Re: Comparing predictors

From: Ross Clement (clemenr_at_wmin.ac.uk)
Date: 06/26/04 

Date: 25 Jun 2004 17:00:26 -0700

clemenr@wmin.ac.uk (Ross Clement) wrote in message news:<e5b5d377.0406250709.4603faaa@posting.google.com>...
> Lionel Barnett <mail@lionelb.com> wrote in message news:<2k24skF16dfdnU1@uni-berlin.de>...
> > The figures are:
> >
> > corr(Y',Y'') = 0.8851
> > corr(Y'-Y,Y''-Y) = 0.9991 (!)
>
> I must say that I agree with the addition of the exclamation mark to
> your figures. I'm not quite sure how you get a correlation for the
> error so much higher than the correlation of the raw values. By that I
> mean that I'm doubting my own knowledge, rather than your results.

After a quick check, it appears that if your residuals are perfectly
correlated, but of different magnitude, that you can get the pattern
of correlations above. Doing a quick test in R:

> y <- rnorm( 1000, 100, 20 )

1000 items of (output) data

> residuals <- rnorm( 1000, 0, 3 )

normally distributed residuals

> pred1 <- y + residuals

one predicator

> pred2 <- y + residuals * 5

second predictor with perfectly correlated, but larger magnitude,
residuals

> cor( pred1, pred2 )
[1] 0.8779349

correlation coefficient of raw predicted values

> cor( pred1-y, pred2-y )
[1] 1

correlation coefficient of residuals.

Cheers,

Ross-c

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