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RE: st: RE: RE: A math question


From   "Feiveson, Alan H. (JSC-SK311)" <alan.h.feiveson@nasa.gov>
To   "statalist@hsphsun2.harvard.edu" <statalist@hsphsun2.harvard.edu>
Subject   RE: st: RE: RE: A math question
Date   Fri, 21 Dec 2012 10:29:38 -0600

You can't optimize a function of median |ei| by setting it's derivative to zero because the partial derivatives don't exist at ei = 0. You would have do something like linear programming (as is done in LAD regression).


-----Original Message-----
From: owner-statalist@hsphsun2.harvard.edu [mailto:owner-statalist@hsphsun2.harvard.edu] On Behalf Of CJ Lan
Sent: Friday, December 21, 2012 10:10 AM
To: statalist@hsphsun2.harvard.edu
Subject: RE: st: RE: RE: A math question

What happens is the following:

Recall the Huber weighting functions in the robust estimator is defined as (k= tuning constant and sigma is the robust estimate of the "scale" used to standardize the errors) w=1, if |e|<=k*sigma w=k*sigma/|e|, if |e|>k*sigma

Sigma (aka scale) in several software including Stata (I believe) is defaulted as MAE/norminv(0.75), where MAE = median of absolute errors = median{|ei|, i=1,...,n}.  n=no. of data points.  "Norminv" is the inverse of standard normal fun.
e = estimation error = y - u(b,x).

Now, what I try to do is to compute the derivative as well as Hessian of the weights, w, with respect to the parameters b.  Both Sigma and errors are a function of b.  

I am stuck at taking derivative of the Sigma wrt b, because it involves with the "median".  

-----Original Message-----
From: owner-statalist@hsphsun2.harvard.edu [mailto:owner-statalist@hsphsun2.harvard.edu] On Behalf Of David Hoaglin
Sent: Friday, December 21, 2012 10:52 AM
To: statalist@hsphsun2.harvard.edu
Subject: Re: st: RE: RE: A math question

The median is a function of a distribution.  You did not specify the distribution of x.  If you explained the context for your question, we might be able to shed better light.

David Hoaglin

On Fri, Dec 21, 2012 at 10:42 AM, CJ Lan <CJ@jupiter.fl.us> wrote:
> Al,
>
> I think the "median" is a function, just like the "mean" is a function, not just a number.
>
> BTW, the derivative of abs(x^3) should be 3*x*abs(x) because the 
> derivative can be derived as
>  abs(x^3)/(x^3)*(3*x^2) = 3*abs(x)*x^2/x = 3*x*abs(x).
>
> I wish others could shed a light too?  Thx.
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