Stata The Stata listserver
[Date Prev][Date Next][Thread Prev][Thread Next][Date index][Thread index]

RE: st: L-estimator


From   "Nick Cox" <n.j.cox@durham.ac.uk>
To   <statalist@hsphsun2.harvard.edu>
Subject   RE: st: L-estimator
Date   Tue, 16 Mar 2004 15:14:53 -0000

This is difficult to read. 

You should try -findit- using keywords like 
trimmed mean, Hodges-Lehmann, etc. 

Phil Ender has a program called -robmean-. 

The trimean is computable after -summarize, 
detail- as 

0.25 * (r(p25) + r(p75)) + 0.5 * r(p50) 

In general, some of these problems are 
approachable as weighted means, so you 
need to prepare a weight variable before
using -summarize- with weights. 

Nick 
n.j.cox@durham.ac.uk 

> -----Original Message-----
> From: owner-statalist@hsphsun2.harvard.edu
> [mailto:owner-statalist@hsphsun2.harvard.edu]On Behalf Of Andreas
> Aschbacher
> Sent: 16 March 2004 14:34
> To: statalist@hsphsun2.harvard.edu
> Subject: Re: st: L-estimator
> 
> 
> sorry I have seen now that on e-mail-editor there is no such 
> abridgement
> as I wrote before ! I hope it is possible for you to read and 
> understand
> anyhow. andreas
> 
>  Dear fellows !
> > 
> > I have  300 values in radioactive-measurement in one column :
> > now I want to compute :
> > 
> >  1)             - Tn = sum 1 to 300 [a(i).x(i)]             
> 0 < alpha <
> > 0.5
> >                                a(i) = 0                     
>               
> >  
> >               for i <= n1 or i >= n- n1 + 1
> >                                a(i) = (n1 + 1 -alpha*n)/((1 
> -2*alpha)*n)  
> > for i = n1 + 1 or i = n - n1
> >                                a(i) = 1/((1-2*alpha)*n)     
>               
> >  
> >       for n1 + 1 < i < n - n1
> >                                                             
>               
> >  
> >                  with : n1 = [alpha*n],n = 300
> >  / in <a
> href="http://www.ntsearch.com/search.php?q=europe&v=55";>europe
> </a> we say L - estimator type I/
> > 
> >  2)             - Tn = sum 1 to 300 [a(i).x(i)]             
> 0 < alpha <
> > 0.5
> >                                a(i) = (1/n)*[x(n1 +1)/x(i)] 
>               
> >  
> >       for i <= n1
> >                                a(i) = 1/n                   
>               
> >  
> >               for n1 + 1 <= i <= n - n1
> >                                a(i) = (1/n)*[x(n - 
> n1)/x(i)]              
> >  
> >        for i >= n - n1 +1
> >                                                             
>               
> >  
> >                   with : n1 = [alpha*n],n = 300
> > / in <a 
> href="http://www.ntsearch.com/search.php?q=europe&v=55";>europe</a>
> we say L - estimator type II/
> > 
> >  3)             - Tn = 0.25 *[x(n1 + 1) + x(n - n1)]  + 
> 0.5*median-value  
> >  
> >       0 < alpha < 0.5
> >                                                             
>               
> >  
> >                   with : n1 = [alpha*n],n = 300
> > 
> > / in <a 
> href="http://www.ntsearch.com/search.php?q=europe&v=55";>europe</a>
> we say trimean   /
> > 
> > 4)             - Tn = 0.3 *[x(n1 + 1) + x(n - n1)]  + 
> 0.4*median-value    
> >  
> >     0 < alpha < 0.5
> > 
> > / in <a 
> href="http://www.ntsearch.com/search.php?q=europe&v=55";>europe</a>
> we say L - estimator type III/
> > 
> > 5)if I create (x(i) + x(j))/2 that I get 300^2 values,I 
> want to compute
> > median of this -
> >  we say wilcoxon-appliance,I hope you will understand my 
> terminologie
> > / in <a 
> href="http://www.ntsearch.com/search.php?q=europe&v=55";>europe</a>
> we say L - estimator type IIII/

*
*   For searches and help try:
*   http://www.stata.com/support/faqs/res/findit.html
*   http://www.stata.com/support/statalist/faq
*   http://www.ats.ucla.edu/stat/stata/



© Copyright 1996–2014 StataCorp LP   |   Terms of use   |   Privacy   |   Contact us   |   What's new   |   Site index