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We have defined expected loss as follows:

$\mathbb{E}_y[C(y,a)|x] = \int C(y,a)P(y|x)dy$

"C(.)" being the cost-function, "y" the class, "a" a taken action. In the case that y can only have two possible values, say y $\in \{0,1\}$, isn't it then the case that if $P(y=1|x) = p$ it follows that $P(y=-1|x) = 1-p$? And hence if two actions are possible, expected loss could be written (in the discrete case) as:

$C(y=1,a=-1)*P(y=1|x)+C(y=-1,a=1)*1-P(y=-1|x)$?

Is this correct? I see that even in cases where there are only two classes, the posteriors P(y=1|x) and P(y=-1|x) would be used, without expressing one in the terms of the other.

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  • $\begingroup$ What is your question? $\endgroup$ – Tim Aug 4 '17 at 19:53
  • $\begingroup$ Edited: Included my question. Particularly I was told here that this is not always the case, and I do not understand why: stats.stackexchange.com/questions/296230/… $\endgroup$ – user24544 Aug 4 '17 at 19:56

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