Thread: Discussion of the VC proof View Single Post
#24
11-09-2016, 05:21 AM
 magdon RPI Join Date: Aug 2009 Location: Troy, NY, USA. Posts: 596
Re: Discussion of the VC proof

Correct.

Quote:
 Originally Posted by CountVonCount I got it by myself. When we have a uniform distribution of P[S] the outcome of the product-sum is simply the average of all P[A|S], since And an average is of course less than or equal to the maximum of P[A|S]. If P[S] is not distributed uniformly we still have an average, but a weighted average. But also here the result is always less than or equal to the maximum. Because you cannot find weighting factors that are in sum 1 but will lead to a higher result as the maximum P[A|S].
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