Discussion of the VC proof

[CountVonCount]

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?

[CountVonCount]

Originally Posted by magdon

Suppose

Then, .

In which case and the bound in Theorem A.1 is trivial.

Hello,

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.

[CountVonCount]

Originally Posted by CountVonCount

...
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?

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].

[magdon]

Correct.

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].

[magdon]

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.

Originally Posted by CountVonCount

Hello,

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.

[CountVonCount]

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

[eh3an2010]

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

[alireza-res]

i agree with you , thank you very much

[sarafox]

i had same question but thanks to your website my question just solved. thanks

[babak]

I just want to say thank you, my problem is solved