From: Hanke, Alex <HankeA_at_mar.dfo-mpo.gc.ca>

Date: Sat 11 Jun 2005 - 03:53:05 EST

R-help@stat.math.ethz.ch mailing list

https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html Received on Sat Jun 11 04:20:31 2005

Date: Sat 11 Jun 2005 - 03:53:05 EST

I am trying to back out the values for the baseline hazard, h_o(t_i), for
each event time or observation time.

Now survfit(fit)$surv gives me the value of the survival function,
S(t_i|X_i,B), using mean values of the covariates and the coxph() object
provides me with the estimate of the linear predictors, exp(X'B).
If S(t_i|X_i,B)=S_o(t_i)^exp(X_iB) is the expression for the survival
function

And

-ln(S_o(t_i) ) is the expression for the cumulative baseline hazard

function, H_o(t_i)

Then by rearranging the expression for the survival function I get the
following:

-ln(S_o(t_i) ) = -ln( S(t_i|X_i,B) ) / exp(X_iB)

- basehaz(fit)/exp(fit$linear.predictors) Am I right so far and is there an easier way? The plot of the cumulative baseline hazard function , H_o(t_i), should be linear across time. Once I have, H_o(t_i), to get at h_o(t_i) I then need to reverse the cumsum operation. The corresponding plot should have a constant baseline hazard over time.

I am aware of cox.zph() for testing the proportionality of hazards assumption.

Thanks

Alex

Alex Hanke

Department of Fisheries and Oceans

St. Andrews Biological Station

531 Brandy Cove Road

St. Andrews, NB

Canada

E5B 2L9

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R-help@stat.math.ethz.ch mailing list

https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html Received on Sat Jun 11 04:20:31 2005

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