Thread: Hw5 Q8 E_out
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Old 05-05-2013, 10:12 PM
marek marek is offline
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Join Date: Apr 2013
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Default Re: Hw5 Q8 E_out

Quote:
Originally Posted by yaser View Post
The above approach is correct. The problem specifies the cross entropy error measure, so E_{\rm out} = {\rm E} [ \ln (1+e^{-y{\bf w}^\top {\bf x}})], where the expectation is w.r.t. both {\bf x},y. The above formula estimates that through a random sample.
I suspected as much. I'll try to figure out why my other approach is wrong tomorrow. I think I've burned out on it today and am probably not seeing something obvious. Thanks for your help!
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