Invariance and minimax. Let a problem remain invariant relative to the groups G, G¯, and G∗ over the spaces X , Ω, and D respectively.

Then a randomized procedure Yx is defined to be invariant if for all x and g the conditional distribution of Yx given x is the same as that of g∗−1Ygx.

(i) Consider a decision procedure which remains invariant under a finite group G = {g1,...,gN }. If a minimax procedure exists, then there exists one that is invariant. (ii) This conclusion does not necessarily hold for infinite groups, as is shown by the following example. Let the parameter space Ω consist of all elements θ of the free group with two generators, that is, the totality of formal products π1 ...πn (n = 0, 1, 2,...) where each πi is one of the elements

a, a−1,

b, b−1 and in which all products aa−1, a−1a, bb−1, and b−1b have been canceled. The empty product (n = 0) is denoted by

e. The sample point X is obtained by multiplying θ on the right by one of the four elements

a, a−1,

b, b−1 with probability 1 4 each, and canceling if necessary, that is, if the random factor equals π−1 n . The problem of estimating θ with L(θ,

d) equal to 0 if d = θ and equal to 1 otherwise remains invariant under multiplication of X, θ, and d on the left by an arbitrary sequence π−m ...π−2π−1(m = 0, 1,...). The invariant procedure that minimizes the maximum risk has risk function R(θ, δ) ≡ 3 4 . However, there exists a noninvariant procedure with maximum risk 1 4 .

[(i): If Yx is a (possibly randomized) minimax procedure, an invariant minimax procedure Y

x is defined by P(Y

x =

d) = N i=1 P(Ygix = g∗

i d)/N.

(ii): The better procedure consists in estimating θ to be π1 ...πk−1 when π1 ...πk is observed (k ≥ 1), and estimating θ to be

a, a−1,

b, b−1 with probability 1 4 each in case the identity is observed. The estimate will be correct unless the last element of X was canceled, and hence will be correct with probability ≥ 3 4 .]

Section 1.7