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RE: st: [Fwd: Heteroskedastic probit]


From   jverkuilen <[email protected]>
To   <[email protected]>
Subject   RE: st: [Fwd: Heteroskedastic probit]
Date   Thu, 30 Apr 2009 12:45:37 -0400

I second Rich's advice. 165 observations is a pretty small N for heteroscedastic probit. But I suspect that even regular probit won't find much either Look at the initial set of iterations to get starting values---the log-likelihood only changes a little. 

-----Original Message-----
From: "Richard Williams" <[email protected]>
To: "[email protected]" <[email protected]>; "[email protected]" <[email protected]>
Sent: 4/30/2009 12:42 PM
Subject: Re: st: [Fwd: Heteroskedastic probit]

At 10:27 AM 4/30/2009, [email protected] wrote:
>Heteroskedastic probit model                    Number of obs     =
>165
>                                                 Zero outcomes     =
>  56
>                                                 Nonzero outcomes  =
>109
>
>                                                 Wald chi2(3)      =
>1.11
>Log likelihood = -102.3917                      Prob > chi2       =
>0.7758
>
>------------------------------------------------------------------------------
>              |      Coef.   Std. Err.      z    P>|z|     [95% Conf.
>Interval]
>-------------+----------------------------------------------------------------
>newproc      |
>   normscapab |   3.722963   3.615273     1.03   0.303    -3.362842
>10.80877
>           e6 |  -.2859454   .6005927    -0.48   0.634    -1.463085
>.8911945
>         d12b |  -.0033544   .0098363    -0.34   0.733    -.0226332
>.0159244
>        _cons |  -.1114548    1.10271    -0.10   0.919    -2.272727
>2.049817
>-------------+----------------------------------------------------------------
>lnsigma2     |
>   normscapab |   1.320455   1.518283     0.87   0.384    -1.655325
>4.296234
>------------------------------------------------------------------------------
>Likelihood-ratio test of lnsigma2=0: chi2(1) =     0.61   Prob > chi2 =
>0.4342
>
>The model now converged but the prob is very high. Should I suppose the
>presence of Heteroskedasticity now and before?

I would suppose just the opposite.  Neither the z value in the 
lnsigma2 equation nor the corresponding LR test indicate that there 
is a problem with hetero, at least with normscapab.  Indeed, having 
the hetero equation may be keeping you from finding significant 
results in the choice equation.  It looks to me like you should just 
do a regular probit model.


-------------------------------------------
Richard Williams, Notre Dame Dept of Sociology
OFFICE: (574)631-6668, (574)631-6463
HOME:   (574)289-5227
EMAIL:  [email protected]
WWW:    http://www.nd.edu/~rwilliam

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