[R] predict.lm

Prof Brian Ripley ripley at stats.ox.ac.uk
Tue May 2 21:54:08 CEST 2006

On Tue, 2 May 2006, Christos Hatzis wrote:

> I think you got it right.
> The mean of the (weighted) sum of a set of random variables is the
> (weighted) sum of the means and its variance is the (weighted) sum of the
> individual variances (using squared weights).  Here you don't have to worry
> about weights.
> So what you proposed does exactly this.

Yes, but the theory has assumptions which are not met here: the random 
variables are correlated (in almost all case).

> -Christos
> -----Original Message-----
> From: r-help-bounces at stat.math.ethz.ch
> [mailto:r-help-bounces at stat.math.ethz.ch] On Behalf Of Bill Szkotnicki
> Sent: Tuesday, May 02, 2006 2:59 PM
> To: 'R-Help help'
> Subject: [R] predict.lm
> I have a model with a few correlated explanatory variables.
> i.e.
>> m1=lm(y~x1+x2+x3+x4,protdata)
> and I have used predict as follows:
>> x=data.frame(x=1:36)
>> yp=predict(m1,x,se.fit=T)

How can this work?  You fitted the model to x1...x4 and supplied x.

>> tprot=sum(yp$fit) # add up the predictions tprot
> tprot is the sum of the 36 predicted values and I would like the se of that
> prediction.
> I think
>> sqrt(sum(yp$se.fit^2))
> is not correct.
> Would anyone know the correct approach?
> i.e. How to get the se of a function of predicted values (in this case sum)

You need to go back to the theory: it is easy to do for a linear function, 
otherwise you will need to linearize.

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

More information about the R-help mailing list