Standard deviation of standard deviation What is an estimator of standard deviation of standard deviation if normality of data can be assumed?
 A: Let $X_1, ..., X_n \sim N(\mu, \sigma^2)$. As shown in this thread, the standard deviation of the sample standard deviation, 
$$ 
s = \sqrt{ \frac{1}{n-1} \sum_{i=1}^{n} (X_i - \overline{X}) }, $$ 
is
$$ {\rm SD}(s) = \sqrt{ E \left( [E(s)- s]^2 \right) } = \sigma \sqrt{  1 - \frac{2}{n-1} \cdot \left( \frac{ \Gamma(n/2) }{ \Gamma( \frac{n-1}{2} ) } \right)^2 } $$ 
where $\Gamma(\cdot)$ is the gamma function, $n$ is the sample size and $\overline{X} = \frac{1}{n} \sum_{i=1}^{n} X_i$ is the sample mean. Since $s$ is a consistent estimator of $\sigma$, this suggests replacing $\sigma$ with $s$ in the equation above to get a consistent estimator of ${\rm SD}(s)$. 
If it is an unbiased estimator you seek, we see in this thread that $ E(s)
= \sigma \cdot \sqrt{ \frac{2}{n-1} }  \cdot \frac{ \Gamma(n/2) }{ \Gamma( \frac{n-1}{2} ) } $, which, by linearity of expectation, suggests 
$$ s \cdot  \sqrt{ \frac{n-1}{2} } \cdot \frac{\Gamma( \frac{n-1}{2} )}{ \Gamma(n/2) } $$
as an unbiased estimator of $\sigma$. All of this together with linearity of expectation gives an unbiased estimator of ${\rm SD}(s)$:  
$$ s \cdot \frac{\Gamma( \frac{n-1}{2} )}{ \Gamma(n/2) } \cdot \sqrt{\frac{n-1}{2} - \left( \frac{ \Gamma(n/2) }{ \Gamma( \frac{n-1}{2} ) } \right)^2 } $$ 
A: Assume you observe $X_1,\dots,X_n$ iid from a normal with mean zero and variance $\sigma^2$. The (empirical) standard deviation is the square root of the estimator $\hat{\sigma}^2$ of $\sigma^2$ (unbiased or not that is not the question). As an estimator (obtained with $X_1,\dots,X_n$), $\hat{\sigma}$ has a variance that can be calculated theoretically. Maybe what you call the standard deviation of standard deviation is actually the square root of the variance of the  standard deviation, i.e. $\sqrt{E[(\sigma-\hat{\sigma})^2]}$?  It is not an estimator, it is a theoretical quantity (something like $\sigma/\sqrt{n}$ to be confirmed) that can be calculated explicitely !
A: @Macro provided a great mathematical explanation with equation to compute. Here is a more general explation for less mathematical people.
I think the terminology "SD of SD" is confusing to many. It is easier to think about the confidence interval of a SD. How precise is the standard deviation you compute from a sample? Just by chance you may have happened to obtain data that are closely bunched together, making the sample SD much lower than the population SD. Or you may have randomly obtained values that are far more scattered than the overall population, making the sample SD higher than the population SD.  
Interpreting the CI of the SD is straightforward. Start with the customary assumption that your data were randomly and independently sampled from a Gaussian distribution. Now repeat this sampling many times. You expect 95% of those confidence intervals to include the true population SD.
How wide is the 95% confidence interval of a SD? It depends on sample size (n) of course. 
n:   95% CI of SD
2:   0.45*SD to 31.9*SD
3:   0.52*SD to 6.29*SD
5:   0.60*SD to 2.87*SD
10:  0.69*SD to 1.83*SD
25:  0.78*SD to 1.39*SD
50:  0.84*SD to 1.25*SD
100: 0.88*SD to 1.16*SD
500: 0.94*SD to 1.07*SD
Free web calculator
