Page 232 - Introduction to Statistical Pattern Recognition

214 Introduction to Statistical Pattern Recognition

- 1 2
-(&--) . (5.96)
2
The first line of (5.96) is (~-1/2)~ from (5.37), and the second and third lines
are each (~-1/2)z from (5.65). Furthermore, the summation of the first,
2
second, third, and fifth lines is (~-1/2)~ + 2(&--1/2)G - (E-1/2)2 = -s where
-
E =E + E. Since is proportional to E,(Ah(X) + joAh2(X)/2} (-1lrL) from
(5.65) and (5.58), E2 is proportional to 1/9* and can be neglected for a large
n. Thus, only the fourth line remains uncancelled. Thus,

Vard(E) ZLjjjjEd(Ah(X)Ah(Y)}
4x2

= 1 Ed(Ah(X)Ah(Y))p(X)F(Y)dXdY
h(X)=o h(Y)=O
Equation (5.97) indicates that the integration is carried out along the
classification boundary where h (X) = 0. When h (X) is the Bayes classifier,
;(X) of (5.38) must be zero at the boundary. Thus, (5.97) becomes 0. Since
we neglected the higher order terms of Ah(X) in the derivation of (5.97), the
actual Vard{;) is not zero, but proportional to 1P according to (5.58). When
h (X) is not the Bayes classifier, F(X) is no longer equal to zero at h (X) = 0.
Thus, we may observe a variance dominated by a term proportional to
E,(Ah(X)Ah(Y)). Since E,(Ah(X)Ah(Y)) is a second order term, it is propor-
tional to 1/17.