#21




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: 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? 
#22




Re: Discussion of the VC proof
Quote:
thanks for the answer. I understand this argument, however this holds also for or for Thus the value 1/4 is somehow magic for me. Edit: I think you choose 1/4 because it is so easy to see, that the RHS of Theorem A.1 gets 1. Nevertheless with a different value you would get a different outcome of the final formula. 
#23




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]. 
#24




Re: Discussion of the VC proof
Correct.
Quote:
__________________
Have faith in probability 
#25




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:
__________________
Have faith in probability 
#26




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.

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