Timeline for The expected square of the determinant of a random row stochastic matrix
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| S Feb 6, 2018 at 20:57 | history | bounty ended | Rodrigo de Azevedo | ||
| S Feb 6, 2018 at 20:57 | history | notice removed | Rodrigo de Azevedo | ||
| Jan 31, 2018 at 15:27 | answer | added | Guillaume Aubrun | timeline score: 9 | |
| Jan 31, 2018 at 1:39 | comment | added | Hans | @RichardStanley: I just checked Exercise 5.64 of your book but have not studied it carefully yet. The matrix there is binary. Are you saying the same technique applies to the current uniform distribution on the simplex? | |
| Jan 31, 2018 at 1:10 | answer | added | Haggai Nuchi | timeline score: 1 | |
| Jan 30, 2018 at 23:31 | comment | added | Richard Stanley | @Hans: The technique is that of Exercise 5.64 in my book Enumerative Combinatorics, vol. 2. | |
| Jan 30, 2018 at 22:30 | comment | added | Hans | @RichardStanley: Would you mind showing your "messy" computation here? Thanks. | |
| S Jan 30, 2018 at 21:52 | history | suggested | Rodrigo de Azevedo |
Added tags.
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| S Jan 30, 2018 at 21:36 | history | bounty started | Rodrigo de Azevedo | ||
| S Jan 30, 2018 at 21:36 | history | notice added | Rodrigo de Azevedo | Canonical answer required | |
| Jan 30, 2018 at 21:34 | review | Suggested edits | |||
| S Jan 30, 2018 at 21:52 | |||||
| Nov 21, 2017 at 23:37 | comment | added | Maesumi | @RichardStanley Have you published your result? | |
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Dec 5, 2016 at 15:40 | comment | added | Victor Kleptsyn | @RichardStanley: that's not an answer to the "reason" question, but I think I see how to make a perhaps nicer computation: by computing the expectation of the $det^2$ for the i.i.d. exponential variables case (see mathoverflow.net/a/13040/31371, though I would do the handling of $Fix$ summation part differently). Then, divide it by $(n(n+1))^n$ that corresponds to the normalization (expectation of $(a_{i1}+\dots+a_{in})^2$ per row), that sends it to the simplex. It provides $\frac{(n+1)!}{(n(n+1))^n}= \frac{(n-1)!}{(n(n+1))^{n-1}}$ for the expectation; perhaps this way is less messy? | |
| Dec 4, 2016 at 3:12 | comment | added | Anthony Quas | Great - so this gives the sort of size that I was expecting. Thanks for taking an interest in my original question. | |
| Dec 4, 2016 at 1:11 | comment | added | Richard Stanley | This is $(n+1)!$ to the $(n-1)$st power. However, this is not the average value of $(\det A)^2$. To get the average value one must divide by the volume, which is $1/(n-1)!^n$. | |
| Dec 3, 2016 at 15:53 | comment | added | Anthony Quas | Interesting. Is this the $(n-1)$st power of the factorial? This looks a lot smaller than I was expecting; or maybe I just misunderstood the random matrix result. | |
| Dec 3, 2016 at 1:18 | history | asked | Richard Stanley | CC BY-SA 3.0 |