 Classification-tree deviance

Maximizing a node's multinomial likelihood gives class probabilities ${\hat{p}}_{tk}=n_{tk}/n_t$. Twice the negative maximized log-likelihood relative to perfect classification is the displayed deviance, with $0\log 0=0$. It equals twice the node size times its Shannon entropy. A split is chosen by maximizing $D(t)-D(t_L)-D(t_R)$, and tree deviance is the sum over terminal nodes. The training misclassification count is instead ${\sum }_t(n_t-{\max }_k{n}_{tk})$; the two criteria are different.