Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
$V_p(X)$ 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

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 $D\leqslant-1$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.