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magdon · #24 11-09-2016, 06:21 AM

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

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

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].