LFD Book Forum Discussion of the VC proof
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#21
10-26-2016, 02:28 PM
 CountVonCount Member Join Date: Oct 2016 Posts: 17
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:

$\sum_S&space;\mathbb{P}[S]\times\mathbb{P}[sup_{h\in&space;H}\vert&space;E_{in}(h)&space;-&space;{E_{in}}'(h))&space;\vert&space;>&space;\frac{\varepsilon&space;}{2}]&space;\leq&space;sup_S&space;\mathbb{P}[sup_{h\in&space;H}\vert&space;E_{in}(h)&space;-&space;{E_{in}}'(h))&space;\vert&space;>&space;\frac{\varepsilon&space;}{2}]&space;\vert&space;S&space;]$

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?
#22
10-26-2016, 02:32 PM
 CountVonCount Member Join Date: Oct 2016 Posts: 17
Re: Discussion of the VC proof

Quote:
 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.
#23
10-28-2016, 12:52 AM
 CountVonCount Member Join Date: Oct 2016 Posts: 17
Re: Discussion of the VC proof

Quote:
 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].
#24
11-09-2016, 05:21 AM
 magdon RPI Join Date: Aug 2009 Location: Troy, NY, USA. Posts: 590
Re: Discussion of the VC proof

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 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].
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Have faith in probability
#25
11-09-2016, 05:23 AM
 magdon RPI Join Date: Aug 2009 Location: Troy, NY, USA. Posts: 590
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:
 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.
__________________
Have faith in probability
#26
11-11-2016, 02:52 AM
 CountVonCount Member Join Date: Oct 2016 Posts: 17
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.
#27
06-14-2017, 11:38 AM
 eh3an2010 Junior Member Join Date: Jun 2017 Posts: 1
Re: Discussion of the VC proof

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