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In his answer to my previous question, @Erik P. gives the expression $$ \mathrm{Var}[s^2]=\sigma^4 \left(\frac{2}{n-1} + \frac{\kappa}{n}\right) \>, $$ where $\kappa$ is the excess kurtosis of the distribution. A reference to the Wikipedia entry on the distribution of the sample variance is given, but the wikipedia page says "citation needed".

My primary question is, is there a reference for this formula? Is it 'trivial' to derive, and if so, can it be found in a textbook? (@Erik P. couldn't find it in Mathematical statistics and data analysis nor I in Statistical Inference by Casella and Berger. Even though the topic is covered.

It would be nice to have a textbook reference, but even more useful to have a (the) primary reference.

(A related question is: What is the distribution of the variance of a sample from an unknown distribution?)

Update: @cardinal pointed out another equation on math.SE: $$ \mathrm{Var}(S^2)={\mu_4\over n}-{\sigma^4\,(n-3)\over n\,(n-1)} $$ where $\mu_4$ is the fourth central moment.

Is there some way that to rearranged the equations and resolve the two, or is the equation in the title wrong?

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I don't think that formula is correct. –  cardinal Jun 6 '12 at 15:45
Related: –  cardinal Jun 6 '12 at 15:46
that related question was asked by @byron-schmuland –  Abe Jun 6 '12 at 16:57
I think you mean answered, not asked. The formula given in this question is incorrect; as Byron's answer nicely demonstrates. :) –  cardinal Jun 6 '12 at 16:58
@cardinal yes, I meant answered. I just wanted to ping him –  Abe Jun 6 '12 at 17:08

2 Answers 2

up vote 10 down vote accepted

Source: Introduction to the Theory of Statistics, Mood, Graybill, Boes, 3rd Edition, 1974, p. 229.

Derivation: Note that in the OP's Wikipedia link, $\kappa$ is not the kurtosis but the excess kurtosis, which is the "regular" kurtosis - 3. To get back to the "regular" kurtosis we have to add 3 in the appropriate place in the Wikipedia formula.

We have, from MGB:

$\text{Var}[S^2] = {1\over{n}}(\mu_4 - {{n-3}\over{n-1}}\sigma^4)$

which, using the identity $\mu_4 = (\kappa + 3)\sigma^4$, can be arranged to (derivation mine, so any errors are too):

$ = {1\over{n}}(\kappa \sigma^4 + {{n-1}\over{n-1}}3\sigma^4 -{{n-3}\over{n-1}}\sigma^4) = \sigma^4\left({\kappa \over{n}}+{3(n-1)-(n-3)\over{n(n-1)}}\right) = \sigma^4\left({\kappa\over{n}} + {{2}\over{n-1}}\right) $

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(+1) Nearly 40 years since the last edition, MGB is still the best beginning/intermediate introduction to math stat. It's a shame it has been out of print in the Western world for so long. –  cardinal Jun 6 '12 at 21:02
Many thanks for figuring this out. –  Abe Jun 6 '12 at 21:47
I found a pdf of MGD, but there is no citation to the original proof. Which is okay, but it would be nice to know where to find it. –  Abe Jun 6 '12 at 22:50
The actual derivation of the result is not in MGB, but rather us relegated to problem 5(b) on page 266. –  cardinal Jun 6 '12 at 22:54
@Abe: You will almost certainly not find an "original" reference for this. It is not the sort of standalone "publishable" result found in academic journals. It is simply a (rather tedious) calculation following from the basic properties of mathematical expectation. Quoting a textbook like MGB is perfectly reasonable and acceptable. –  cardinal Jun 6 '12 at 22:57

It is not clear if this will suit your needs for a definitive reference, but this question comes up in the exercises of Casella and Berger:

(page 364, exercise 7.45 b):

enter image description here

With reference to exercise 5b that provides another variant, in which $\Theta_2$ and $\Theta_4$ are the second and fourth moments ($\sigma^2$ and $\kappa$), respectively:

enter image description here

These are equivalent to the equation given in an answer on math.SE:

$\mbox{Var}(S^2)={\mu_4\over n}-{\sigma^4\,(n-3)\over n\,(n-1)}$

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It is interesting that your link and my link (in the comments to the OP) are different, but point to the same place. –  cardinal Jun 6 '12 at 16:30
@cardinal - I just copy-pasted from the OP - but the last digits are the user id of the person who copies the link, e.g. my link would be –  David LeBauer Jun 6 '12 at 16:33
Aha! (+1) I did not notice that the last part of the link was one's own id! Thanks for pointing that out. We're being followed... –  cardinal Jun 6 '12 at 16:39
it is good to have a trustworthy reference, but would still be nice to track down the original. +1 for looking through the exercises. –  Abe Jun 6 '12 at 16:39
@cardinal one justification for / use of tracking is the badges for sharing links (announcer, booster, publicist) –  David LeBauer Jun 6 '12 at 16:44

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