Given a sample of the variable $X$ = ($X_1$, .., $X_n$).

If the $X$ follows a normal distribution, I know that we can estimate the confidence interval of the variance of $X$ by using the $\chi^{square}$ distribution.

However, now I have a sample of an variable $Y$ which is not normally distributed (in fact, unknow distribution). Could you please tell me how could I calculate the confidence interval for the variance ? If there is any relevant article, I am very appreciated.

Thank you very much for your help!

  • $\begingroup$ Finite sample, no. Asymptotically, yes, if we assume some mild regularity conditions. Variance is just estimating the 2nd moment. Use the law of large numbers and central limit theorem. $\endgroup$ Commented Jul 27, 2021 at 21:28
  • $\begingroup$ Does this answer your question? Bootstrap data in order to get an estimate of the variance You can ignore the parts about the "block bootstrap" and time series; bootstrapping is a standard way to do this, although sometimes it can fail. $\endgroup$
    – EdM
    Commented Jul 27, 2021 at 21:28
  • $\begingroup$ @Shang Zhang: I don't think we could you the central limit theorem because it will calculate directly the variance, instead of the expectation of variance. $\endgroup$ Commented Jul 27, 2021 at 21:33
  • $\begingroup$ @Edm: sadly no because i want to calulcate the confidence interval of the variance (other than just estimate it). $\endgroup$ Commented Jul 27, 2021 at 21:36
  • $\begingroup$ You didn't understand my point. If we can asymptotically estimate the first moment, and such estimator's asymptotic variance, without distributional assumptions, then what's there to stop you from doing the same on the second moment? Further hint: compute the sample 4th moment. $\endgroup$ Commented Jul 27, 2021 at 21:40

2 Answers 2


It would be better to say "construct" a confidence interval instead of "estimate" a confidence interval. The confidence interval is the interval estimate of the unknown population parameter. I believe @Shang Zhang is suggesting you estimate the variance of the sample variance and use this to construct an asymptotic confidence interval. Here is a thread that derives the variance of the sample variance, "Variance of sample variance?". You could use this to invert a Wald test to construct a confidence interval for the unknown population variance. That is,

$$\mbox{Var}(S^2)={\mu_4\over n}-{\sigma^4\,(n-3)\over n\,(n-1)},$$ where $\mu_4=\mathbb{E}[(X-\mu)^4]$ is the fourth central moment of $X$. The resulting confidence interval would be

$$s^2\pm1.96\cdot\hat{\text{se}},$$ where $\hat{\text{se}}=\sqrt{{\hat{\mu}_4\over n}-{s^4\,(n-3)\over n\,(n-1)}}$. If the sample size is small or the sample variance estimate is small (or both), using a log link function will help to maintain the coverage probability of the interval, i.e.

$$\text{exp}[\text{log}\{s^2\}\pm1.96\cdot\hat{\text{se}}/s^2],$$ where $\hat{\text{se}}/s^2$ is the estimated standard error of $\text{log}\{s^2\}$ based on the delta method.


Here is the basic idea.

Let $E(x) = m$ and $E(x^2) = v$. You want to estimate $v-m^2$ with an estimator, and also estimate the variance of the estimator. (What a mouthful.)

Let $\bar{x}$ and $\bar{x^2}$ be the sample average of $x$ and $x^2$ respectively.

From CLT, we have $\sqrt{n}(b-\beta) \rightarrow N(0, V)$ where vector $b = (\bar{x}, \bar{x^2})'$ and vector $\beta = (m, v)'$.

$V$ can be consistently estimated from the data (involving up to 4th order sample moments).

$v - m^2$ is a non-linear scalar function of $\beta$ defined as: $v - m^2 = g(\beta) = g(m, v) = v - m^2$.

Then $\sqrt{n}[g(b) - g(\beta)] \rightarrow N[0, (-2m, 1) V (-2m, 1)']$.


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