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  #21  
Old 10-26-2016, 02:28 PM
CountVonCount CountVonCount is offline
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Default 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?
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  #22  
Old 10-26-2016, 02:32 PM
CountVonCount CountVonCount is offline
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Default Re: Discussion of the VC proof

Quote:
Originally Posted by magdon View Post
Suppose e^{-\frac12\epsilon^2N}\ge\frac14

Then, e^{-\frac18\epsilon^2N}\ge e^{-\frac12\epsilon^2N}\ge\frac14.

In which case 4 m(2N) e^{-\frac18\epsilon^2N}\ge 4m(2N)/4\ge 1 and the bound in Theorem A.1 is trivial.
Hello,

thanks for the answer. I understand this argument, however this holds also for

e^{-\frac12\epsilon^2N}\ge\frac18

or for

e^{-\frac12\epsilon^2N}\ge\frac{1}{16}

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.
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  #23  
Old 10-28-2016, 12:52 AM
CountVonCount CountVonCount is offline
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Default Re: Discussion of the VC proof

Quote:
Originally Posted by CountVonCount View Post
...
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

\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].
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  #24  
Old 11-09-2016, 05:21 AM
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magdon magdon is offline
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Default Re: Discussion of the VC proof

Correct.

Quote:
Originally Posted by CountVonCount View Post
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].
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  #25  
Old 11-09-2016, 05:23 AM
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magdon magdon is offline
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Default 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 View Post
Hello,

thanks for the answer. I understand this argument, however this holds also for

e^{-\frac12\epsilon^2N}\ge\frac18

or for

e^{-\frac12\epsilon^2N}\ge\frac{1}{16}

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.
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  #26  
Old 11-11-2016, 02:52 AM
CountVonCount CountVonCount is offline
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Default 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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  #27  
Old 06-14-2017, 11:38 AM
eh3an2010 eh3an2010 is offline
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Default 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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  #28  
Old 07-29-2017, 12:13 AM
alireza-res alireza-res is offline
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Default Re: Discussion of the VC proof

i agree with you , thank you very much
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