[R] rlm/M/MM-estimator questions
Kjetil Brinchmann Halvorsen
kjetil at acelerate.com
Thu Jul 7 03:30:49 CEST 2005
Christian Hennig wrote:
>Hi list,
>
>1) How can the MM-estimator method="MM" in function rlm be tuned to 85%
>efficiency? It seems that there is a default tuning to 95%. I presume, but
>am not sure, that the MM-estimator uses phi=phi.bisquare as default and
>the tuning constant could be set by adding a parameter c=...
>Is this true? Which value to use for 85%?
>(In principle I should be able to figure that out theoretically, but it
>would be much easier if somebody already knew the constant or a
>straightforward way to compute it.)
>
>
I have done this once, but cannot find the code or remember the
constant. But given the constant, it is easy to do this
in R. rlm has an argument psi with default psi huber:
> psi.huber
function (u, k = 1.345, deriv = 0)
{
if (!deriv)
return(pmin(1, k/abs(u)))
abs(u) <= k
}
<environment: namespace:MASS>
Use this argument with psi=function(u, k= your.value, deriv=0)
psi.huber(u,k,deriv)
>2) The M-estimator with bisquare uses "rescaled MAD of the residuals" as
>scale estimator according to the rlm help page. Does this mean that a
>scale estimator is used which is computed from least squares residuals? Are
>M-estimator and residual scale estimator iterated until convergence of
>them both? (Does this converge?)
>
Not sure about this at the moment.
> Or what else? What does "rescaled" mean?
>
>
"rescaled" means multiplied with the constant which makes it a
consistent estimator
under the normal model, default in the R mad function
Kjetil
>Thank you,
>Christian
>
>
>*** NEW ADDRESS! ***
>Christian Hennig
>University College London, Department of Statistical Science
>Gower St., London WC1E 6BT, phone +44 207 679 1698
>chrish at stats.ucl.ac.uk, www.homepages.ucl.ac.uk/~ucakche
>
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--
Kjetil Halvorsen.
Peace is the most effective weapon of mass construction.
-- Mahdi Elmandjra
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