# Re: [R] rlm/M/MM-estimator questions

From: Matthias Kohl <Matthias.Kohl_at_uni-bayreuth.de>
Date: Thu 07 Jul 2005 - 17:49:03 EST

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.)
>
>
Hi Christian,

I have not calculated the efficiency myself ... But the thesis of Matias Salibian-Barrera (SB 2000) might help you to find the answer
(cf. Chapter 4).
See: http://mathstat.math.carleton.ca:16080/~matias/thesis.pdf

As far as I understand the choice k0=1.548 is to obtain a breakdown point 0.5 whereas k0=1.988 leads to a breakdown point of 0.4 - at least in the location case; confer p. 60 of SB 2000.

In the article "Optimal robust \$M\$-estimates of location" by Fraiman, Yohai and Zamar (Ann. Stat. 29(1): 194 - 223) which is, of course, concerned with the location case, the authors recommend to use k0=1.988 instead of k0=1.548 (cf. p. 206).

Hope that helps!
Matthias

>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?) Or what else? What does "rescaled" mean?
>
>Thank you,
>Christian
>
>
>Christian Hennig
>University College London, Department of Statistical Science
>Gower St., London WC1E 6BT, phone +44 207 679 1698
>chrish@stats.ucl.ac.uk, www.homepages.ucl.ac.uk/~ucakche
>
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