# Re: [R] Need to fit a regression line using orthogonal residuals

From: <Bill.Venables_at_csiro.au>
Date: Wed 31 Jan 2007 - 01:36:47 GMT

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 - b*x)^2/(1+b^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.

Jonathon Kopecky
University of Michigan

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