[R] comparing lm(), survreg( ... , dist="gaussian") and survreg( ... , dist="lognormal")
Prof Brian Ripley
ripley at stats.ox.ac.uk
Tue May 3 08:54:39 CEST 2005
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 $se.fit for the dist="lognormal" are quite obviously different
> from the other two methods, and although I have tried everything I could
> imagine I cannot reconcile the differences.
How did you do this? (BTW, I assume you mean upper and lower confidence
>limits< for the predicted means.) For the predictions and standard
errors are (or should be) on the response scale, a non-linear function of
the parameters. In that case it is normal to form confidence limits on
the linear predictor scale and transform.
> I believe that the confidence bounds for both models should agree. After
> all, both calls to survreg() produce identical parameter estimates.
They will, if computed on the same basis. On log-scale (to avoid large
numbers)
pr1 <- predict(lognormal.survreg, se.fit=T)
log(cbind(pr1$fit - 1.96*pr1$se.fit, pr1$fit + 1.96*pr1$se.fit))
pr2 <- predict(gaussian.survreg, se.fit=T)
cbind(pr2$fit - 1.96*pr2$se.fit, pr2$fit + 1.96*pr2$se.fit)
are really pretty close. The main difference is a slight shift, which
comes about because the mean of a log(X) is not log(mean(X)). Note that
the second set at the preferred ones. Transforming to log scale before
making the confidence limits:
cbind(log(pr1$fit) - 1.96*pr1$se.fit/pr1$fit, log(pr1$fit) + 1.96*pr1$se.fit/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, ripley at stats.ox.ac.uk
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UK Fax: +44 1865 272595
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