5
$\begingroup$

Assume $X_1,\ldots,X_n$ are i.i.d. with unknown distribution $\mathcal D$ - we only know it is not normal and has finite variance.

Is there a way to give confidence intervals for the variance of $\mathcal D$? Can we base the confidence interval on the sample variance $\hat \sigma^2 = \frac1{n-1}\sum_{i=1}^n (X_i - \bar X)^2$ with $\bar X = \frac1n\sum_{i=1}^n X_i$ the sample mean?

If it helps, we can restrict ourselves to distributions with mean $0$. (From my application, I can compute the exact mean $\mu$ of $\mathcal D$ and then consider $Y_i := X_i - \mu$ to estimate the variance.)

I know the approaches via $\chi^2$-distribution if we assume a normal distribution $\mathcal D = \mathcal N(\mu,\sigma^2)$, but I as noted above $\mathcal D$ is not normal.

$\endgroup$

3 Answers 3

Reset to default
2
$\begingroup$

In practical terms, if the distribution is unknown and one has a lot of data, one can assume that the distribution of the sample variance converges to a Gaussian one (e.g. see here). Then the confidence interval can be computed from this.

One can also do bootstrapping to approximate the statistic's distribution, and use it to estimate the confidence interval. This is (asymptotically) quite accurate.

Another method might be to use a Bayesian approach (see: A Bayesian perspective on estimating mean, variance, and standard-deviation from data by Oliphant). (This method is built into scipy already :].) Essentially, it finds that, with an "ignorant" Bayesian prior, the sample variance follows an inverted Gamma distribution, from which confidence intervals can be constructed.

See also this question about the distribution of the sample variance and this one, which is also related.

$\endgroup$
1
  • $\begingroup$ The Bayesian approach in scipy for a distribution from [0, 1] produces confidence intervals outside of [0, 1], so doesn't seem appropriate for that type of distribution $\endgroup$
    – endolith
    Jul 21, 2022 at 16:07
1
$\begingroup$

If the distribution is unknown then it will be a problem of Non-parametric or Distribution-Free.And in non-parametric we go with median or p-th quartile. So, I think there is no confidence intervals for the variance you define.

$\endgroup$
2
  • $\begingroup$ >I think there is no confidence intervals for the variance you define. Do you mean, it can proven that one cannot give such intervals (in general) or are there just no results known? $\endgroup$
    – Sebastian
    Mar 24, 2013 at 16:28
  • $\begingroup$ @Sebastian:I am not familiar with such result. $\endgroup$
    – A.D
    Mar 24, 2013 at 17:43
1
$\begingroup$

This is late, but I was also looking into this and figured it would be worth answering the question for other people who is also googling for the answer. You can derive concentration inequalities (and thus confidence intervals) for the sample variance through the use of $U$-statistics since the sample variance is a $U$-statistic. See https://arxiv.org/abs/1712.06160 for more details.

$\endgroup$
2
  • 1
    $\begingroup$ Your answer basically comes down to a link to an off-site source. Such answers are generally frowned upon here, as answers on MSE should be as self-contained as possible, and off-site links can suffer from "rot". Please take the time to explain why that link is relevant and to summarize the major results or ideas. $\endgroup$
    – Xander Henderson ♦
    Jan 22, 2019 at 1:52
  • $\begingroup$ @Ojash Thanks for digging this out; since you seem to have gone through this, would you summarize the result here? $\endgroup$
    – Sebastian
    Feb 18, 2019 at 8:31

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .