Thread: Hw5 Q8 E_out
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Old 05-05-2013, 10:07 PM
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yaser yaser is offline
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Default Re: Hw5 Q8 E_out

Quote:
Originally Posted by marek View Post
I am struggling to understand how to calculate E_{out} in this question. I have two competing theories, which I will describe below. Any help is greatly appreciated.

Once the algorithm terminates, I have w^{(t)}. I now generate a new set of data points \{X_i\}_{i=1}^M. Using my original target function to generate the corresponding Y_i = f(X_i).

Case 1. Just use the same cross entropy error calculation but on this new data set.

E_{out} = \frac{1}{M} \sum_{i=1}^M \ln (1+e^{-Y_i w^\top X_i})
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.
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