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Old 01-27-2015, 03:46 PM
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magdon magdon is offline
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Join Date: Aug 2009
Location: Troy, NY, USA.
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Default Re: Hoeffding inequality for multiple hypothesis

Hoeffding for a single hypothesis h_1 tells you that, with high probability,

|E_{in}(h_1)-E_{out}(h_1)|<\epsilon.

As you point out, "The only thing that remains is to find hypothesis with minimal in-sample rate." Why would one do this? Because one is confident that Ein is close to Eout for every hypothesis, and so if we find the the hypothesis with minimum Ein, it will likely have minimum Eout. So, to be justified in picking the hypothesis with minimum Ein, we require that

\forall h_i, |E_{in}(h_i)-E_{out}(h_i)|\le\epsilon.

Equivalently,

for no h_i, |E_{in}(h_i)-E_{out}(h_i)|>\epsilon.

The factor of M comes from using the union bound

P[for\ no\ h_i, |E_{in}(h_i)-E_{out}(h_i)|>\epsilon]\le P[|E_{in}(h_1)-E_{out}(h_1)|>\epsilon]+P[|E_{in}(h_2)-E_{out}(h_2)|>\epsilon]+\cdots.


Quote:
Originally Posted by kostya3312 View Post
It's clear for me how inequality works for each hypothesis separately. But I don't understand why we need Hoeffding inequality for multiple hypothesis. If i have training data set of size 'N' then (for fixed tolerance 'e') Hoeffding upper bound is determined for each hypoyhesis. The only thing that remains is to find hypothesis with minimal in-sample rate. Why do we need to consider all hypothesis simultaneously? What information gives us Hoeffding inequality with factor 'M' in it? I undetstand example with coins but I can not relate it to learning problem.

Sorry for my english and thanks.
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