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

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up vote 32 down vote accepted

Let $X_1, ..., X_n \sim N(\mu, \sigma^2)$. As shown in this thread, the standard deviation of the sample standard deviation,

$$ s = \sqrt{ \frac{1}{n-1} \sum_{i=1}^{n} (X_i - \overline{X}) }, $$


$$ {\rm SD}(s) = \sqrt{ E \left( [E(s)- s]^2 \right) } = \sigma \sqrt{ 1 - \frac{2}{n-1} \cdot \left( \frac{ \Gamma(n/2) }{ \Gamma( \frac{n-1}{2} ) } \right)^2 } $$

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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This should really be a comment. – Silverfish Oct 29 '15 at 0:34
@Silverfish, it should have been originally, but might be worth grandfathering in this case. – gung Oct 29 '15 at 0:57
@gung It is a link to a Wikipedia section rather than a page, and these are especially vulnerable to link rot (because an edit to the wiki article - e.g. retitling the section to "Sampling distribution" or similar - can lead to the link simply pointing to the "variance" article, which wouldn't add much value for a future reader). I suppose one advantage to leaving this as an answer rather than a comment is that if any future user notices such a change, they might edit this answer. – Silverfish Oct 29 '15 at 1:04

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

I think both the concept and the terminology of "SD of SD" is too slippery to tackle. But it is easier to think about the confidence interval of a SD. You compute the SD from a sample of data and want to compute a confidence interval that is 95% (or some other confidence level) likely to contain the true SD of the population from which the data were sampled.

Detailed explanation

Free web calculator

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I can do Monte Carlo, I just wanted to do in a more 'sciency' way; still you're right that the distribution is not normal, so this sd will be useless for testing. – mbq Jul 26 '10 at 19:50
For what it's worth, I'm uncomfortable with the statement "a confidence interval that is 95%... likely to contain the true SD" (or, stated more explicitly in the linked page: "you can be 95% sure that the CI computed from the sample SD contains the true population SD"). I think these statements flirt w/ reinforcing a popular misconception, see here, eg, for a related discussion on CV. – gung May 17 '12 at 16:37
What is "I think both the concept and the terminology of "SD of SD" is too slippery to tackle" supposed to mean? The sample standard deviation is a random variable that has a standard deviation. – Macro Jun 8 '12 at 12:18

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