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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/

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