Re: Discussion of the VC proof
I have also another question on the same page (190):
At the end of the page there is the formula: https://latex.codecogs.com/gif.latex...pace;S&space;] I don't understand why the RHS is greater or equal to the LHS. The only legitimation I see for this is, that the distribution of P[S] is uniform, but this has not been stated in the text. Or do I oversee here anything and this is also valid for all kinds of distribution? 
Re: Discussion of the VC proof
Quote:
When we have a uniform distribution of P[S] the outcome of the productsum is simply the average of all P[AS], since And an average is of course less than or equal to the maximum of P[AS]. 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[AS]. 
Re: Discussion of the VC proof
Correct.
Quote:

Re: Discussion of the VC proof
You are correct. We could have made other assumptions.
But there is a special reason why we DO want 1. Because we are bounding the probability, and so there is nothing to prove if we claim that a probability is less equal to 1. So, whenever the RHS (i.e. the bound) evaluates to 1 or bigger, there is nothing to prove. So we only need to consider the case when the bound evaluates to less than 1. Quote:

Re: Discussion of the VC proof
Thank you very much for your answers. This helps me a lot to understand the intention of the single steps of this prove.

Re: Discussion of the VC proof
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Re: Discussion of the VC proof
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Re: Discussion of the VC proof
at the beginning everything was complicated, so at first, I gave it a thought, but I could take in nothing. by the way that was pretty easy. thank you so much.

Re: Discussion of the VC proof
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