Re: [math-fun] Average distance on the unit sphere puzzle
Trouble is, there are so many equators ... —Dan ----- ... almost all the hypervolume is near the equator. -----
That is disturbing, isn't it? I still think what I said is true. If you pick any unit vector in N-space as a reference, my intuition is that the dot product of a second, randomly chosen unit vector with the reference vector will follow a distribution more and more tightly centered on 0 as N increases. If this is true, then as Dan points out, for a large enough N, even if you pick a bunch of k different reference vectors, the random vector will tend to be near 90 degrees away from *all* of them. It may be that N has to increase dramatically to accommodate small increases in k, though. I don't think there is a logical contradiction here. Or my intuition may be completely misguided. On Thu, May 31, 2018 at 1:12 AM, Dan Asimov <dasimov@earthlink.net> wrote:
Trouble is, there are so many equators ...
—Dan
----- ... almost all the hypervolume is near the equator. -----
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Yes, almost all of the hypervolume is near the intersection of your favourite set of boundedly many equators. A union bound makes this easy to see: if N is such that 99.99% of the volume is within epsilon from 'the' equator, then 99% of the volume is within epsilon from every one of 100 equators. B\'{e}la Bollob\'{a}s gave an excellent course on probabilistic combinatorics, which talked a lot about concentration of measure and its various incarnations. -- APG.
Sent: Thursday, May 31, 2018 at 2:00 PM From: "Allan Wechsler" <acwacw@gmail.com> To: "Dan Asimov" <dasimov@earthlink.net>, math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Average distance on the unit sphere puzzle
That is disturbing, isn't it? I still think what I said is true. If you pick any unit vector in N-space as a reference, my intuition is that the dot product of a second, randomly chosen unit vector with the reference vector will follow a distribution more and more tightly centered on 0 as N increases. If this is true, then as Dan points out, for a large enough N, even if you pick a bunch of k different reference vectors, the random vector will tend to be near 90 degrees away from *all* of them. It may be that N has to increase dramatically to accommodate small increases in k, though. I don't think there is a logical contradiction here.
Or my intuition may be completely misguided.
On Thu, May 31, 2018 at 1:12 AM, Dan Asimov <dasimov@earthlink.net> wrote:
Trouble is, there are so many equators ...
—Dan
----- ... almost all the hypervolume is near the equator. -----
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participants (3)
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Adam P. Goucher -
Allan Wechsler -
Dan Asimov