 Solution

The inequality in the printed rule is reversed. With the stated definition of $L$, a larger score favors group one. For example, in one dimension take ${\mu }_1=1$, ${\mu }_2=-1$, $\Sigma =1$, equal priors and $x=1$. The score is $2$, and $f_1(1)>f_2(1)$, whereas the printed rule assigns group two. The correct Bayes classifier follows from minimizing the error cost as follows.

Let $R$ be the region assigned to group one, let $f_i$ be its multivariate normal density, and let each misclassification cost $c>0$, with zero cost for correct classification. The expected cost isThe permitted integral-minimization result therefore selects $R=\{x:{\pi }_1{f}_1(x)>{\pi }_2{f}_2(x)\}$. Points of equality may be assigned to either group without changing the risk. Equivalently, the Bayes classifier chooses the larger posterior probability.

For positive priors and a common positive-definite covariance matrix, the normalization constants cancel in the likelihood ratio. Expanding the two quadratic forms givesThus the correct Gaussian Bayes classifier, choosing group one on ties, isThis proves the intended classification result with the necessary correction. Equal error costs remove $c$ from the decision, while prior odds determine the threshold.