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Say you are running a regression: $Y_i$= $X_i$$\beta$ + $\eta_i$

And we are not assuming normality of $\eta_i$.

My understanding is that as long as your sample size is relatively large (and i know how large is arbitrary), you can rely on the CLT to justify using the same formula of the t -statistic anyways as a close approximation to the actual underlying sampling distribution, i.e. $\hat{\beta}$ converges in distribution to a normal distribution, and using consistent estimators of the asymptotic covariance matrix, you can use a formulation that looks like a t statistic, i.e.

$\frac{(\hat{\beta}-\beta)}{s_\hat{\beta}}$

where $s_i$ is the consistent estimator of the derived asymptotic variance of the estimator, for hypothesis testing. With this, is it accurate to say, that

1). this is NOT a t-test (I have heard it described as an 'asymptotic t test' before) 2). this is a test statistic that is 'asymptotically z' 3). given 2), we use the z table for p-values, essentially assuming that we can use the z distribution as an approximation of the underlying sampling distribution?

If the above are true, then is it correct than to just use z tables in this case as approximation of the sampling distribution? is the effect no different than just using a t table with large b, as that converges to standard normal too?

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  • $\begingroup$ Should your denominator be $s_{\hat{\beta}}$? (Welcome to CV, also!) $\endgroup$
    – Alexis
    May 25, 2020 at 21:58
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    $\begingroup$ Yes It should, I meant it as a (lazy) short hand for that. I can edit it to clear any confusion $\endgroup$
    – Steve
    May 25, 2020 at 21:59
  • $\begingroup$ What do you mean that it is asymptotically z' ? Do you mean normal ? $\endgroup$
    – Pohoua
    May 25, 2020 at 23:58
  • $\begingroup$ Yes I mean asymptotically normal ( I believe asymptotically N(0,1)? which is why I used 'z') $\endgroup$
    – Steve
    May 26, 2020 at 0:06
  • $\begingroup$ @Pohoua I take that to mean that the t-stat is asymptotically $t_{\infty}=N(0,1)$, so if we’re assuming convergence to a t-distribution, why not take the convergence all the way to standard normal? $\endgroup$
    – Dave
    May 26, 2020 at 0:06

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Since you are considering large sample theory where $n$ tends to infinity, there are some additional assumptions you may need in order to make the assertion.

(1) $(\eta_1,\ldots,\eta_n)$ are uncorrelated with equal variance

(2) The design matrix $X$ grows as $n$ becomes larger. We should have something like: $\frac1n X'X$ tends to a finite limit in some way. For example, if we assume that $X_1,\ldots,X_n$ are iid from a distribution with mean 0 and finite variance, then $\frac1n X'X$ tends to the variance matrix in probability.

These are the things that you need to say that the test statistics are "asymptotically" normal.

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  • $\begingroup$ So when you have these (these are conditions for the CLT, correct?) at is is 'asymptotically' normal, you just use the z statistic/distribution for hypothesis testing? so it is more accurately described as an 'asymptotic z test' rather than an 'asymptotic t test?' $\endgroup$
    – Steve
    May 26, 2020 at 17:33
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    $\begingroup$ Yes, this is exact. This test statistic does not follow a t-distribution. I would agree with you that the name "asymptotic z-test" would be relevant. $\endgroup$
    – Pohoua
    May 26, 2020 at 21:08
  • $\begingroup$ When n tends to infinity, t-distribution with n-k degrees of freedom will converge in distribution to z-distribution (std normal distribution), so they are asymptotically the same. It is more customary to use z-distribution but theoretically they should yield similar results for large n. $\endgroup$
    – L Y
    May 27, 2020 at 1:09
  • $\begingroup$ Just to clarify what this implies - this 'asymptotic' z test I described above, doesn't necessarily have a defined sampling distribution, but we approximate it with Z when n is large. But it should still give equivalent or near equivalent results if we assumed normality of errors, and used a t distribution with a large n, sinc e the t distribution will also converge to z? so one may be more 'technically' correct but with large enough n, they both would yield the same inference? $\endgroup$
    – Steve
    May 27, 2020 at 2:52
  • $\begingroup$ Yes. You are right. $\endgroup$
    – L Y
    May 28, 2020 at 1:55

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