Timeline for Two geometric probability questions (one answered, one more to go)
Current License: CC BY-SA 2.5
9 events
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| Nov 12, 2010 at 17:52 | history | edited | zhoraster | CC BY-SA 2.5 |
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| Nov 12, 2010 at 16:41 | comment | added | John Jiang | Thanks again! Your geometric ansatz really helped. I'd be glad to hear how to go directly from I to J, as I am still interested in the entire J sequence, and hopefully prove something nice about them. | |
| Nov 12, 2010 at 10:09 | comment | added | zhoraster | @John Jiang: I see, I had another formula initially (from $E_k/E$), just didn't see that it can be reduced to this. I even discovered now that it is quite straightforward to go from the distribution of $I$ to the one of $J$ omitting $E$. Anyway, glad that you've found it, congratulations! | |
| Nov 12, 2010 at 9:26 | comment | added | John Jiang | @zhoraster: actually I was able to use the formula you gave above to compute the exact distribution of the minimum: $P(\min J_i > y) = (1-y n(n+1)/2)^{n-1}$, so it doesn't seem bad at all. What I did is a pretty geometric argument. Notice that $y$ ranges between $0$ and $2/(n(n+1))$, as expected from its being the smallest gap of gaps of $n$ points on the circle. Using that formula, we just need to integrate $P(\min J_i > y)$ for $y \in [0,2/(n(n+1))]$ to get the expected value, which is $2/((n-1)n(n+1))$. | |
| Nov 12, 2010 at 8:01 | comment | added | zhoraster | @John Jiang: You're warmly welcome. Precise formulas useless here, if you want, I can look into asymptotics. | |
| Nov 12, 2010 at 6:16 | comment | added | John Jiang | Thank you for the great answer. I am always scared of exact formulas. | |
| Nov 12, 2010 at 6:09 | vote | accept | John Jiang | ||
| Nov 12, 2010 at 6:09 | |||||
| Nov 11, 2010 at 10:54 | history | edited | zhoraster | CC BY-SA 2.5 |
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| Nov 11, 2010 at 10:47 | history | answered | zhoraster | CC BY-SA 2.5 |