# Standard Error of Sample Variance

I have a time-series of values $X_1, X_2, \ldots, X_t$, for which I compute sample variance:

$$\hat{\sigma}^2 = \operatorname{var}(X_1, \ldots, X_t)$$

(unabiased estimator using $\frac{1}{t-1})$.

In a subsequent calculation, I would like to use to shrink this variance estimate in proportion to its precision. How can I empirically estimate the standard error of this variance estimate?

In theory, the variance of sample variance (for normal distribution) is:

$$\operatorname{var}(\hat{\sigma}^2) = \frac{2}{t - 1} (\sigma^2)^2$$

where $\sigma^2$ is the true variance.

For the purporse of this calculation, can I safely assume that $\hat{\sigma}^2 = \sigma^2$, and thus let:

$$\operatorname{var}(\hat{\sigma}^2) = \frac{2}{t - 1} (\hat{\sigma}^2)^2$$

be an estimate of the variance of the sample variance?

Any help would be appreciated! Thanks!

• Your "theoretical estimate" is not an estimate because it's not a statistic, but a function of an unobservable parameter. You're finding the variance of a chi-square random vairable with $t-1$ degrees of freedom and then multiplying that random variable by $\sigma^2/(t-1)$. But the thing you say you can safely assume is an estimate. ${}\qquad{}$ Nov 10, 2014 at 16:42
• Thanks, I have corrected the wording in the post. Nov 10, 2014 at 16:45
• You should define which formula for sample variance you are using ( $\frac{1}{n}$ or $\frac{1}{n-1}$). Nov 10, 2014 at 16:52
• I have added the information to the post. Nov 10, 2014 at 16:53
• It's a valid estimate (assuming certain things about the population from which it was taken). Nov 10, 2014 at 17:57