[R] Need to fit a regression line using orthogonal residuals
davidr at rhotrading.com
davidr at rhotrading.com
Wed Jan 31 17:04:26 CET 2007
This problem also comes up in financial hedging problems,
but usually the 'errors' need not be of comparable size, so Errors in
Variables or Total Least Squares might be used.
David L. Reiner
Rho Trading Securities, LLC
Chicago IL 60605
312-362-4963
-----Original Message-----
From: r-help-bounces at stat.math.ethz.ch
[mailto:r-help-bounces at stat.math.ethz.ch] On Behalf Of Prof Brian Ripley
Sent: Wednesday, January 31, 2007 1:57 AM
To: Bill.Venables at csiro.au
Cc: jkopecky at umich.edu; r-help at stat.math.ethz.ch
Subject: Re: [R] Need to fit a regression line using orthogonal
residuals
Just to pick up
> Third, the resulting line is not optimal for either predicting y for a
> new x or x from a new y. It's hard to see why it is ever of much
> interest.
It is not a regression (and hence the subject line was misleading), but
it
does come up in errors-in-variables problems. Suppose you have two sets
of measurements of the same quantity with the same variance of
measurement
error and you want a line calibrating set 2 to set 1. Then the optimal
(in the sense of MLE, for example) line is this one, and it is
symmetrical
in the two sets.
Now those are rather specific assumptions but they do come up in some
problems in physics and analytical chemistry, and the result goes back
to
the 19th century. In the 1980s I implemented a version which allowed
for
unequal (but known) heteroskedastic error variances which is quite
popular
in analytical chemistry.
The literature is patchy: Fuller's `Measurement Error Models' covers
the
general area, and I recall this being in Sprent's (1969) book
`Models in Regression and Related Topics'.
See also the thread starting at
http://tolstoy.newcastle.edu.au/R/help/00a/0285.html
almost 7 years ago. If that is the thread Jonathon Kopecky refers to
(how
are we to know?) then he is misquoting me: I said it was the same thing
as
using the first principal component, not an alternative proposal.
On Wed, 31 Jan 2007, Bill.Venables at csiro.au wrote:
> Jonathon Kopecky asks:
>
> -----Original Message-----
> From: r-help-bounces at stat.math.ethz.ch
> [mailto:r-help-bounces at stat.math.ethz.ch] On Behalf Of Jonathon
Kopecky
> Sent: Tuesday, 30 January 2007 5:52 AM
> To: r-help at stat.math.ethz.ch
> Subject: [R] Need to fit a regression line using orthogonal residuals
>
> I'm trying to fit a simple linear regression of just Y ~ X, but both X
> and Y are noisy. Thus instead of fitting a standard linear model
> minimizing vertical residuals, I would like to minimize
> orthogonal/perpendicular residuals. I have tried searching the
> R-packages, but have not found anything that seems suitable. I'm not
> sure what these types of residuals are typically called (they seem to
> have many different names), so that may be my trouble. I do not want
to
> use Principal Components Analysis (as was answered to a previous
> questioner a few years ago), I just want to minimize the combined
noise
> of my two variables. Is there a way for me to do this in R?
> [WNV] There's always a way if you are prepared to program it. Your
> question is a bit like asking "Is there a way to do this in Fortran?"
> The most direct way to do it is to define a function that gives you
the
> sum of the perpendicular distances and minimise it using, say,
optim().
> E.g.
> ppdis <- function(b, x, y) sum((y - b[1] - b[2]*x)^2/(1+b[2]^2))
> b0 <- lsfit(x, y)$coef # initial value
> op <- optim(b0, ppdis, method = "BFGS", x=x, y=y)
> op # now to check the results
> plot(x, y, asp = 1) # why 'asp = 1'?? exercise
> abline(b0, col = "red")
> abline(op$par, col = "blue")
> There are a couple of things about this you should be aware of, though
> First, this is just a fiddly way of finding the first principal
> component, so your desire not to use Principal Component Analysis is
> somewhat thwarted, as it must be.
> Second, the result is sensitive to scale - if you change the scales of
> either x or y, e.g. from lbs to kilograms, the answer is different.
> This also means that unless your measurement units for x and y are
> comparable it's hard to see how the result can make much sense. A
> related issue is that you have to take some care when plotting the
> result or orthogonal distances will not appear to be orthogonal.
> Third, the resulting line is not optimal for either predicting y for a
> new x or x from a new y. It's hard to see why it is ever of much
> interest.
> Bill Venables.
--
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 272866 (PA)
Oxford OX1 3TG, UK Fax: +44 1865 272595
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