Standard error of variance estimation

Question

We know that a mean estimation is the following

$\bar{x}=\frac{1}{n}\sum_{n=1}^{N}x_{n}$ and the standard error of the mean estimation is

$Var(\bar{x})=\sigma^{2}/n$.

Also the estimation of variance is

$s^{2}=\frac{1}{n-1}\sum_{n=1}^{N}(x_{n}-\bar{x})^{2}$

Is there a way to calculate the standard error of variance ??

$Var(s^{2})=\frac{1}{n-1}nVar((x_{n}-\bar{x})^{2})\frac{n}{n-1}Var(x_{n}^{2}+\bar{x}^{2}-2x_{n}\bar{x})$

How we are supposed to handle the term $Var((x_{n}-\bar{x})^{2})$?

Lastly $Var(x_{n})=\sigma^{2}$ and $\mathbb{E}(x_{n})=\mu$.

Answer 1

Confidence Intervals for Population Mean and Variance of Normal Data

Normal Data. Suppose you have vector x with $n=25$ observations, sampled using R, from $\mathsf{Norm}(\mu = 100, \sigma=25)$ and seek 95% confidence intervals for $\mu$ and $\sigma^2.$

set.seed(104)
x = round(rnorm(25, 100, 25), 2)
summary(x); var(x) 
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  58.94   86.98   96.02   99.94  115.69  136.20 
[1] 416.2467   # sample variance

95% CI for Population Mean. The t.test procedure in R provides the computation of a 95% CI ${91.52, 108.36}$ for $\mu:$

t.test(x)$conf.int 
[1]  91.51802 108.36118
 attr(,"conf.level")
 [1] 0.95

Here the CI does cover $\mu = 100,$ as should be the case, over the long run, for 95% of datasets of size 25 sampled from $\mathsf{Norm}(\mu = 100, \sigma=25).$

The formula is $\bar X \pm t^*2S/\sqrt{n},$ where $t^*$ cuts probability 0.25 from the upper tail of Student's t distribution with $n-1$ degrees of freedom. This formula is explained in most elementary statistics texts. You can find values of $t^*$ in printed tables of Student's t distribution.

(5% CI for Population Variance and SD. To get a 95% CI for $\sigma^2$ one uses the relationship $(n-1)S^2/\sigma^2 \sim \mathsf{Chisq}(n-1)$ to find $l$ and $U$ such that $$0.95 = P\left(L \le \frac{(n-1)S^2}{\sigma^2} \le U\right) = P\left(\frac{(n-1)S^2}{U} \sigma^2 \le \frac{(n-1)S^2}{U}\right),$$ where $L$ and $U,$ respectively,cut probability 0.025 from the lower and upper tails of (the asymmetrical) distribution $\mathsf{Chisq}(n-1).$ [You can find values of $ L$ and $U$ in printed tables of the chi-squared distribution. Their values from R are $L = 12.40$ and $U = 39.36.]$

qchisq(c(.025,.975), 24)
[1] 12.40115 39.36408

For the data in x we have the 95% CI $(253,8,805,6).$

24*var(x)/qchisq(c(.975,.025), 24)
[1] 253.7827 805.5640

A 95% CI $(15.93, 28,38)$ for $\sigma$ is found by taking square roots of endpoints. So this is one of the 95% of cases in which the CI includes $\sigma=25.$

sqrt(24*var(x)/qchisq(c(.975,.025), 24))
[1] 15.93056 28.38246

Notice that neither the CI for $\sigma^2$ nor the CI for $\sigma$ is of the form

$$\text{Point Estimate} \pm \text{Margin of Error}.$$ That is, the point estimate is typically not at the center of the CI.