Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 47 / 2 / b

There are two disjoint ways to make an error: an individual from class one falls outside $R_1$, or an individual from class two falls inside $R_1$. The law of total probability givesSince $f_1$ integrates to one, this also equalsThis form isolates the only term changed by choosing the decision region.

Apply the minimum-integral decision region principle to $g(x)={\pi }_2{f}_2(x)-{\pi }_1{f}_1(x)$. The optimal Bayes classifier chooses class one when ${\pi }_1{f}_1(x)>{\pi }_2{f}_2(x)$, and class two for the reverse inequality. Ties can be assigned either way. An everywhere valid discriminant score isWhere both weighted densities are positive one may equivalently use $\log ({\pi }_1{f}_1(x)/({\pi }_2{f}_2(x)))$ and threshold zero. This chooses the largest posterior class probability, because both posterior probabilities have the same positive mixture-density denominator.

With zero cost for correct assignments, the total expected misclassification cost isThe part depending on $R_1$ is the integral of $c(1\mid 2){\pi }_2{f}_2-c(2\mid 1){\pi }_1{f}_1$. Therefore the cost-sensitive Bayes classifier isFor positive costs and weighted densities, the log posterior-odds score from part (ii) now has threshold $\log (c(1\mid 2)/c(2\mid 1))$. Larger cost of sending class one to class two expands the region assigned to class one. The weighted-density comparison also handles zero costs without taking an undefined logarithm.