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Old 04-19-2013, 06:56 PM
manish manish is offline
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Default The other side of |Eout - Ein|<= epsilon

I don't understand how Eout(g) >= Ein(g) - epsilon
implies that there is no other hypothesis better than g?
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Old 04-19-2013, 08:11 PM
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yaser yaser is offline
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Default Re: The other side of |Eout - Ein|<= epsilon

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Originally Posted by manish View Post
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 g" is E_{\rm out} (h) \ge E_{\rm in} (h) - \epsilon for all h\in{\cal H}. Since one of the h's is g, one can also conclude the same relationship for g, 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 g only since this is sufficient for the conslusion we need in that direction.
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