8.5.∗Minimax and Classification Procedures 509
whereβ=1−γ(θ′′) is the probability of the type II error.
Let us now see if we can find a minimax solution to our problem. That is, we
want to find a critical regionCso that
max[R(θ′,C),R(θ′′,C)]
is minimized. We shall show that the solution is the region
C=
{
(x 1 ,...,xn):
L(θ′;x 1 ,...,xn)
L(θ′′;x 1 ,...,xn)
≤k
}
,
provided the positive constantkis selected so thatR(θ′,C)=R(θ′′,C). That is, if
kis chosen so that
L(θ′,θ′′)
∫
C
L(θ′)=L(θ′′,θ′)
∫
Cc
L(θ′′),
then the critical regionCprovides a minimax solution. In the case of random vari-
ables of the continuous type,kcan always be selected so thatR(θ′,C)=R(θ′′,C).
However, with random variables of the discrete type, we may need to consider an
auxiliary random experiment whenL(θ′)/L(θ′′)=kin order to achieve the exact
equalityR(θ′,C)=R(θ′′,C).
To see thatCis the minimax solution, consider every other regionAfor which
R(θ′,C)≥R(θ′,A). A regionAfor whichR(θ′,C)<R(θ′,A) is not a candidate for
a minimax solution, for thenR(θ′,C)=R(θ′′,C)<max[R(θ′,A),R(θ′′,A)]. Since
R(θ′,C)≥R(θ′,A)meansthat
L(θ′,θ′′)
∫
C
L(θ′)≥L(θ′,θ′′)
∫
A
L(θ′),
we have
α=
∫
C
L(θ′)≥
∫
A
L(θ′);
that is, the significance level of the test associated with the critical regionAis less
than or equal toα.ButC, in accordance with the Neyman–Pearson theorem, is a
best critical region of sizeα.Thus
∫
C
L(θ′′)≥
∫
A
L(θ′′)
and
∫
Cc
L(θ′′)≤
∫
Ac
L(θ′′).
Accordingly,
L(θ′′,θ′)
∫
Cc
L(θ′′)≤L(θ′′,θ′)
∫
Ac
L(θ′′),