# What is the precision of standard deviation estimates with small samples?

I was asked today to "quantify" the precision of an estimated the standard deviation from a small sample, I was not sure how to answer.

The case is quite simple, I have a sample of $n=25$ measures (returns as you would have guessed). I used the classic unbiased estimator for the standard deviation:

$$\sigma_x = \sqrt{\frac{1}{N-1}\sum_{n=1}^n (x_i-\bar{x})^2}$$

The underlying question was : how much data do we need for the standard deviation to be statistically meaningful.

I read here that computing the standard error of the standard deviation is difficult to estimate, but I wanted to know if there was a common procedure used by you guys in general?

## 2 Answers

Treat the estimate of standard deviation as a random variable. Then you can bootstap the sample estimate and generate t-statistics and associated confidence intervals for your statistics. I describe a generic boostrap process on this post.

• Right but if I have only 25 measures, does it make sense to do the bootstrapping with sample of size 15 for example? – SRKX Dec 6 '11 at 8:01
• In each re-sample I would sample at least 'n' times (i.e. 25 in your example) - or sample 2n or 3n times - with replacement from your 25 measures. Another words take, say, 75 draws from the empirical distribution and calculate the statistic of interest. Then repeat this procedure say a 1,000 times so you can generate a confidence interval. – Ram Ahluwalia Dec 6 '11 at 15:40
• Bootstrap is the way to go. – Samik R Dec 6 '11 at 17:54

Actually you should be interested by the Berry Essen's theorem which precises the rate of convergence of the central limit theorem.

Given i.i.d. $X_1,\dots, X_n \sim X$

1) GLN : assuming $E(X)<\infty$ then $\overline{X}_n-E(X)\to 0$

2) CLT ("rate" of the GLN) : assuming $E(X^2)<\infty$ then $\frac{\sqrt{n}}{\sigma^2} \big(\overline{X}_n-E(X)\big)\to N(0,1)$

3) Berry Essen ("rate" of the CLT) : assuming $E(X^3)<\infty$ , then

$\sup_{x\in \mathbb{R}}\bigg| \,F_{\frac{\sqrt{n}}{\sigma^2} \big(\overline{X}_n-E(X)\big)}(x) - F_{N_{0,1}}(x) \bigg| \leq \frac{0.34445 E|X|^3 + 0.16844}{\sqrt{n}}$

Where $F_{}$ holds for the CDF.

This is an upper bound (of the order $\sqrt{n}$) usable for your CLT approximation.

• I'm really sorry but I'm not really how to use this for my problem? Could you explain a bit more? – SRKX Dec 6 '11 at 7:58
• Sorry @SRKX I have not responded to your question I thought you were interested by the rate of convergence of the CLT estimate. To be clear, you are interested by the standard error of your standard deviation..? – Beer4All Dec 6 '11 at 9:41
• Exactly, or the confidence interval of the estimation of my SD. – SRKX Dec 6 '11 at 10:37