# RE: st: predict options in heckprob

 From "FEIVESON, ALAN H. (AL) (JSC-SD) (NASA)" <[email protected]> To "'[email protected]'" <[email protected]> Subject RE: st: predict options in heckprob Date Tue, 20 Aug 2002 08:14:44 -0500

Bobby -

So suppose y(select)=1 if a customer enters the store and y(probit)=1 if the
customer buys one of the store's widgets. Then p10 is the (in general,
nonzero) probability that the customer buys one of the store's widgets
without entering the store. Are you then suggesting that the heckprob model
is not appropriate for this situation?

Al

-----Original Message-----
From: [email protected] [mailto:[email protected]]
Sent: Monday, August 19, 2002 4:21 PM
To: [email protected]
Subject: Re: st: predict options in heckprob

> Hello - I notice in the writeup for heckprob under "Options for predict"
> (Stata 7 manual H-P, p.33), "p10" is defined as the predicted joint
> probability that y(probit)=1 and y(select)=0. I thought that under this
> model, y(probit) is only observed if y(select)=1. Therefore the joint
> probability should be zero.

> (1): y(probit) as originally constructed in the example (i.e. all 0's and
> 1's, no missing) and (2): y(probit) set to missing when y(select)=0.

> I was relieved to see that I got the same estimation results whether or
not
> y(probit) was set to missing or zero when y(select)=0. Using the "p10"
> option, however, I obtained non-zero values. Apparently, what is
calculated
> is the joint probability of y1*>0 and y2* >0 where y1* and y2* are the
> underlying latent normally distributed variables for y(probit) and
> y(select). While I agree that P(y1*>0 , y2* < 0) may be of some interest,
> the writeup should be changed to reflect what is really being calculated.

y(probit) and the event "y1*>0" are one in the same, they either both
evaluate to true (1) or false (0).  Therefore P{y(probit)==1} and
P(y1*>0) are equal.

However, there are situations where y(probit) is left unobserved (when
y(select)=0), but that doesn't change the fact that the (unobserved) value
exists somewhere out there and is equal to 0 or 1.  To see this, think about
performing the following experiment:  flip a fair coin and walk away while
the
coin is still in the air.  I contend that the probability of flipping a head
is 0.5 even though you were never around to see it.

As such, I think the terminology in the manual is correct as it stands.

--Bobby
[email protected]
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