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.
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.