# Follow-up: [R] Inverse prediction with R?

Prof Brian Ripley ripley at stats.ox.ac.uk
Tue Feb 22 14:02:45 CET 2000

```> Date: Tue, 22 Feb 2000 13:28:41 +0100
> Subject: Follow-up: [R] Inverse prediction with R?
>
> (message from 22.2.2000 13:04 Uhr):
>
> >
> >
> > Why don't you inverse the modelling instead:
> >
> > t.m.i <- lm((x~y)
> >
>
> Jan,
>
> thanks for the tip, but it's not just the same. The coefficients come out
> differently, since the squared y residuals are minimized. Orthogonal
> regression would be symmetric, but least squares is not, I'm afraid. And,
> what's more, I have to state my model that way since the x values I have got
> are fixed.
>
> Perhaps I have to explain where my problem comes from: I want to measure the
> age of trees. I cannot cut them and I can take core samples only from the
> bigger ones. What I can measure is the number of bud rings and branchings,
> which is proportional to the age, although the measurement is rather
> inaccurate. So, I 'd like to model that number based on the core samples
> I've got (N—40), which are accurate, and the inversely "predict core
> samples" for the rest of my trees (N—300).
>
> Maybe there is another way to do what I intend, but at the moment I've no
> clue. I'm grateful for any hint.

So, you only have one x? Then this is a calibration problem, and the
terms you mentioned before are not relevant.

Assuming that the trees are a random sample x, is not fixed but (y,x)
are random pairs, and inverse regression is the right way to solve
the problenm. E.g. P.J. Brown (1993), `Measurement, Regression and
Calibration', OUP, page 22.  If somehow you managed to choose the
trees to have specific x's, then there would be better ways.

--
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 272860 (secr)
Oxford OX1 3TG, UK                Fax:  +44 1865 272595

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