R help - Jan 2007 - loess

Hi,

I have tried

 > for (i in 1:100) L[,i] <- loess((i = =(1:100))~I(1:100), span=.5, 
degree=1)$fit

to create a matrix which gives me the smoothing weights (correctly as 
far as I have experienced), eg.

 > yhat <- loess(y~I(1:100), span=.5,degree=1)$fit
 > yhat[30]
[1] -0.2131983
 > L[30,]%*%y
           [,1]
[1,] -0.2131983

But,  L[30,] has 56 nonzero coefficients, not 50 that I expect with span 
= 0.5. Actually the number of nonzero elements on rows varies being 49, 
50, 55 or 56.

Does anyone know why?

Jukka Nyblom

On Thu, 4 Jan 2007, Jukka Nyblom wrote:
> Hi,
>
> I have tried
>
> > for (i in 1:100) L[,i] <- loess((i = =(1:100))~I(1:100), span=.5,
> degree=1)$fit
>
> to create a matrix which gives me the smoothing weights (correctly as
> far as I have experienced), eg.
>
> > yhat <- loess(y~I(1:100), span=.5,degree=1)$fit
> > yhat[30]
> [1] -0.2131983
> > L[30,]%*%y
>           [,1]
> [1,] -0.2131983
>
> But,  L[30,] has 56 nonzero coefficients, not 50 that I expect with span
> = 0.5. Actually the number of nonzero elements on rows varies being 49,
> 50, 55 or 56.
>
> Does anyone know why?
loess is a complicated algorithm, and you need to study the background 
references in depth to fully understand it.  In particular, the default is 
not to do direct fitting (as I guess you are assuming) but interpolation. 
See ?loess.control.

Most descriptions, including the help page, are simplifications.
> Jukka Nyblom
-- 
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