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

Prof Brian Ripley ripley at
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
Professor of Applied Statistics,
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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