# Expectation of Reciprocals Involving Permutations

Let $a_i$ be $n$ distinct real numbers. What is the expectation:

$$\mathbb E_\sigma \left[ \sum_{i=1}^{n} \frac {1} {a_{\sigma(i)} - \sum_{j=1}^{i-1}a_{\sigma(j)}} \right]$$

where the expectation is taken over all $n!$ permutations $\sigma:[1,2,..,n]\rightarrow[1,2,..,n]$

I appreciate your help.

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Not sure what here is random that admits taking expected values? – gt6989b Dec 20 '12 at 20:49
nothing is random, it's just the sum of all these n! terms, divided by n!. maybe average is a more appropriate word. – Troy McClure Dec 20 '12 at 20:52
I don't see how it can be made any simpler than the expression you already have. – Robert Israel Dec 20 '12 at 21:01
can't it be independant of $\sigma$? or any way to calculate it more efficiently than o(n!)? – Troy McClure Dec 20 '12 at 21:02
For every subset $J\subseteq \{1,2,\cdots,n\}$ denote $$R_J:=\sum_{\ell\in J}\frac{1}{x_\ell-\sum\limits_{t\in J\setminus \{\ell\}}x_t}.$$ Intuitively then the average should be $$\sum_{J\subseteq\{1,\cdots,n\}}(n-|J|)!(|J|-1)!R_J.$$ This might help, if it is correct. – anon Dec 20 '12 at 21:12