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Old 11-09-2016, 06:21 AM
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magdon magdon is offline
Join Date: Aug 2009
Location: Troy, NY, USA.
Posts: 597
Default Re: Discussion of the VC proof


Originally Posted by CountVonCount View Post
I got it by myself.
When we have a uniform distribution of P[S] the outcome of the product-sum

\sum_S P[S] \times P[A|S]

is simply the average of all P[A|S], since

\sum_S P[S] = 1

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].
Have faith in probability
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