# 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?

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, 2011 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. Dec 6, 2011 at 15:40
• Bootstrap is the way to go. Dec 6, 2011 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, 2011 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..? Dec 6, 2011 at 9:41
• Exactly, or the confidence interval of the estimation of my SD.
– SRKX
Dec 6, 2011 at 10:37