Asymptotic t test question- regression when you do not assume normality of errors

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?

• Should your denominator be $s_{\hat{\beta}}$? (Welcome to CV, also!) May 25, 2020 at 21:58
• Yes It should, I meant it as a (lazy) short hand for that. I can edit it to clear any confusion May 25, 2020 at 21:59
• What do you mean that it is asymptotically z' ? Do you mean normal ? May 25, 2020 at 23:58
• Yes I mean asymptotically normal ( I believe asymptotically N(0,1)? which is why I used 'z') May 26, 2020 at 0:06
• @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?
– Dave
May 26, 2020 at 0:06

1 Answer

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.

• 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?' May 26, 2020 at 17:33
• 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. May 26, 2020 at 21:08
• 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.
– L Y
May 27, 2020 at 1:09
• 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? May 27, 2020 at 2:52
• Yes. You are right.
– L Y
May 28, 2020 at 1:55