Dear Mark,
Thanks for your comments!
Sorry for my late response. Here are the results reported by Stata:
First I run: xtreg earned_inc t1 t2 t3 t4 t5 cadre membership edu2 edu3 edu4, re i(xhn)
Then I store the results of the random effect model: estimates store random
Then I run fixed effect model: xtreg earned_inc t3 t4 t5 t6 t7 cadre membership edu2 edu3
edu4, fe i(xhn)
Finally I do the hausman test: hausman . random, sigmaless
The results reported by Stata are as follows:
---- Coefficients ----
(b) (B) (b-B) sqrt(diag(V_b-V_B))
. random Difference S.E.
t3 -27.42538 -26.08845 -1.336927 .
t4 -28.17475 -31.47857 3.303822 .
t5 -64.7231 -63.43336 -1.289746 .4918778
cadre 44.64596 57.50894 -12.86298 7.411722
membership -10.44655 2.795032 -13.24158 8.840997
edu2 24.21366 57.38761 -33.17395 6.615227
edu3 83.88996 129.8966 -46.00666 7.334445
edu4 132.2866 207.6196 -75.33295 11.79708
b = consistent under Ho and Ha; obtained from xtreg
B = inconsistent under Ha, efficient under Ho; obtained from xtreg
Test: Ho: difference in coefficients not systematic
chi2(8) = (b-B)'[(V_b-V_B)^(-1)](b-B)
= 256.13
Prob>chi2 = 0.0000
(V_b-V_B is not positive definite)
It says that V_b-V_B is not positive definite, although chi2 is positive.
I pretty much figure out what is going on. In my re model i use dummies t1 t2 t3 t4 and t5 while
in my fe model i use dummies t3 t4 t5 t6 t7. This means that the two models have different sets
of left-hand variables. I did this on purpose to figure out why stata reported a postive chi2 but
states that V_b-V_B is not positive .
What I found is that When stata states V_b-V_B are not positve definite, the variance-covariance
matrix (V_b-V_B) that stata refers to is the varaiance-covariance matrix for ALL coefficients (in
my case: t1 t2 t3 t4 t5 t6 t7 cadre membership edu2 edu3 edu4). Since in the re model, there
are no variables t6 and t7; in the fe model there are no t1 and t2. This leads to v(b)-v(B) for ALL
coefficients not positive definite. However the variance-covariance matrix v(b)-v(B) that stata
used for calculating chi2 is the one for the coefficients that appear in BOTH models (in my case:
t3 t4 t5 cadre membership edu2 edu3 edu4). This one is in fact positive definite. I calculated
the chi2 by hand using this variance-covariance matrix. I got the exactly same chi2 as stata
reported.
In my real case, since I have so many dummies( about 400 ), when I do re regression or fe
regressions, due to collinearity or other reasons, some variables are droped. This leads to that
the two regressions do not have a same set of left-hand variables. Thus v(b)-v(B) for ALL
coefficients is not positive definite. However, the v(b)-v(B) that stata uses for calculating Chi
square is the variance covariance matrix of the coefficients that exist in both re and fe model.
Best regards,
Jian
> Jian Zhang,
>
> Can you post the output of -hausman- to the list? I think I know what
> might be going on, but it's easiest to see by looking at the output.
>
> Cheers,
> Mark
>
> > -----Original Message-----
> > From: owner-statalist@hsphsun2.harvard.edu
> > [mailto:owner-statalist@hsphsun2.harvard.edu] On Behalf Of
> > Rodrigo A. Alfaro
> > Sent: 19 July 2006 04:06
> > To: statalist@hsphsun2.harvard.edu
> > Subject: Re: st: RE: a question on testing for random effect
> > model against fixed effect model
> >
> > Is the Hayashi proof valid for unbalanced-panel?
> > R
> >
> > ----- Original Message -----
> > From: "Jian Zhang" <jzh@ucdavis.edu>
> > To: "Schaffer, Mark E" <M.E.Schaffer@hw.ac.uk>;
> > <statalist@hsphsun2.harvard.edu>;
> <statalist@hsphsun2.harvard.edu>
> > Sent: Monday, July 17, 2006 8:44 PM
> > Subject: RE: st: RE: a question on testing for random effect
> > model against fixed effect model
> >
> >
> > Thanks, Mark! It seems that the CALCULATED standard hausman
> > test statistic is always positive even in FINITE samples
> > (i.e., calculated V(b)-v(B) is positive definite) as long as
> > one uses same variance estimates (mathematically this is
> > proved by Hayashi, 2000, as you
> > mentioned:
> >
> > "This appendix proves that the Avar(q_hat) in (5.2.21) is
> > positive definite and the Hausman statistic (5.2.22) is
> > guaranteed to be nonnegative in any finite samples."
> > (Hayashi, Econometrics (2000), Appendix 5.A, pp. 346-349 and 334-335.)
> >
> > So by adding option -sigmamore- or -sigmaless-, I did get a
> > positive standard hausman test (Chi square).
> >
> > However, confusing to me is that at the end of the results of
> > implementing hausman test in stata there is one line saying
> > (V(b)-V(B) is not positive definite) despite that I added option
> > -sigmamore- and got a positive Chi square. Any thoughts why
> > stata said that? From what I understand, calculated
> > V(b)-V(B) should be ALWAYS positive definite as long as one
> > uses option
> > -sigmamore- or -sigmaless-. The statement made by stata
> > results seems to
> > contradict the
> > mathematical argument made by Hayoshi.
