Re: [R] comparing lm(), survreg( ... , dist="gaussian") and survreg( ... , dist="lognormal")

From: Prof Brian Ripley <>
Date: Tue 03 May 2005 - 16:54:39 EST

On Mon, 2 May 2005, Charles Annis, P.E. wrote:

> I have tried everything I can think of and hope not to appear too foolish
> when my error is pointed out to me.
> I have some real data (18 points) that look linear on a log-log plot so I
> used them for a comparison of lm() and survreg. There are no suspensions.
> survreg.df <- data.frame(Cycles=c(2009000, 577000, 145000, 376000, 37000,
> 979000, 17420000, 71065000, 46397000, 70168000, 69120000, 68798000,
> 72615000, 133051000, 38384000, 15204000, 1558000, 14181000), stress=c(90,
> 100, 110, 90, 100, 80, 70, 60, 56, 62, 62, 59, 56, 53, 59, 70, 90, 70),
> event=rep(1, 18))
> sN.lm<- lm(log(Cycles) ~ log10(stress), data=survreg.df)
> and
> vvvvvvvvvvv
> gaussian.survreg<- survreg(formula=Surv(time=log(Cycles), event) ~
> log10(stress), dist="gaussian", data=survreg.df)
> produce identical parameter estimates and differ slightly in the residual
> standard error and scale, which is accounted for by scale being the MLE and
> thus biased. Correcting by sqrt(18/16) produces agreement. Using predict()
> for the lm, and predict.survreg() for the survreg model and correcting for
> the differences in stdev, produces identical plots of the fit and the upper
> and lower confidence intervals. All of this is as it should be.

I trust you called predict() on both and let R choose the method.

> And,
> vvvvvv
> lognormal.survreg<- survreg(formula=Surv(time=(Cycles), event) ~
> log10(stress), dist="lognormal", data=survreg.df)
> produces summary() results that are identical to the earlier call to
> survreg(), except for the call, of course. The parameter estimates and SE
> are identical. Again this is as I would expect it.
> But since the call uses Cycles, rather than log(Cycles) predict.survreg()
> returns $fit in Cycles units, rather than logs, and of course the fits are
> identical when plotted on a log-log grid and also agree with lm()
> Here is the fly in the ointment: The upper and lower confidence intervals,
> based on the $ for the dist="lognormal" are quite obviously different
> from the other two methods, and although I have tried everything I could
> imagine I cannot reconcile the differences.

How did you do this? (BTW, I assume you mean upper and lower confidence
>limits< for the predicted means.) For the predictions and standard
errors are (or should be) on the response scale, a non-linear function of the parameters. In that case it is normal to form confidence limits on the linear predictor scale and transform.

> I believe that the confidence bounds for both models should agree. After
> all, both calls to survreg() produce identical parameter estimates.

They will, if computed on the same basis. On log-scale (to avoid large numbers)

pr1 <- predict(lognormal.survreg, log(cbind(pr1$fit - 1.96*pr1$, pr1$fit + 1.96*pr1$ pr2 <- predict(gaussian.survreg, cbind(pr2$fit - 1.96*pr2$, pr2$fit + 1.96*pr2$

are really pretty close. The main difference is a slight shift, which comes about because the mean of a log(X) is not log(mean(X)). Note that the second set at the preferred ones. Transforming to log scale before making the confidence limits:

cbind(log(pr1$fit) - 1.96*pr1$$fit, log(pr1$fit) + 1.96*pr1$$fit)

does give identical answers.

Consider care is needed in interpreting what predict() is actually predicting in non-linear models. For both glm() and survreg() it is closer to the median of the uncertainty in the predictions than to the mean.

Brian D. Ripley,        
Professor of Applied Statistics,
University of Oxford,             Tel:  +44 1865 272861 (self)
1 South Parks Road,                     +44 1865 272866 (PA)
Oxford OX1 3TG, UK                Fax:  +44 1865 272595

______________________________________________ mailing list
PLEASE do read the posting guide!
Received on Tue May 03 17:02:57 2005

This archive was generated by hypermail 2.1.8 : Fri 03 Mar 2006 - 03:31:32 EST