The optional tolerance argument [Design Issues]

posted by ElMaestro  – Belgium?, 2019-12-23 13:37 2001:2012:306:2500:187a:d176:60b:ac29 – Posting: # 21021
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Hi Hötzi,

» » » For observation 8 we have -3.608225e-16, I think,
» » I think, is around the "effective zero" for fits in R at default settings on 64- and 32-bit systems.
»
» Yes, it is. ;-)
»

x    <- -3.608225e-16
» zero <- .Machine$double.eps
» all.equal(x, zero)
» [1] TRUE
» zero
» [1] 2.220446e-16


This comparison in your context is just a test if the difference is less than about 10-8 since there is an implied tolerance argument for all.equal, the square root of .Machine$double.eps

Effective zero residuals will be somewhat better than 10-8 in practice. They will depend on the approach used to find the solution; in lm I believe the approach is via a qr decomposition of the model matrix, and R by defualt has a tol argument in that function of 10-7 which lm may be leaning on.

Here's an example of a perfect fit, therefore having effective zero residuals:

a=c(rep(1,5), rep(2,5), rep(3,5))
b=c(rep("A",5), rep("B",5), rep("C",5))
M=lm(a~0+b)
resid(M)


It may actually not be the best example since the dependents are all representable inernally in R's (and computer's) binary.

Perhaps this makes a better point:

a=c(rep(pi,5), rep(sin(1.5+pi),5), rep(log(pi),5))
b=c(rep("A",5), rep("B",5), rep("C",5))
M=lm(a~0+b)
resid(M)

Le tits now.

Best regards,
ElMaestro

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