Timeline for There is mathematics behind the 1989 Tour de France !
Current License: CC BY-SA 2.5
6 events
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| Jan 7, 2011 at 13:28 | comment | added | Tony Huynh | @Douglas: I think the issue is the domain of the uniform distribution. Indeed, it is more convenient to work with $U(-1/2,1/2)$ instead of $U(0,1)$, because as you say we can just take the convolution of $U(-1/2,1/2)$ 22 times (because it is symmetric about zero). I only used $U(0,1)$ to be consistent with Alex's answer. @fedja: If you don't mind to post your comment as an answer, I would vote it up. | |
| Jan 6, 2011 at 17:16 | comment | added | fedja |
Impossibility to edit or preview comments is really irritating. Of course, it should be $(3/2)^{22}$ and $\left(\frac{3e}{22}\right)^{22}$. Sorry for stupid typos :).
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| Jan 6, 2011 at 17:13 | comment | added | fedja |
Which means that you take a unit cube in 22 dimensions and want the volume of the part given by $x_1+\dots+x_{22}\le 3$. I'm not sure what MATLAB has to do with it, but the answer is $\frac 1{22!}(1^{22}+(2^{22}-22\cdot 1^{22})+(3^{22}-22\cdot 2^{22}+231\cdot 1^{22}))$. Since $(3/2)^22>2^{11}>2000$, we can reduce it to $3^{22}/{22!}\approx \frac 1{\sqrt 44\pi}\left(\frac{3e}{22}\right)^{-22}$. Now that is almost $0.1\cdot 8^{22}/22^{22}\approx \frac 8{27}10^{-10}$ give or take a factor of $1.2$. Not a fat chance, if you ask me :).
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| Jan 6, 2011 at 16:10 | comment | added | Douglas Zare | Instead of taking the convolution of 11 distributions which are each convolutions of 2 identical distributions, you can just take a convolution of 22. | |
| Jan 6, 2011 at 14:07 | history | edited | Tony Huynh | CC BY-SA 2.5 |
added 21 characters in body
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| Jan 6, 2011 at 13:44 | history | answered | Tony Huynh | CC BY-SA 2.5 |