037159980 0.036186931 0.034645002 0.06485394
> [10,] 0.044181716 0.044621159 0.043221145 0.040872939 0.07931028
> [11,] 0.044181716 0.044621159 0.043221145 0.040872939 0.07931028
> [12,] 0.055452631 0.056018832 0.054236881 0.051586391 0.10800413
> [13,] 0.070665160 0.071363749 0.069208056 0.066655730 0.14976433
> [14,] 0.124140400 0.125564637 0.121281571 0.118002021 0.30971860
> [15,] 0.173132357 0.175309455 0.168821266 0.164860523 0.46393111
>
> survfit(fit, newdata=data.frame(time=time,status=status,x=x)[1:5,],
> conf.type='log')$time
> ?[1]? 1? 2? 3? 4? 5? 6? 7? 8? 9 10 11 12 13 14 15
>
> ??? ### method 2
>
> summary(survfit(fit, newdata=data.frame(time=time,status=status,x=x)[1:5,],
> conf.type='log'),time=10)$std.err
>
> ????????????? 1????????? 2????????? 3????????? 4????????? 5
> [1,] 0.04061384 0.04106186 0.03963184 0.03715246 0.06867532
>
> By reading the help of ?survfit.object and ?summary.survfit, the standard
> error provided in the output of method 1 (survfit()) was for cumulative
> hazard-log(survival), while the standard error provided in the output of
> method 2 (summary.survfit()) was for survival itself, regardless of how you
> choose the value for "conf.type" ('log', 'log-log' or 'plain'). This explains
> why the standard error output is different between method 1 (10th row) and
> method 2.
>
> My question is how do I get standard error estimates for log(-log(survival))?
>
> Thanks!
>
> John
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