The other side of |Eout - Ein|<= epsilon

manish · #1 04-19-2013, 07:56 PM

I don't understand how Eout(g) >= Ein(g) - epsilon
implies that there is no other hypothesis better than g?

yaser · #2 04-19-2013, 09:11 PM

Quote:

Originally Posted by manish

I don't understand how Eout(g) >= Ein(g) - epsilon implies that there is no other hypothesis better than g?

You are right. The statement that is relevant to the conclusion that "there is no other hypothesis better than

" is $E_{\rm out} (h) \ge E_{\rm in} (h) - \epsilon$ for all $h\in{\cal H}$ . Since one of the

's is

, one can also conclude the same relationship for

, but that says something else. In the other direction, it is also true that $E_{\rm out} (h) \le E_{\rm in} (h) + \epsilon$ for all $h\in{\cal H}$ , but we write it in terms of

only since this is sufficient for the conslusion we need in that direction.

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