Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

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′′),