Page 220 - Introduction to Statistical Pattern Recognition

202 Introduction to Statistical Pattern Recognition

(5.16) and (5.17) show

where Ed indicates the expectation with respect to the design samples, and 97. is
the number of design samples (while N indicates the number of test samples).
Therefore, from (5.56) and (5.57),

Assuming that ri is reasonably large, we can eliminate E ( Ahm(X) ] for rn larger
than 2.

From (5.37), the error of a random classifier for given test distributions is
expressed by

(5.59)
When Ah is small, we may use the following approximation

ejoh(X) = joh(X)ejwAh(X) E - ejoll(x)[ l+jmAh(x)+- tio)2 Ah2(X)] . (5.60)
e
2
A
Then, A& = 8: - E can be approximated by
1
A& 2- -jj{Ah(X) + ~Ah2(X)Jc’W”‘X’p(X)dwdX (5.61)
2
2x
Bayes classifier: When h(X) is the Bayes classifier for the given test
distributions, F(X) = 0 at h (X) = 0. In this case, the Bayes error, E, is the
minimum error and A& of (5.61) must be always positive. In order to confirm
the validity of the error expression (5.59), let us prove AE 2 0 as an exercise.
The first step to prove A& 2 0 is to show that the first order variation of
(5.61) is zero regardless of Ah(X), as follows.