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.
[email protected]
-----Original Message-----
From: Wiji Arulampalam [mailto:[email protected]]
Sent: Friday, September 20, 2002 8:29 AM
To: [email protected]; [email protected]
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: [email protected]
http://www.warwick.ac.uk/Economics/arulampalam/
RES2003: http://www.warwick.ac.uk/res2003/
>>> [email protected] 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
[email protected]
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