In article <9r4nbg$dka$[EMAIL PROTECTED]>,
David B <[EMAIL PROTECTED]> wrote:
>"Radford Neal" <[EMAIL PROTECTED]> a crit dans le message news:
>> >If the explanatory variable is generated by an integrated process, it
>won't
>> >work, even if the error term is an iid process.
>> This is what I am disputing. What basis do you have for claiming that
>> it won't work? And in what sense do you mean that "it won't work"?
>> I suspect that you've encountered a claim that is somewhat like this
>> in some reference book, and have mis-interpreted it.
>> Radford Neal
>Well, I may have not explained myself very clearly, or understood what you
>really meant, in which case I apologize in advance.
>Now, here is what I mean when I say that standard procedures shouldn't work
>with integrated processes.
>If X is non stationary, and if the regression equation is true, Y is non
>stationary too.
>The OLS slope estimator is (X'X)(-1) X'Y
>If the X is generated by an integrated process, (X'X) will not be convergent
>in probability, nor will X'Y.
Neither is it in general. For consistency of the estimator,
the inverse of X'X needs to converge to 0. But this is not
generally a problem because of using integrated processes..
>In the case of Y and X being two independent random walks, the mean of
>(XX)(-1)X'Y can be calculated using Wiener distribution theory however, and
>it is not zero (it looks very bad). The t-stat for slope is not zero either.
>The variance of both slope estimator and t-stats are much higher than
>standard theory forecast, and, what is even worse, do not decrease as sample
>size increase.
If they are independent random walks with mean 0, or even if
E(Y|X)=0, the mean of this will have to be 0.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Dept. of Statistics, Purdue Univ., West Lafayette IN47907-1399
[EMAIL PROTECTED] Phone: (765)494-6054 FAX: (765)494-0558
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