Yes. f(x) is not the output. The output is either 0 or 1, and f(x) is the probability of 1.
Rémi On 6 mars 2013, at 09:04, Chin-Chang Yang wrote: > Thank you, Olivier. > > Let the observable function value be o(x). It can be defined as: > > o(x) = 1, with probability f(x); > o(x) = 0, with probability (1 - f(x)). > > where f(x) = 1 / (1 + e(-r(x))) has been defined in the paper. Also, we can > see that the expected value is f(x). > Did I get this correct? > > Best regards, > Chin-Chang Yang, 2013/03/06 > 2013/3/6 Olivier Teytaud <[email protected]> > It's a Bernoulli noise. > define f (x) = 1/ (1 + e(−r(x)) ) > and the objective function at x is 1 with probability f(x). > So the expected value at x is f(x), but the values you get are noisy. > > Best regards, > Olivier > > Since the functions are not noise-free, they should be defined in terms of > some noise. I really need the definition of the noise for comparison between > CLOP and other optimizers. > > I have downloaded the source codes, but I cannot find the codes related to > the noise currently. > > > _______________________________________________ > Computer-go mailing list > [email protected] > http://dvandva.org/cgi-bin/mailman/listinfo/computer-go > > > > -- > Chin-Chang Yang _______________________________________________ > Computer-go mailing list > [email protected] > http://dvandva.org/cgi-bin/mailman/listinfo/computer-go _______________________________________________ Computer-go mailing list [email protected] http://dvandva.org/cgi-bin/mailman/listinfo/computer-go
