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Re: st: Test of ordered probit vs ordinary probits


From   Partha Deb <partha.deb@hunter.cuny.edu>
To   statalist@hsphsun2.harvard.edu
Subject   Re: st: Test of ordered probit vs ordinary probits
Date   Wed, 31 Oct 2007 16:01:55 -0400

Mark,

Schaffer, Mark E wrote:
<snip>
"Ordered probit amounts to estimating (1.3) and (1.2) simultaneously, and with the constraint, that b in (1.3) equals b in (1.2). Ergo, ordered probit amounts to estimating the standard binary probit models

        Pr(outcome==3) = Pr( X*b + (−/cut2) + u > 0)                 (1.3)

and

        Pr(outcome>=2) = Pr( X*b + (−/cut1) + u > 0)                 (1.2)

with the constraint that the cofficients, but not the INTERCEPTS, are equal."

My question: Say we estimate two probits as above and test the constraint that the coefficients are equal.  Can we interpret this test as a test of probit vs. ordered probit?  That is, if the test fails to reject the null that the coefficients are different, would estimated an ordered probit be a logical next step?

<snip>

The test for probit versus ordered probit is a test of
H0: /cut1 = /cut2

against the alternative

Ha: /cut1 < /cut2.

But b needs to be the same in 1.2 and 1.3 in H0 and Ha. It is a maintained assumption in the ordered probit.

As I understand it, Richard's test is of ordered versus unordered probits / logits, rather than binary probit versus ordered multinomial probit.

Cheers,

Partha

--
Partha Deb
Department of Economics
Hunter College
ph: (212) 772-5435
fax: (212) 772-5398
http://urban.hunter.cuny.edu/~deb/

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