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Let $b_i, i=1,\ldots,m$ be real numbers.

Let $r_i, i=1,\ldots,m$ be random variables with $P(r_i=1)=P(r_i=-1)=1/2$.

Consider group $\Pi_m$ of all permutations of the set $\{1,\ldots,m\}$. On the group $\Pi_m$ consider the normalized counting measure $\mu_m(A)=\operatorname{card}(A)/m!$ for $A\subset \Pi_m$.

Let $f:\Pi_m\longrightarrow R$ as $f(\pi)=|\sum_{i=1}^{m/2} b_{\pi(i)}-\sum_{i=m/2+1}^{m} b_{\pi(i)}|$, where $$ \pi(\cdot)\longleftrightarrow r_i= \begin{cases} 1,&\quad \text{if} \quad \pi(i)\leq \frac m2\\[4mm] -1,&\quad \text{if} \quad \pi(i)>\frac m2 \end{cases}. $$

Is it possible to calculate $\mu_m(f=0)$?

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I am not sure if I understand your notation, is this the same as $\mu_m(g = 0)$ for $g(\pi) = \sum_{i=1}^{m/2}b_{\pi(i)} - \sum_{i=m/2+1}^{m}b_{\pi(i)}$? –  dtldarek Mar 12 '12 at 8:11
    
@ dtldarek: Yes. its the same coubting measure. –  David Mar 12 '12 at 13:34
    
@ Didier Piau: I see now. Thank you very much. –  David Mar 12 '12 at 13:35
    
In your definition of $f$ as $f(\pi)=|\sum_{i=1}^m b_ir_{\sigma(i)}|$, what is $\sigma$? –  Nate Eldredge Mar 12 '12 at 17:37
    
@ Nate Eldredge: Sorry, its typo- it should be $\pi$. –  David Mar 13 '12 at 1:57

1 Answer 1

up vote 0 down vote accepted

As written, the question is difficult to answer. Here are two remarks to help you reach a more suitable formulation.

  • Let $B=\sum\limits_{i=1}^mb_i$ and, for every $I\subseteq\{1,2,\ldots,m\}$, $B_I=\sum\limits_{i\in I}b_i$. If $B\ne2B_I$ for every $I\subseteq \{1,2,\ldots,m\}$, then $[f=0]=\varnothing$ hence $\mu_m(f=0)=0$. This happens, for example, if the numbers $b_i$ are linearly independent.
  • If $b_i=b_j$ for every $i\ne j$, then $[f=0]=\Pi_m$ hence $\mu_m(f=0)=1$ (if $m$ is even, but I guess $m$ is even since each random variable $r_i$ is said to be centered).
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