Standard error of the sample standard deviation

**Taqman** · 06-09-2010 08:33 AM

Hi, all

Does anyone let me know the formula for "the standard error of the sample standard deviation" when no specific distribution is assumed for an i.i.d random sequence? I also want to know a (famous if possible) reference book or paper.

Thanks a lot for your help.

**Dason** · 06-09-2010 10:32 AM

There is some helpful info here... http://mathworld.wolfram.com/SampleV...tribution.html

If you really need a reference in a book... off the top of my head I know that in an exercise at the end of (chapter 4? maybe 5?) in Casella and Berger you're asked to show that the var(s^2) is what it is so you'd have the final result there.

**BGM** · 06-09-2010 11:52 AM

Thanks Dason for reminding the useful textbook

it is in the problem 5.8, page 257
In part a) we are asked to show
$S^2 = \frac {1} {2n(n-1)} \sum_{i=1}^n\sum_{j=1}^n(X_i - X_j)^2$ , and use this result to derive
$Var[S^2] = \frac {1} {n} \left( E[(X - E[X])^4] - \frac {n - 3} {n - 1} E[(X - E[X])^2]^2 \right)$

Also note that since $S^2$ is an unbiased estimator of $\sigma^2$ ,
the mean squared error and the variance is the same. See
http://en.wikipedia.org/wiki/Mean_sq...error#Examples

I have gone through a quite tedious way, which lead to the same result:
(you can expand the answer above to see)

$S^2 = \frac {1} {n - 1} \sum_{k=1}^n(X_k - \bar{X})^2$

$= \frac {1} {n - 1} \left(\sum_{k=1}^nX_k^2 - n\bar{X}^2 \right)$

$= \frac {1} {n - 1} \left[\sum_{k=1}^nX_k^2 -\frac {1} {n} \left( \sum_{k=1}^n X_k^2 + 2\sum_{i=1}^{n-1}\sum_{j=i+1}^nX_iX_j \right) \right]$

$= \frac {1} {n} \sum_{k=1}^nX_k^2 -\frac {2} {n(n-1)} \sum_{i=1}^{n-1}\sum_{j=i+1}^nX_iX_j$

Therefore, $Var[S^2]$

$= Var\left[\frac {1} {n} \sum_{k=1}^nX_k^2 -\frac {2} {n(n-1)} \sum_{i=1}^{n-1}\sum_{j=i+1}^nX_iX_j \right]$

$= \frac {1} {n} Var[X_1^2]+ \frac {4} {n^2(n-1)^2} \sum_{i=1}^{n-1}\sum_{j=i+1}^n\sum_{k=1}^{n-1}\sum_{l=k+1}^nCov[X_iX_j, X_kX_l]$ $-\frac {4} {n^2(n-1)} \sum_{k=1}^n\sum_{i=1}^{n-1}\sum_{j=i+1}^nCov[X_k^2, X_iX_j]$

$= \frac {1} {n} E[X_1^4] - \frac {1} {n} E[X_1^2]^2 + \frac {4} {n^2(n-1)^2} \frac {n(n-1)} {2}$ $\left( Cov[X_1X_2, X_1X_2] + 2(n-2)Cov[X_1X_2, X_1X_3] + \frac {n^2 - 5n + 2} {2} Cov[X_1X_2,X_3X_4] \right)$
$-\frac {4} {n^2(n-1)} n \left( (n-1)Cov[X_1^2, X_1X_2]+ \frac {(n-1)(n-2)} {2} Cov[X_1^2, X_2X_3] \right)$

$= \frac {1} {n} E[X_1^4] - \frac {1} {n} E[X_1^2]^2 + \frac {2} {n(n-1)}$ $\left( E[X_1^2]E[X_2^2] - E[X_1]^2E[X_2]^2 +2(n-2)(E[X_1^2]E[X_2]E[X_3] - E[X_1]^2E[X_2]E[X_3]) \right)$
$-\frac {4} {n(n-1)} \left( (n-1)(E[X_1^3]E[X_2] - E[X_1^2]E[X_1]E[X_2]) \right)$

$= \frac {1} {n} E[X^4] - \frac {1} {n} E[X^2]^2$ $+ \frac {2} {n(n-1)} E[X^2]^2 - \frac {2} {n(n-1)} E[X]^4 + \frac {4(n-2)} {n(n-1)} E[X^2]E[X]^2 - \frac {4(n-2)} {n(n-1)} E[X]^4$
$- \frac {4} {n} E[X^3]E[X] + \frac {2} {n} E[X^2]E[X]^2$

$= \frac {1} {n} E[X^4] - \frac {4} {n} E[X^3]E[X]+ \frac {3 - n} {n(n-1)} E[X^2]^2 + \frac {4(2n - 3)} {n(n-1)} E[X^2]E[X]^2$ $+ \frac {2(3 - 2n)} {n(n - 1)}E[X]^4$

**Dragan** · 06-09-2010 12:49 PM

This short article may also be of some help:

http://www.eecs.umich.edu/~fessler/p.../tr/stderr.pdf

**Dason** · 06-09-2010 12:57 PM

I'm giving you a thanks BGM because I remember doing that problem and there's no way I would subject myself to that again.

**Taqman** · 06-10-2010 08:50 PM

Thank you for your perfect lecture.
I would like to appreciate it very much.

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