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.

  • $\begingroup$ What do you mean by uncertainty in the standard deviation? $\endgroup$
    – velut luna
    Apr 30, 2015 at 16:44
  • $\begingroup$ @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"? $\endgroup$ Apr 30, 2015 at 18:11
  • $\begingroup$ You said about a hundred points. Can use central limit theorem? $\endgroup$
    – velut luna
    Apr 30, 2015 at 18:26
  • $\begingroup$ @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. $\endgroup$ Apr 30, 2015 at 20:01

4 Answers 4


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

which matches the other answers.


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


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


[3] - WolframMathWorld h-Staatistic


[4] - StatLect Point estimation of the variance


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


  • $\begingroup$ 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! $\endgroup$ Sep 25, 2020 at 16:46
  • 2
    $\begingroup$ @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) $\endgroup$ 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!

  • 9
    $\begingroup$ It would be cool if you could give a citation for this! $\endgroup$
    – Charphacy
    Jul 23, 2019 at 21:04

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


  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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