# How to calculate unbiased higher central co-moments?

Given four time series $W_i$, $X_i$, $Y_i$ and $Z_i$ for $i=1,2,3,...,T$ observations we define

$\bar W:= \frac{1}{T}\sum_{i=1}^T w_i$, $\phantom{ff}$ $\bar X:= \frac{1}{T}\sum_{i=1}^T x_i$, $\phantom{ff}$ $\bar Y:= \frac{1}{T}\sum_{i=1}^T y_i$, $\phantom{ff}$ $\bar Z:= \frac{1}{T}\sum_{i=1}^T z_i$

$m_n^W:= \frac{1}{T} \sum_{i=1}^T (w_i-\bar W)^n$ $\phantom{ffffffff}$ $m_n^X:= \frac{1}{T} \sum_{i=1}^T (x_i-\bar X)^n$ $\phantom{ffffffff}$ $m_n^Y:= \frac{1}{T} \sum_{i=1}^T (y_i-\bar Y)^n$ $\phantom{fffffffff}$ $m_n^Z:= \frac{1}{T} \sum_{i=1}^T (z_i-\bar Z)^n$

From standard statistics we know that for the variable $X$ the following holds:

$M_2^X=\frac{1}{T-1} \sum_{i=1}^T (x_i-\bar X)^2 = \frac{T}{T-1} m_2^X$ $\phantom{f}$ is an unbiased estimator for $\mathbb{E}[(X-\mathbb{E}(X))^2]$

$M_3^X=\frac{T}{(T-1)(T-2)} \sum_{i=1}^T (x_i-\bar X)^3 = \frac{T^2}{(T-1)(T-2)} m_3^X$ is an unbiased estimator for $\mathbb{E}[(X-\mathbb{E}(X))^3]$

$M_4^X = \frac{T^2}{(T-2)(T-3)}\bigg(\frac{T+1}{T-1}m_4^X - 3(m_2^X)^2 \bigg)$ is an unbiased estimator for $\mathbb{E}[(X-\mathbb{E}(X))^4]$.

Now what would be an unbiased estimator for the central comoments

$\mathbb{E}[(X-\mathbb{E}(X))(Y-\mathbb{E}(Y))(Z-\mathbb{E}(Z))]$ $\phantom{ff}$

and especially for

$\mathbb{E}[(W-\mathbb{E}(W))(X-\mathbb{E}(X))(Y-\mathbb{E}(Y))(Z-\mathbb{E}(Z))]$?

I can't imagine the answer is just

$M_4^{WXYZ}=\frac{T}{(T-2)(T-3)} \sum_{i=1}^T (w_i-\bar W)(x_i-\bar X)(y_i-\bar Y)(z_i-\bar Z)$

EDIT:

The question has to do with the estimation of financial return moments. In finance you normally assume IID data and you calculate the sample Covariance matrix in the following way:$$\sigma_{ij} =\frac{1}{T-1} \sum_{i=1}^T (x_i-\bar X)(y_i-\bar Y)$$ for all $i,j$. As I know it is not necessary to assume a specific distribution.

Have a look at the Paper of Ledoit and Wolf.They do only assume IID Data: http://www.ledoit.net/ole2.pdf. (Notice that Ledoit and Wolf define the sample estimator differently.)

Question: what would be an unbiased estimator for the central comoments $$\mathbb{E}\big[ \; (X-\mathbb{E}[X]) \; (Y-\mathbb{E}[Y]) \; (Z-\mathbb{E}[Z]) \;\big]$$

Answer: You seek the multivariate $h$-statistic $h_{a,b,c}$ which, by definition, is an unbiased estimator of the trivariate central moment $\mu_{a,b,c}$. That is:

$$\mathbb{E}[h_{a,b,c}] = \mu _{a,b,c} \quad \text{where} \quad \mu _{a,b,c}=\mathbb{E}\big[ \; (X-\mathbb{E}[X])^a \; (Y-\mathbb{E}[Y])^b \; (Z-\mathbb{E}[Z])^c \;\big]$$

In your case, $\{a,b,c\}=\{1,1,1\}$.

That's the easy part. The difficult part is the actual calculation of the $h$-statistics. Doing this by hand is fraught with computational difficulties and the literature is peppered with published results that are incorrect.

This problem can be solved using the HStatistic function in the mathStatica package for Mathematica which automates the calculation.

For your first problem, the h-statistic $h_{1,1,1}$ is:

where each trivariate power sum $s_{a,b,c}$ is defined by:

$$s_{a,b,c} =\sum _{i=1}^n X_i^a \; Y_i^b \; Z_i^c$$

and $n$ is the sample size (your $T$).

For the 4 variable case, the unbiased estimator you seek is:

While this may look complicated, given a data set and a computer, one can readily apply it. One can also operate symbolically on the estimator, for example to show that $\mathbb{E}[sol] = \mu_{1,1,1}$, where the expectation operator is just the $1^\text{st}$ Raw Moment of sol:

This has just calculated the mathematical expectation of the estimator sol derived above, and returned the product central moment $\mu_{1,1,1}$ .

More detail

There is an extensive discussion of such unbiased estimator problems in Chapter 7 of our book:

• Rose and Smith, "Mathematical Statistics with Mathematica", Springer, NY

As disclosure, I should note that I am an author of the HStatistic function used above.