# Question on central limit theorem

I have a question regarding central limit theorem! I understand that as the sample size increases and gets large enough, the sampling distribution of the sample mean can said to be approximated by a normal distribution.

But is that also the same for the sampling distribution of sample median and sample variance?

• Possible duplicate of Central limit theorem for sample medians Oct 6 '18 at 11:53
• Btw, it is unrealistic to assume sample variance going to normal since $S^2$ is always positive. Oct 6 '18 at 11:55
• @Gunes That remark is misleading. Indeed, the sample variance does become asymptotically normal provided the fourth moment of the underlying distribution is finite. You might have forgotten about the effects of the standardization in the statement of the CLT.
– whuber
Oct 6 '18 at 15:37
• Hi: It's true for any "statistic" provided that certain assumptions hold. These assumptions vary depending on which CLT one is using. I think there are probably 4 or 5 different CLT variants that differ with respect to dependence assumptions and existence of finite moment assumptions. Oct 12 '18 at 10:27
• Hi: The link below talks about variations and has some references. It may be helpful. johndcook.com/blog/central_limit_theorems Oct 12 '18 at 10:43

Yes, the central limit theorem can be used for sample medians. For the second question about sample variance, yes, assuming some extra conditions (fourth moments of the underlying distribution you are sampling from must exist), the central limit theorem can be used to get a limiting distribution for the sample variance. This is a similar question using the delta method

In fact, there are multiple versions of the central limit theorem, using somewhat different assumptions. But for most statistics (mean, median, sample variance, sample interquartile range, ...) one of the central limit theorems can be used to get some normal limiting distribution for the statistic. One important exception is extreme value statistics, like sample maximum and similar, which do not have normal limiting distributions.

Asymptotic distribution of the sample variance: The sample variance does converges in distribution to the normal (under mild conditions), but there are better asymptotic approximations to the distribution, which respect the support of the sample variance. You can find relevant results and proof in O'Neill (2014) (Result 14, pp. 285, 293-4). So long as the kurtosis $$\kappa$$ of the underlying distribution is finite, you have:

$$\sqrt{n} \Bigg( \frac{S_n^2}{\sigma^2} - 1 \Bigg) \overset{\text{Dist}}{\longrightarrow} \text{N}(0, \kappa-1).$$

(The result is usually proved by writing out a decomposition of the sample variance and then using a combination of the classical central limit theorem and Slutsky's theorem.) For large $$n$$, this gives the large-sample approximating distribution:

$$S_n^2 \overset{\text{Approx}}{\sim} \text{N}(\sigma^2, n(\kappa-1) \sigma^4).$$

This approximation has the drawback that its support extends over negative values, which cannot be sample variance values. A better (asymptotically equivalent) aproximating distribution is:

$$S_n^2 \overset{\text{Approx}}{\sim} \sigma^2 \cdot \frac{\text{Chi-Sq}(DF_n)}{DF_n} \quad \quad \quad DF_n \equiv \frac{2n}{\kappa - (n-3)/(n-1)}.$$

This latter approximation respects the support of the sample variance, and it corresponds to the exact distribution of the sample variance in the special case where the underlying distribution is normal. For further discussion, see the linked paper.