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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 $n = 2,3,4,5,6$). ?

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$V_p(X)$ is not a function of $X$, it is a number. –  Dilip Sarwate Sep 28 '12 at 17:46
    
Well, I see $V_p(X)$ as a function of the $X_i$, so I write $V_p(X)$, as a shortcut, but it means $V_p(X_1, X_2, ....,X_n)$ –  Trimok Sep 28 '12 at 18:00
    
@Trimok: I think this is what Dilip is getting at: $V_p$ is a function of $X$, but $V_p(X)$ is a number. When you apply a function to an argument, it is replaced with its value. –  Snowball Sep 28 '12 at 19:36
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1 Answer

If $D\leqslant-1$ and $n=2$, then $\langle V(x_1,x_2)^D\rangle$ diverges.

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 $D\leqslant-1$.

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Thanks, did. In fact, I realize that the interesting point was the analyse of the divergences, that is obtaining a finite result after regularising the integral. But this is an other question. –  Trimok Sep 29 '12 at 12:47
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