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BojanVujatovic · #3 08-03-2014, 09:23 PM

Thank you for your reply.

The only part I don't quite follow is:

Quote:

Originally Posted by magdon

$y_n\textbf{w}^T\textbf{x}_n<1$ 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

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 $E_n(\textbf{w}) = 0$ .