Thread: Problem 3.5
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Old 08-03-2014, 09:23 PM
BojanVujatovic BojanVujatovic is offline
Join Date: Jan 2013
Posts: 13
Default Re: Problem 3.5

Thank you for your reply.

The only part I don't quite follow is:
Originally Posted by magdon View Post
y_n\textbf{w}^T\textbf{x}_n<1 in which case E(\mathbf{x}_n)=1-y_n\textbf{w}^T\textbf{x}_n, which is differentiable.
As I see, and please correct me if I'm wrong, for 0 \leq y_n\textbf{w}^T\textbf{x}_n<1, the signal \textbf{w}^T\textbf{x}_n and the output y_n agree.
(e.g. \textbf{w}^T\textbf{x}_n = -0.5 and y_n = -1 \implies y_n\textbf{w}^T\textbf{x}_n = 0.5).

Therefore, \text{sign}(\textbf{w}^T\textbf{x}_n)=y_n and [\!\![\text{sign}(\textbf{w}^T\textbf{x}_n)\neq y_n]\!\!] = 0, meaning also E_n(\textbf{w}) = 0.
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