The exponential-loss minimiser

Topic: Ensembles, random forests, and boosting · more questions on this topic
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If \tau_1(x)=0.8, find f^*(x) and check that 1/(1+e^{-2f^*(x)}) returns 0.8. What would the minimiser be if \tau_1(x)=0.5?

Worked answer (click to show)

f^*=\tfrac12\log\frac{0.8}{0.2}=\tfrac12\log 4=0.693.

Then e^{-2f^*}=e^{-1.386}=0.25 and \tau=1/(1+0.25)=0.8. For \tau_1=0.5 the log odds are 0, so f^*=0, the decision boundary of the classifier, where the sign changes.