In many statistics papers, authors suggest a new data analysis methodology and prove its properties such as consistency or asymptotic normality. I think it's a kind of tradition or custom. I understand that consistency is important, but I don't understand why asymptotic normality is so important.

It is large sample property. (written 'large', read 'infinite') In real data analysis, we never have infinite sample. Even though the estimator is asymptotically normal, its distribution in a realistic sample size may be far from normal distribution.

  • 5
    $\begingroup$ "Even though the estimator is asymptotically normal, its distribution is far from normal distribution in a realistic sample size." --- what is the basis for this claim? Sometimes, it's true that approximate asymptotic behaviors don't "kick in" until very large sample sizes, but in fact you do see it in practice, sometimes at quite modest sample sizes. $\endgroup$
    – Glen_b
    Nov 27, 2015 at 1:02
  • $\begingroup$ @Glen_b It was unclear, so I edited. I meant it 'may not be normal' $\endgroup$
    – user67275
    Nov 27, 2015 at 1:58
  • 2
    $\begingroup$ From experience, it can be really problematic if the estimator is not asymptotically normal: you may have a good estimate, but it can be really hard to put any sort of degree of certainty on that estimate. The next go-to option is often the bootstrap. Turns out that doesn't work well either for these problems. Very unpleasant. $\endgroup$
    – Cliff AB
    Nov 27, 2015 at 5:51

2 Answers 2


It is for example useful to do so in order to be able to quantify the sampling uncertainty of an estimator, or the null distribution of a test.

Recall that normal random variables take 95% of their realizations in the interval $\mu\pm1.96\sigma$. So if you can demonstrate that (typically, a scaled version of) an estimator is asymptotically normal, then you know it behaves normally at least in large samples, so you can easily construct confidence intervals, for example.

Whether or not the approximation is useful to settings in which (as always in practice) your sample is finite is in general unfortunately indeed not known analytically - if could derive the finite-sample distribution analytically, that is what we would work with. Unfortunately, that only works in very rare cases (for example, when sampling from a normal distribution, the t-statistic follows a t distribution).

Typically, simulations are then used to at least get an idea of the usefulness of the approximation in relevant cases.


Knowing the distribution of the random variable is knowing everything that is knowable about it. The estimators are random variables because they're functions of random samples. Therefore, it's not just a tradition but an ultimate goal of statistical analysis to establish the probability distribution of the metric. Often, small sample properties of estimators are hard to determine, in these cases asymptotic properties are the next best things. So, this is not much about normality, but the asymptotic distribution of the variable. If it's normal, then it's even better. Due to CLT and laws of large numbers, in many cases the asymptotic distributions end up being normal.

  • 2
    $\begingroup$ The OP is correct to question this practice. It is good to know large-sample normality to help you feel comfortable, but only one one direction. Small sample performance can be arbitrarily bad. So overall the practice of studying asymptotics is overdone. $\endgroup$ Jan 24, 2022 at 17:29
  • $\begingroup$ You can’t argue against studying asymptotic properties. It Is just a part of understanding the estimator. If anything is overdone here then it is using the assumption properties on small samples, but that’s a different concern $\endgroup$
    – Aksakal
    Jan 24, 2022 at 20:00
  • 1
    $\begingroup$ I think it's a good exercise for mathematical frequentist statisticians. Not so much for applied statisticians. $\endgroup$ Jan 24, 2022 at 22:08
  • $\begingroup$ Disagreed. It’s like saying that you don’t want to know what is $f(\infty)$ because your data is in finite range. $\endgroup$
    – Aksakal
    Jan 24, 2022 at 22:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.