Calculating uncertainty in standard deviation

I have a distribution with literally an infinite number of potential data points. I need the standard deviation. I generate about a hundred points and take the standard deviation of the points. This gives a hopefully good approximation of the true standard deviation, but it won't, of course, be exact. How do I estimate the uncertainty in the standard deviation? This seems like a very basic question, but web searching hasn't provided any solution. If I missed it somehow, my apologies.

• What do you mean by uncertainty in the standard deviation? Apr 30, 2015 at 16:44
• @KYson I mean that the distribution has some value of standard deviation. I can get some approximation to that value by pulling a number of points out of it and taking their standard deviation. But the number that I get won't be the exact S.D. of the entire distribution. But surely, I am thinking, it must be good to within some "window". What is "the width of that window"? Apr 30, 2015 at 18:11
• You said about a hundred points. Can use central limit theorem? Apr 30, 2015 at 18:26
• @Kyson I've looked it over. I don't see how I can use it. I'm admittedly not the sharpest tool in the shed sometimes. Any details you are thinking about will be most welcome. Apr 30, 2015 at 20:01

the answer to OP's question depends on whether or not the mean of the distribution is known. if the mean is known ( for example if you know that the mean of you sampled population should eventually average out to be zero) than the problem is a little different, not by much but I did not do the research to find out to what extent, [4] might help. I am assuming the mean is not known.

so you have a sample of 100 values, for which you don't know the mean or variance. you can calculate the unbiased variance estimator:[1] $$S^2 = variance\ estimator = \frac{1}{n-1}\sum_i\left(x_i- \frac{\sum x}{n}\right)^2 = \frac{1}{n(n-1)}\sum_{i,j}\frac{(x_i-x_j)^2}{2}$$

but you also want to know how accurate this estimation of the sample variance is. so in other words you want the variance of the variance estimator. $$Var\left(S^2\right)$$ this is shown in [2] to be: $$Var\left(S^2\right)=\frac{1}{n}\left(\mu_4-\frac{n-3}{n-1}\mu_2^2\right)$$

$$where\ \ \mu_k := E[(X-E[X])^k]$$ ($$\mu_k$$ are the centered moments) and so you get: $$\sigma^2:=\mu_2 = S^2 \pm \sqrt{\frac{1}{n}\left(\mu_4-\frac{n-3}{n-1}\mu_2^2\right)}$$ but regrettably this is not given as a function of you're data points (it's a function of $$\mu_4,\mu_2$$ both of which are unknown), what you really want is an unbiased estimator for $$Var\left(S^2\right)$$. I couldn't completely find the right way to achieve this. unbiased estimators of nonlinear function are in general not easy to find (in this case I think it's probably impossible) so as far as I know you will have to deal with some bias. in attempt to minimise this bias you could just find good estimators for $$\mu_4,\mu_2$$, and plug them in to $$\sqrt{\frac{1}{n}\left(\mu_4-\frac{n-3}{n-1}\mu_2^2\right)}$$ and ignore the bias that arises from the nonliniearity. the unbiased estimators for centered moments ($$\mu_4,\mu_2$$) are called the H-statistics, they are pretty easy to find online or in books and are not too complex to calculate. for my uses the H-statistic for $$\mu_4$$ is a pretty terrible expression [3], and as I already said, using it is not without bias, so what i decided to do was assume Xi are close enough to gaussian so that $$\mu_4=3\mu_2^2$$ and thus I got: $$Var\left(S^2\right)= \frac{1}{n}\left(\mu_4-\frac{n-3}{n-1}\mu_2^2\right)= \frac{1}{n}\left(3\mu_2^2-\frac{n-3}{n-1}\mu_2^2\right)= \frac{1}{n}\left(3-\frac{n-3}{n-1}\right)\mu_2^2= \frac{1}{n}\left(\frac{2n}{n-1}\right)\mu_2^2= \frac{2\mu_2^2}{n-1}$$

and so now (assuming $$\mu_4=3\mu_2^2$$): $$\sigma^2:=\mu_2 = S^2 \pm \sqrt{\frac{2}{n-1}} \sigma^2\approx S^2 \pm \sqrt{\frac{2}{n-1}} S^2$$

to finish up, OP asked for the uncertainty in S and not in $$S^2$$. so if you use propagation of uncertainty [5] to evaluate how the uncertainty is affected by taking the square root:

($$SE$$ stands for Standard Error) $$SE[\sqrt{Y}]\approx\frac{1}{2\sqrt{E[Y]}}SE[Y]$$ $$\sigma = S \pm \frac{1}{2\sqrt{S^2}}\sqrt{\frac{2}{n-1}}S^2= S \pm \frac{S}{\sqrt{2n-2}}$$

references:

[1] - A few properties of sample variance By Eric Benhamou

https://arxiv.org/pdf/1809.03774.pdf

[2] - Variance of Simple Variance By Eungchun Cho & Moon Jung Cho

http://www.asasrms.org/Proceedings/y2008/Files/300992.pdf

[3] - WolframMathWorld h-Staatistic

https://mathworld.wolfram.com/h-Statistic.html

[4] - StatLect Point estimation of the variance

https://www.statlect.com/fundamentals-of-statistics/variance-estimation

[5] - Wikipedia Propagation of uncertainty 26/09/2020

https://en.wikipedia.org/wiki/Propagation_of_uncertainty

• Thanks for your great work here! One question: Your final result looks like it disagrees with the accepted answer. Is that the way you read it? Thanks again! Sep 25, 2020 at 16:46
• @bob.sacamento all the answers match, I added a bit to the end of the answer to show the answers match. (the conversion between the expression for sigma squared and sigma is what was missing earlier) Sep 25, 2020 at 23:18

If you want to find out the uncertainty or standard error (SE) in the standard deviation of a chosen sample, then you can simply use $SE(\sigma) = \frac{\sigma}{\sqrt{2N - 2}}$, where $N$ is the number of data points in your sample.

Hope that helps!

• It would be cool if you could give a citation for this! Jul 23, 2019 at 21:04

If you're allowed to take that sample repeatedly, it's basically bootstrapping.

Procedure:

1. Draw 100 points

2. Calculate standard deviation

3. Repeat Steps 1 & 2 a lot of times (empirically, I've found 5-10,000 to be enough), keeping track of the results of step 2.

4. Examine the distribution of estimates from Step 2 with whatever tools you'd like -- histograms, sample moments, etc.

This is pretty standard and can be answered by searching "Confidence interval of a standard deviation." Here are the steps:

Step 1) Pick a confidence level. The confidence level is the probability of your interval estimate containing the actual population standard deviation. Common choices for confidence levels are 90%, 95%, 99%. I'll work through the steps for a 90% confidence interval.

Step 2) Use a chi-squared distribution to find the left and right critical values $$\chi^2_L, \chi^2_R$$ for your chosen confidence level. The degrees of freedom are the sample size minus one, in this case, $$99$$. For your example, the critical values for 90% confidence would be approximately $$\chi^2_L = 77.93$$, $$\chi^2_R = 124.32$$

Step 3) Use your sample standard deviation $$s$$ and sample size $$n$$ to find the left and right endpoints of the confidence interval for the population standard deviation $$\sigma$$ via the formula: $$s\sqrt{ \frac{n-1}{\chi^2_R}} < \sigma < s\sqrt{ \frac{n-1}{\chi^2_L}}.$$ In your example, whatever your value for $$s$$ was, you can be 90% confident that the true value of $$\sigma$$ is between $$s \sqrt{ \frac{99}{124.32}} = 0.892s$$ on the low end, and $$s \sqrt{ \frac{99}{77.93}} = 1.127s$$ on the high end.
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