# Standard deviation of standard deviation

What is an estimator of standard deviation of standard deviation if normality of data can be assumed?

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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 ${\rm SD}(s)$:

$$s \cdot \frac{\Gamma( \frac{n-1}{2} )}{ \Gamma(n/2) } \cdot \sqrt{\frac{n-1}{2} - \left( \frac{ \Gamma(n/2) }{ \Gamma( \frac{n-1}{2} ) } \right)^2 }$$

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+1 It's nice to see not only a better reply come along after almost two years, but a reply that provides more useful detail than the references elsewhere in this thread. –  whuber May 16 '12 at 21:55

I suppose that you are looking for the distribution of the sample variance.

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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 !

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Isn't that a function of estimator is still an estimator? I still don't know \sigma, only X_i. –  mbq Jul 26 '10 at 16:45
ok, then you will possibly estimate the square root of the variance of the estimation of the square root of the variance... right :) should be something like $\hat{\sigma}/n$ ? –  robin girard Jul 26 '10 at 17:09
What Srikant found (and what seems confirmed at PhysicsForums) there should be $\sqrt{2}$, so rather $\hat{\sigma}\frac{\sqrt{2}}{2n}$. –  mbq Aug 1 '10 at 15:34
Aww, those comments locks; $\frac{\hat{\sigma}}{\sqrt{2n}}$. At least this one gives the result in agreement with bootstrap. –  mbq Aug 1 '10 at 15:58