# Re: [R] Weighted least squares

From: John Fox <jfox_at_mcmaster.ca>
Date: Tue, 08 May 2007 11:19:16 -0400

I think that the problem is that the term "weights" has different meanings, which, although they are related, are not quite the same.

The weights used by lm() are (inverse-)"variance weights," reflecting the variances of the errors, with observations that have low-variance errors therefore being accorded greater weight in the resulting WLS regression. What you have are sometimes called "case weights," and I'm unaware of a general way of handling them in R, although you could regenerate the unaggregated data. As you discovered, you get the same coefficients with case weights as with variance weights, but different standard errors. Finally, there are "sampling weights," which are inversely proportional to the probability of selection; these are accommodated by the survey package.

To complicate matters, this terminology isn't entirely standard.

I hope this helps,
John

John Fox, Professor
Department of Sociology
McMaster University
Hamilton, Ontario
905-525-9140x23604
http://socserv.mcmaster.ca/jfox

> -----Original Message-----
> From: r-help-bounces_at_stat.math.ethz.ch
> [mailto:r-help-bounces_at_stat.math.ethz.ch] On Behalf Of hadley wickham
> Sent: Tuesday, May 08, 2007 5:09 AM
> To: R Help
> Subject: [R] Weighted least squares
>
> Dear all,
>
> I'm struggling with weighted least squares, where something
> that I had assumed to be true appears not to be the case.
> Take the following data set as an example:
>
> df <- data.frame(x = runif(100, 0, 100)) df\$y <- df\$x + 1 +
> rnorm(100, sd=15)
>
>
> summary(lm(y ~ x, data=df, weights=rep(2, 100))) summary(lm(y
> ~ x, data=rbind(df,df)))
>
> would be equivalent, but they are not. I suspect the
> difference is how the degrees of freedom is calculated - I
> had expected it to be sum(weights), but seems to be
> sum(weights > 0). This seems unintuitive to me:
>
> summary(lm(y ~ x, data=df, weights=rep(c(0,2), each=50)))
> summary(lm(y ~ x, data=df, weights=rep(c(0.01,2), each=50)))
>
> What am I missing? And what is the usual way to do a linear
> regression when you have aggregated data?
>
> Thanks,
>
>
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