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From |
"Sayer, Bryan" <BSayer@s-3.com> |

To |
"'statalist@hsphsun2.harvard.edu'" <statalist@hsphsun2.harvard.edu> |

Subject |
RE: st: Random effects probit |

Date |
Fri, 20 Sep 2002 14:13:07 -0400 |

rho-hat has a skewed distribution, and the distribution of the test statistic is not asymptotically normal. Either a transformation is needed or I believe David's tables can be used. The transfomation is somewhat complex, and I cannot remember it off the top of my head. Bryan Sayer Statistician, SSS Inc. bsayer@s-3.com -----Original Message----- From: Wiji Arulampalam [mailto:Wiji.Arulampalam@warwick.ac.uk] Sent: Friday, September 20, 2002 8:29 AM To: statalist@hsphsun2.harvard.edu; ddrukker@stata.com Subject: Re: st: Random effects probit Dear David, Am I right in thinking that one could actually do the test for rho=0 using the coefficient est and its std error? The statistic coeff/se will be asymp distributed sort of like a std. normal. The actual distribution has a mass of 0.5 at 0 and the rest is Normal on the positive side due to the fact that the null is on the boundary of the parameter space. Essentially a 5% test requires one to use a 10% critical level (same in the LR chisq test as specified below). But Likelihood ratio is better since it is invariant to the way one specifies the parameters where as the above Wald type test is not because of the non-linear transformations involved. Is this wrong? many thanks best wiji ================================= Professor Wiji Arulampalam, Department of Economics, University of Warwick, Coventry, CV4 7AL, UK. Tel: +44 (24) 7652 3471 Sec. Tel: +44 (24) 7652 3202 Fax: +44 (24) 7652 3032 email: wiji.arulampalam@warwick.ac.uk http://www.warwick.ac.uk/Economics/arulampalam/ RES2003: http://www.warwick.ac.uk/res2003/ >>> ddrukker@stata.com 09/19/02 12:57AM >>> Now, let's look into quesiton iv). The test of the null that there is no heterogeneity is a test of the null hypothesis that sigma_u = 0. This is a test on the boundary of the parameter space for sigma_u. (See help j_chibar for more on this topic and a reference.) The computation of the test statisitic remains the same, but its asymptotic distribution is different than that of a stardard likelihood ratio test. Instead of converging to chi-squared with one degree of freedom it converges to another distribution which is basically .5*(chi-squared with one degree of freedom). Let's compute the value of the likelihood ratio test and its p-value. . scalar lr = 2*(ll1-ll0) . scalar p = .5*chi2tail(1,lr) And display the results: . di "hand lr is " lr " with p-value " p hand lr is .22917373 with p-value .31606859 This result indicates that we fail to reject the null hypothesis of no heterogeneity, or equivalently, that sigma_u =0 I hope that this helps. David ddrukker@stata.com * * 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/ * * 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/ * * 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/ * * 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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