Re: [R] Weighted least squares

From: hadley wickham <h.wickham_at_gmail.com>
Date: Wed, 09 May 2007 08:22:26 +0200

On 5/9/07, Adaikalavan Ramasamy <ramasamy_at_cancer.org.uk> wrote:
> http://en.wikipedia.org/wiki/Weighted_least_squares gives a formulaic
> description of what you have said.

Except it doesn't describe what I think is important in my case - how do you calculate the degrees of freedom/n for weighted linear regression?

> I believe the original poster has converted something like this
>
> y x
> 0 1.1
> 0 2.2
> 0 2.2
> 0 2.2
> 1 3.3
> 1 3.3
> 2 4.4
> ...
>
> into something like the following
>
> y x freq
> 0 1.1 1
> 0 2.2 3
> 1 3.3 2
> 2 4.4 1
> ...

Exactly! Thanks for providing that example.

>
> Now, the variance of means of each row in table above is ZERO because
> the individual elements that comprise each row are identical. Therefore
> your method of using inverse-variance will not work here.
>
> Then is it valid then to use lm( y ~ x, weights=freq ) ?
>
> Regards, Adai
>
>
>
> S Ellison wrote:
> > Hadley,
> >
> > You asked
> >> .. what is the usual way to do a linear
> >> regression when you have aggregated data?
> >
> > Least squares generally uses inverse variance weighting. For aggregated data fitted as mean values, you just need the variances for the _means_.
> >
> > So if you have individual means x_i and sd's s_i that arise from aggregated data with n_i observations in group i, the natural weighting is by inverse squared standard error of the mean. The appropriate weight for x_i would then be n_i/(s_i^2). In R, that's n/(s^2), as n and s would be vectors with the same length as x. If all the groups had the same variance, or nearly so, s is a scalar; if they have the same number of observations, n is a scalar.
> >
> > Of course, if they have the same variance and same number of observations, they all have the same weight and you needn't weight them at all: see previous posting!
> >
> > Steve E
> >
> >
> >
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> >
>
>



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