Let $X$ be a real random variable, with real values $X_i$ associated with probabilities $p_i \ge0, p_1 \le 1$, $i=1$ to $n$. The variance $V_p(X)$, is, as usual:
$$V_p(X) = \sum^n_{i=1} p_i X_i^2 - (\sum^n_{i=1} p_i X_i)^2$$
where the '$_p$' of $V_p$ make reference for a particular probability law defined by its probabilities $p_i$.
I am looking for the mathematical meaning (probabilistic, geometrical: polytopes, etc...), and possibly practical applications, for the following expression :
$$<(V(X))^D> = \frac{\int dp \space (Vp(X))^D}{\int dp}$$
where $D$ is an integer, positive or negative.
That is:
$$<(V(X))^D> = \frac{\int (\prod^n_{i=1} \space dp_i) \space \delta(\sum^n_{i=1} p_i - 1) \space(Vp(X))^D}{\int (\prod^n_{i=1} \space dp_i) \space \delta(\sum^n_{i=1} p_i - 1)}$$
As a bonus for very smart minds, one more question:
Is there a general formula, or specific formula, or a recursive formula, for $<(V(X))^D>$, as a function of the $X_i$, especially when $D$ is negative (say $-1,-2,-3,-4,-5$), and $n$ being not to big (say $n = 2,3,4,5,6$). ?
To see this, note that, for $p=(u,1-u)$ and $x=(x_1,x_2)$, $V_p(x)=u(1-u)(x_1-x_2)^2$ hence the numerator of $\langle V(x)^D\rangle$ is $(x_1-x_2)^{2D}\cdot I_D$ with $I_D=\int\limits_0^1u^D(1-u)^D\mathrm du$. Since $I_D$ diverges for every $D\leqslant-1$, so does $\langle V(x)^D\rangle$ as soon as $x_1\ne x_2$.
The same divergence might occur for every $n\geqslant2$ and $D\leqslant-1$.