Timeline for Inverse moment of the number of inversions of a permutation
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 9, 2014 at 19:32 | comment | added | Gerhard Paseman | You might also consider E(1/Inv(pi)) where you draw pi from nonidentical permutations. If you look at those generated by a few neighboring transpositions (12),(23),(34),... you will find a small number of permutations with small number of inversions, and the expected value will not just be less than 1/n but tend to something over n^2, where I would not be surprised if the something were n^{1/2}. Gerhard "Dividing By Zero: Not Good" Paseman, 2014.05.09 | |
| May 9, 2014 at 13:11 | comment | added | PeterR | One approach may be to use the fact that the components of the inversion table sum to the number of inversions, and the components are independent random variables, with the kth component having uniform distribution on {0,1,2,...,k-1}. Note this is assuming your permutations are selected uniformly (which you havent said). | |
| May 9, 2014 at 9:10 | history | edited | Davide Giraudo | CC BY-SA 3.0 |
Formatting improvements.
|
| May 8, 2014 at 21:32 | comment | added | Richard Stanley | For Inv' you are asking about $\int_0^1 (1+x)(1+x+x^2)\cdots (1+x+\cdots+x^{n-1})dx $, though I don't see how to estimate this integral. | |
| May 8, 2014 at 12:35 | review | First posts | |||
| May 8, 2014 at 12:38 | |||||
| May 8, 2014 at 12:17 | history | asked | user50460 | CC BY-SA 3.0 |