Which measure of dispersion is this function related to? Consider a sample $x=\{x_1,...,x_n\}$. Define the average as $\bar x$. Consider the following formula:
$$ \dfrac{\sum_{i=1}^n\left(\dfrac{x_i}{\bar x} \right)^c}{n} $$
or equivalently:
$$ \dfrac{\sum_{i=1}^n\left(1 + \dfrac{\epsilon_i}{\bar x} \right)^c}{n} $$
where $\epsilon_i = x_i - \bar x$ and $c$ is a constant.
To me, these formulas "look like" a measure of dispersion from the mean. But I have not found to which known measure they resemble (at least nothing from this long list). So, my questions:


*

*Do they measure dispersion? Maybe for particular values of $c$ only, e.g. $c=2$ or $c=1$?

*If so, do these have a name?
 A: To facilitate analysis, define the $c$-raw-moment estimator:
$$m_c \equiv \frac{1}{n} \sum_{i=1}^n x_i^c.$$
With a little algebra, your measure can be rewritten as:
$$r_c =\frac{m_c}{m_1^c} $$
Hence, your measure is the ratio of the $c$th raw sample moment, divided by the $c$th power of the first raw sample moment (the sample mean).  This does not have any special name, since it is not in common use for any particular problem.  (It is best identified descriptively as I have done above.)  This would not be a good measure of dispersion (even for $c=2$), since the raw sample moments do not capture the dispersion well.
A: In my opinion it depends how you define the term measure of dispersion. I would define it as any statistics $S(x)$ with this property:
$$
S(cx) = c^2S(x) \quad \lor \quad
S(cx) = |c|S(x)
$$
for any $c \in R.$
If you use this definition, your statistic is not a measure of dispersion.
A: As it turn out, using Taylor expansion of second degree can reveal a variance term. We have that:
$$ x_i \equiv \bar x + \epsilon_i$$
Which is:
$$ x_i = \bar x \left(1 + \frac{\epsilon_i}{\bar x}\right)$$
Define the second order Taylor expansion of $x_i^c$ around $\bar x$:
$$ x_i^c \approx \bar x^c \left(1 + c\frac{\epsilon_i}{\bar x} + \frac{c(c-1)}{2}\frac{\epsilon^2_i}{\bar x^2}\right)$$
Thus, it follows that:
$$ \sum_i x_i^c \approx \bar x^c \left(1 + \frac{c(c-1)}{2}\frac{nV(x)}{\bar x^2}\right)$$
since by definition $\sum_i \epsilon_i = 0$, and $\sum_i \epsilon_i^2 = nV(x)$.
This means, the formula I was interested in is then equivalent to:
$$ \frac{1}{n} + \frac{c(c-1)}{2} CV^2$$ 
where $CV^2 = \frac{V(x)}{\bar x^2}$, which is a nice measure of dimensionless dispersion!
