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Old 06-16-2012, 04:19 PM
mic00 mic00 is offline
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Default Re: Support vectors in three dimensions

I like this problem, though I'm a little confused about this part of the proof:

Originally Posted by Yellin View Post
If, at the other extreme, there is no possible orientation change of the parallel planes without a support vector separating from one of them, there must be two more events on the planes, one for each degree of rotational freedom to be prevented. That gives the maximum number of support vectors, four.
I see it like this: for P>0 we should expect 30 degrees of freedom. If there are 4 SVs, then we have the following degrees of freedom: rotational orientation of the planes (2 degrees), distance between the planes (1 degree), offset from the origin (1 degree), and 2 degrees for each of the 4 SVs. That's 12 degrees for the SVs, plus 6*3 for the other points -- 30 total.

Originally Posted by Yellin View Post
Any additional event on a plane could only be there with zero probability, because the volume within planes is zero fraction of the cube's volume.
That's clear - restricting any other event to be an SV reduces its degrees of freedom by one.

Incidentally, now I'm a little bit curious about the number of SVs in general. I think I'd previously assumed that #SV < d+1 also had P=0, but that doesn't seem to be the case, and I wonder about problems that could be written around testing for that (the relationship between #SVs, number of dimensions, and size of training set)...
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