ABSTRACT
Theorem 7.2.5. If R satisfies (7.59) and (7.61) and if for sufficiently large ? there exist ?0,? and ?1,? satisfying (7.60), then R is asymptotically logarithmically minimax
minimax of level a for testing H0: d 2=0 against H1: d
2=? so that ?? 8 ; that is,
Proof. Suppose, contrary, to (7.62), that there is an e>0 and an unbounded sequence ? of values ? with corresponding tests in Qa for which
There are two cases: (7.64) and (7.67). If and
consider the a priori distribution given by ?i,? and by satisfying
The integrated risk of any Bayes procedure B? must satisfy
by (7.63) and (7.65). But from (7.60) a Bayes critical region is
Hence if ? is so large that supx|B(x, ?)/H(?)|<e/C2, we get from (7.64)
The assumption (7.61) implies that
with e'>0, contradicting (7.66) for large ?. On the other hand, if and
let
Then by (7.60)
Hence if supx|B(x, ?)/H(?)R(?)|<e/2C2, we conclude from (7.67) that so that, by (7.59) and (7.68),
But
which contradicts (7.69) for sufficiently large ?. Q.E.D. Theorem 7.2.6. For every p, N, and a, Hotelling’s T2-test is locally minimax for
testing H0: d 2=0 against H1: d
2=? as ?? 0. Proof. In our search for a locally minimax test as ?? 0 we look for a level a test
which is almost invariant under GT and which minimizes among all level a tests the minimum power under H1 (as discussed in the case of the genuine minimax property
of the T2-test). So we restrict our attention to the space of R=(R1,…, Rp)', the
maximal invariant under GT in the space of We now verify the assumption of
Theorem 7.2.4 with x=r, ?=?=(?1,…, ?p)', and We can take h(?)=b? with b a positive constant. Of course, P?,?{R} does not depend on ?. From (6.66)
where B(r, ?, ?)=o(?) uniformly in r and ?. Here the set {d2=0} is a single point. Also the set {d2=?} is a convex finite-dimensional Euclidian set where in each
component ?i is 0(h(?)). If there exists any ?1,? satisfying (7.51), the degenerate which assigns measure 1 to the mean of ?1,? also satisfies (7.51), and (7.51) is satisfied by letting ?0,? give measure 1 to the single point ?=0, whereas ?1,? gives
measure 1 to the single point ?* (say) whose jth coordinate is (N-j)- 1(N-j+1) - 1p-1N(N-p), so that for all j. Applying Theorem 7.2.4 we get the result. Q.E.D.
