Page 135 - Introduction to Statistical Pattern Recognition

3 Hypothesis Testing 117

Thus, the inequalities of the above equations can be replaced by equalities, and
A and B are approximately determined by (1 and &,/(I -E~). Or,
and €2 can be expressed in terms of A and B as

B(A - 1)
E1 ”= (3.196)
A-B ’
1-B
E2Z-. (3.197)
A-B

A few remarks concerning the properties of the Wald sequential test are
in order.
(1) For the derivation of (3.192) and (3.193), Xl,X2,. . . do not need to
be independent and identically distributed.
(2) It can be proved that the Wald test terminates with probability 1
[201.
(3) The Wald test minimizes the average number of observations to
achieve a given set of errors, E! and ~2 [21].

Expected number of observations: In the Wald sequential test, the
average number of observations varies, depending on the distributions and the
error we would like to achieve. In order to discuss this subject, let m be the
number of observations needed to reach the upper or lower threshold value.
The term m is a random variable. Equation (3.175) is rewritten as

m
s = Ch(Xj) (3.198)
j=l

Then s should be either a or b of (3.184), with
s = a (accept q) with probability 1 - E, wlhen X’s E 01 ,
s = a (accept 0,) with probability e2 when X’s E 02 ,
(3.199)
s = b (accept w2) with probability when X’s E w1 ,
s = b (accept 02) with probability 1 - ~2 when X’s E 02 .

Therefore,
E(slw,) = a(l -el) + he, , (3.200)