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An unbiased estimator of the variance is

$$ \frac{\sum_{i=1}^N (X_i - Mean(X)) ^2}{N-1} $$

where $X_i$ is observation $i$ and $N$ is the number of observations. Am I right that an unbiased estimator of the third central moment can be written as:

$$ \frac{\sum_{i=1}^N (X_i - Mean(X)) ^3}{N-1} $$

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Unbiased estimators of central moments are known as h-statistics $h_r$. That is, $$E[h_r] = \mu_r$$ where $\mu_r$ denotes the $r^\text{th}$ central moment of the population. They can be found in books like Stuart and Ord Kendall's Advanced Theory of Statistics (for simple cases), or can be generated on demand using the HStatistic function in the mathStatica package for Mathematica.

For your case, the unbiased estimator of the 3rd central moment $\mu_3$ is:

enter image description here

where the answer is expressed using power sum notation $s_r=\sum _{i=1}^n X_i^r$. We can also check that the solution is in fact unbiased, by wrapping the expectation operator over it. In the following, we find the 1st RawMoment of the solution just derived, and express the solution in terms of Central Moments:

enter image description here

For a brief introduction, see, for instance, Section 7.4 A and C of Chapter 7 of our book, "Mathematical Statistics with Mathematica". A free download of the chapter is available here:

http://www.mathStatica.com/book/Rose_and_Smith_2002edition_Chapter7.pdf

... which also provides many references. As to theory, in books, the best reference would be to Stuart and Ord: Kendall's Advanced Theory of Statistics (volume 1 - Distribution Theory) ... Chapters 12 and 13 ... although this is set up mostly in terms of k-statistics rather than h-statistics. Even here, after 6 editions, there are still some mistakes in the listed tables of formulae, and deriving one's own example by hand can be tough work.

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  • $\begingroup$ According to Section 7.4 A and C of Chapter 7 of your book, "Mathematical Statistics with Mathematica". $m_3$ is defined as $$m_{3} = \frac{\sum_{i=1}^N (X_i - Mean(X)) ^2}{N} $$ and $h_3$ as $$h_{3} = \frac{n^{2} m_{3}} {(n-2) (n-1)}$$ where $h_3$ is an unbiased estimator of the $\mu_3$. Threfore, finally one has as an unbiased estimator of $\mu_3$ $$ \frac{N\sum_{i=1}^N (X_i - Mean(X)) ^2}{(N-2) (N-1)} $$ Is this correct? $\endgroup$ – Ad van der Ven Oct 31 '19 at 9:15
  • $\begingroup$ Yes - except that your defn of $m_3$ should have the power raised to 3 (not 2) and same for your final line. I would also use lower case $n$ rather than capital $N$, as the convention is the latter denotes a finite population which can be more complicated. Also, the reference to the results you mention is Section 7.2B (not 7.4) $\endgroup$ – wolfies Oct 31 '19 at 15:27
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No, your estimator is biased downward.

$E[\frac{(X_i-\bar{X})^3}{n-1}]=\frac{n-2}{n}\mu_3$.

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  • $\begingroup$ @wolfies yes, I did, thanks. That's what I had written down on my piece of paper $\endgroup$ – Glen_b -Reinstate Monica Oct 31 '19 at 0:40

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