[R] Cox Regression : Spline Coefficient Interpretation?
Bert Gunter
bgunter.4567 at gmail.com
Wed Nov 1 23:11:08 CET 2017
??
It is unclear to me what "How to interpret the result" means. Note that the
survival package is very well documented and there is a vignette
specifically on the topic of the use of "Spline terms in a Cox model." Have
you studied it?
If you want to discuss the statistical issues, e.g. of survival modeling
or the technical details of penalized smoothing splines, that is mostly OT
here: stats.stackexchange.com would probably be a better place to post for
that. This list is mostly about R programming rather than statistics,
although they do sometimes intersect.
If I have misunderstood your question, you might wish to clarify exactly
what it is that you are seeking in another post.
Finally, as you can see from the below, post in PLAIN TEXT ONLY, as html
can get mangled by the server on this plain text mailing list.
Cheers,
Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along and
sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Wed, Nov 1, 2017 at 1:12 PM, Kosta S. <kosmirnov at gmail.com> wrote:
> Hi,
>
> I'm using a Cox-Regression to estimate hazard rates on prepayments.
>
> I'm using the "pspline" function to face non-linearity, but I have no clue
> how to interpret the result.
> Unfortunately I did not find enough information on the "pspline" function
> wether in the survival package nor using google..
>
> I got following output:
>
> * library(survival)*
>
>
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> > *> Option.test2<-coxph(Surv(START,STOP,ZEROBAL==1)~pspline(OPTION),
> > data=FNMA)coxph(formula = Surv(START, STOP, ZEROBAL == 1) ~
> > pspline(OPTION), data = FNMA)> > Option.test2> Call:> coxph(formula =
> > Surv(START, STOP, ZEROBAL == 1) ~ pspline(OPTION), > data = FNMA)>
> > coef se(coef) se2 Chisq DF
> > p> pspline(OPTION), linear -0.1334 0.0131 0.0131
> 104.4325
> > 1.00 <0.0000000000000002> pspline(OPTION), nonlin
> > 1747.1295 3.05 <0.0000000000000002> Iterations: 8 outer, 19
> > Newton-Raphson> Theta= 0.991 > Degrees of freedom for terms= 4 >
> > Likelihood ratio test=2136 on 4.05 df, p=0 n= 3390429 > *
>
>
> Thanks,
>
> KS
>
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>
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