# Can bootstrapping be used to find the standard error of variance?

I am a bit new to the field and had a question regarding bootstrapping. As bootstrapping gives an idea of the standard deviation of the sampling distribution of the mean, does it also give an idea of the standard deviation of the variances of the samples ? If yes, then how? Also, if anyone can provide any reading resources for the topic, that would be very helpful.

I would be grateful for any help.

• Do you have a particular dataset in mind? For what purpose do you want the standard error of the SD? Making a CI for population SD? Testing a hypothesis? Please be specific. Jul 12 at 18:19
• I want to make a CI for population SD. Please let me know if you need any more information. Thanks ! Jul 12 at 18:22
• It is not customary (nor IMHO appropriate) to use the standard error of the sample SD in making such a CI. See my bootstrap CI below. Jul 12 at 19:35

I suppose your data is not normal. If data were normal, there is an exact procedure for a CI of $$\sigma$$ based on pivoting the quantity $$\frac{(n-1)S^2}{\sigma^2} \sim \mathsf{Chisq}(\nu = n-1).$$

Suppose you have right-skewed data with the following summary descriptions:

summary(x);  length(x);  sd(x)
Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
16.55   34.09   48.53   51.87   67.09  104.60
 40
 23.04401
boxplot(x, horizontal=T) There are many styles of 95% nonparametric bootstrap confidence intervals. One of them is illustrated below, giving the CI $$(19.9,\, 28.6).$$ Because $$\sigma > 0,$$ this style of bootstrap CI is based on ratios rather than difference. [Thus, it does not use a 'standard error' for the sample standard deviation.]

s.obs = 23.04401
set.seed(1212)
d.re = replicate(5000, sd(sample(x,40,rep=T))/s.obs)
UL.re = quantile(d.re, c(.975,.025))
CI = s.obs/UL.re;  CI
97.5%     2.5%
19.90666 28.55302


Note: The fictitious data for this demonstration was sampled in R as shown below. The population standard deviation for the distribution $$\mathsf{Gamma}(5, .1)$$ is $$\sigma = \sqrt{500} = 22.3607.$$

set.seed(721)
x = rgamma(40, 5, .1)


Incorrectly assuming normal data would have given the bogus "95% CI" $$(18.88, 29.59)$$ for $$\sigma.$$ This CI also happens to include the true value of $$\sigma$$ (which would remain unknown in a real application).

sqrt(39*var(x)/qchisq(c(.975,.025),39))
 18.87676 29.58933