Quote:
Originally Posted by yaser
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}})]](/vblatex/img/5db35b1931732431fb2272f05bb3e75e-1.gif "E_{\rm out} = {\rm E} [ \ln (1+e^{-y{\bf w}^\top {\bf x}})]") , where the expectation is w.r.t. both  . 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!