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:
$$ \frac{\sum_{i=1}^N (X_i - Mean(X)) ^3}{N-1} $$
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:
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:
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.
No, your estimator is biased downward.
$E[\frac{(X_i-\bar{X})^3}{n-1}]=\frac{n-2}{n}\mu_3$.
