Discussion of the VC proof - Page 3

CountVonCount · #21 10-26-2016, 03:28 PM

I have also another question on the same page (190):
At the end of the page there is the formula:

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

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 · #22 10-26-2016, 03:32 PM

Quote:

Originally Posted by magdon

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.

CountVonCount · #23 10-28-2016, 01:52 AM

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

$\sum_S P[S] \times P[A|S]$

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].

magdon · #24 11-09-2016, 06:21 AM

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

$\sum_S P[S] \times P[A|S]$

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].

magdon · #25 11-09-2016, 06:23 AM

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

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

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.

CountVonCount · #26 11-11-2016, 03:52 AM

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

sarafox · #27 08-28-2017, 10:13 AM

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

samyar1 · #28 04-23-2018, 09:00 AM

thank you for your answer. i like it.

delinglish · #29 07-04-2018, 05:52 AM

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.

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