Re: [R] predict.lm

From: Prof Brian Ripley <ripley_at_stats.ox.ac.uk>
Date: Wed 03 May 2006 - 05:54:08 EST

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@stat.math.ethz.ch
> [mailto:r-help-bounces@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@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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Received on Wed May 03 06:04:21 2006

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