> >
> > Best regards,
> > Jian Zhang
> >
> > > Jian,
> > >
> > > > -----Original Message-----
> > > > From: owner-statalist@hsphsun2.harvard.edu
> > > > [mailto:owner-statalist@hsphsun2.harvard.edu] On Behalf Of Jian
> > > > Zhang
> > > > Sent: 16 July 2006 08:36
> > > > To: statalist@hsphsun2.harvard.edu
> > > > Subject: st: a question on testing for random effect
> > model against
> > > > fixed effect model
> > > >
> > > > Thanks, Clive and Rodrigo!
> > > >
> > > > I wonder if there is an alternative test for random
> > effect against
> > > > fixed effect or a robust form of hausman test if the assumptions
> > > > made for Hausman test do not hold (one of the assumptions for
> > > > hausman test is the homoskedasticity and uncorrelation of the
> > > > idiosyncratic errors.
> > > > But this is often invalid.)
> > >
> > > Sorry to come in late on this, but I have three suggestions
> > relating
> > > to your original question.
> > >
> > > First, in a standard (i.e., non-robust) Hausman test, you can
> > > guarantee a positive test statistic by using the -sigmamore- or
> > > -sigmaless- options; the former is more traditional. Second,
> > > including the constant isn't traditional in a fixed vs.
> > random effects
> > > hausman test. Third, if you want to do a heteroskedastic- or
> > > cluster-robust version of the test, you can use the artificial
> > > regression version of the test described in Wooldridge's 2002 book
> > > (and I believe discussed in Statalist last year by Vince
> > Wiggins, if
> > > I'm not mistaken) and use robust or cluster-robust standard
> > errors in
> > > the artificial regression. The artificial regression
> > version will also guarantee a positive test statistic (of course!).
> > >
> > > Cheers,
> > > Mark
> > >
> > > Prof. Mark Schaffer
> > > Director, CERT
> > > Department of Economics
> > > School of Management & Languages
> > > Heriot-Watt University, Edinburgh EH14 4AS tel
> > +44-131-451-3494 / fax
> > > +44-131-451-3296
> > > email: m.e.schaffer@hw.ac.uk
> > > web: http://www.sml.hw.ac.uk/ecomes
> > >
> > >
> > > >
> > > > Jian
> > > >
> > > >
> > > > On Sat, 15 Jul 2006, Rodrigo A. Alfaro wrote:
> > > >
> > > > > Jian,
> > > > >
> > > > > Try -xtreg, re sa- instead of -xtreg, re- the additional
> > > > option takes
> > > > > care "more carefully" the unbalanced issue using
> > Swamy-Arora method.
> > > > >
> > > > > Read Method and Formulas in the manual, for version 8:
> > > > > http://www.stata-press.com/manuals/stata8/xtreg.pdf and
> > version 9:
> > > > > http://www.stata.com/bookstore/pdf/xtreg.pdf
> > > > >
> > > > > Rodrigo.
> > > > >
> > > > >
> > > > > ----- Original Message -----
> > > > > From: "Clive Nicholas" <Clive.Nicholas@newcastle.ac.uk>
> > > > > To: <statalist@hsphsun2.harvard.edu>
> > > > > Sent: Saturday, July 15, 2006 4:19 AM
> > > > > Subject: Re: st: a question on testing for random effect
> > > > model against
> > > > > fixed effect model
> > > > >
> > > > >
> > > > > Jian Zhang wrote:
> > > > >
> > > > > > I have a question on testing random effect model
> > against fixed
> > > > > > effect model. Hope that you can help me out. Here is the
> > > > > > question;
> > > > > >
> > > > > > I am applying random effect model and fixed effect
> > model to an
> > > > > > unbanlanced panel data (use xtreg, re and xtreg, fe). To
> > > > test which
> > > > > > model is more appropriate, I run a hausman test.
> > > > However, the test
> > > > > > statistics (the chi square) is negative. This makes
> > > > hausman testing
> > > > > > impossible, since chi square cann't be negative. The reason
> > > > > > that hausman test doesn't work is that the model's error
> > > > structure does
> > > > > > not meet the assumptions made for the hausman test.
> > > > >
> > > > > [...]
> > > > >
> > > > > Did you run the following:
> > > > >
> > > > > xtreg ..., fe
> > > > >
> > > > > est store fixed
> > > > >
> > > > > xtreg ..., re
> > > > >
> > > > > hausman fixed ., alleqs constant
> > > > >
> > > > > If not, see if that works. Works for me every time I
> > have to use it.
> > > > >
> > > > > CLIVE NICHOLAS |t: 0(044)7903 397793
> > > > > Politics |e: clive.nicholas@ncl.ac.uk
> > > > > Newcastle University |http://www.ncl.ac.uk/geps
> > > > >
> > > > > Whereever you go and whatever you do, just remember this. No
> > > > > matter how many like you, admire you, love you or adore
> > you, the
> > > > > number of people turning up to your funeral will be largely
> > > > determined by local
> > > > > weather conditions.
> > > > >
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