Re: st: Another method of squaring an instrumental variable

[Stas Kolenikov <skolenik@gmail.com>]

Yes, it will give some coefficients and standard errors, but squaring
the predicted value won't give you a good estimate of the square of
the variable. You would still want to think about x and x^2 as
separate variables, and by any account you would have to agree that
you don't have enough instruments.

On Sat, Jun 5, 2010 at 9:50 AM, Nir Regev <regev.nir@gmail.com> wrote:
> Hi Stas, Thank you for your answer.
>
> I know I have a case of under-identification. I wish I had another
> instrument but I don't.
>
> The question is, maybe there is someway I can fix for the standard
> errors of my forced "second stage".
> What I am thinking is using the Variance of x,
> meaning, Var(x) = E(x^2) - (E(x))^2, since I have two of the three,
> maybe I can derive the right SE for x_hat^2 (This is just a hunch)
> Alas, Econometrics is not my strongest point and I am looking for a
> reference on this matter.
>
> By the way, I ran the procedure I thought of and technically, it works.
> It is not a case of perfect collinearity since squaring is non-linear.
>
> Nir
>
>
>
> On  Sat, Jun 5, 2010 at 5:27 PM, Stas Kolenikov <skolenik@gmail.com> wrote:
>> I don't think this is the right approach. Recall that E[x^2] is not
>> equal to ( E[x] )^2 (Jensen's inequality). You need to have E[
>> regressor | instrument ] in the RHS, so you should've run two
>> regressions of x and x^2 on your instrument(s). But since you have
>> only one, that would create perfect collinearity between your
>> instrumented variables in the main regression -- in other words, you
>> don't have enough instruments for your endogenous variables x and x^2.
>>
>> On Sat, Jun 5, 2010 at 7:47 AM, Nir Regev <regev.nir@gmail.com> wrote:
>>> I am trying to estimate an equation of     y = x + x^2+u
>>> x is endogenous (which makes x^2 endogenous too) and I have only one
>>> instrumental variable z, which is a dummy.
>>>
>>> On a regular case, I would square z and let z and z^2 be the
>>> instrumentals, however, since z is a dummy this is irrelevant.
>>>
>>> I believe it is possible to run the first stage of 2SLS where:
>>> x=z+u,
>>> predict x_hat,
>>> square x_hat
>>> and run the second stage both on x_hat and x_hat^2.
>>>
>>> I think this is a consistent estimate but my standard errors and
>>> hypothesis tests will no doubt be miscalculated.
>>>
>>> Has anyone ever heard of such a method and knows how to fix the standart errors?
>>>
>>
>> --
>> Stas Kolenikov, also found at http://stas.kolenikov.name
>> Small print: I use this email account for mailing lists only.
>>
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-- 
Stas Kolenikov, also found at http://stas.kolenikov.name
Small print: I use this email account for mailing lists only.

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