A sequence of estimators Un for a parameter θ is asymptotically normal if √n(Un−θ)→N(0,v). (source) We then call v the asymptotic variance of Un. If this variance is equal to the Cramer-Rao bound, we say the estimator/sequence is asymptotically efficient.

Question:Why do we use √n in particular?I know that for the sample mean, Var(ˉX)=σ2n and so this choice normalizes it. But since the definitions above apply to more than the sample mean, why do we still choose to normalize by √n.

**Answer**

We don’t get to *choose* here. The “normalizing” factor, in essence is a “variance-stabilizing to something finite” factor, so as for the expression not to go to zero or to infinity as sample size goes to infinity, but to maintain a distribution at the limit.

So it has to be whatever it has to be in each case. Of course it is interesting that in many cases it emerges that it *has* to be √n. (but see also @whuber’s comment below).

A standard example where the normalizing factor has to be n, rather than √n is when we have a model

yt=βyt−1+ut,y0=0,t=1,...,T

with ut white noise, and we estimate the unknown β by Ordinary Least Squares.

If it so happens that the true value of the coefficient is |β|<1, then the the OLS estimator is consistent and converges at the usual √n rate.

But if instead the true value is β=1 (i.e we have in reality a pure random walk), then the OLS estimator is consistent but will converge "faster", at rate n (this is sometimes called a "superconsistent" estimator -since, I guess, so many estimators converge at rate √n).

In this case, to obtain its (non-normal) asymptotic distribution, we *have* to scale (ˆβ−β) by n (if we scale only by √n the expression will go to zero). Hamilton ch 17 has the details.

**Attribution***Source : Link , Question Author : clueless , Answer Author : Alecos Papadopoulos*