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                            160                   Parametric Families of Distributions

                            that is, if and only if

                                                         T                     ∗

                                             x 1 = x 2 +  1 (x 1 − x 2 ) /n 1 n = x 2 + b 1 n
                                                         n
                                       T
                                  ∗
                            where b = [1 (x 1 − x 2 )]/n.It follows from part (iii) of Theorem 5.6 that T is a maximal
                                       n
                            invariant.
                              Consider an invariant statistic T . According to part (ii) of Theorem 5.6, in order to show
                            that T is not a maximal invariant, it is sufficient to find x 1 , x 2 in X that lie on different orbits
                            such that T (x 1 ) = T (x 2 ).

                                                                                        4
                                                                          4
                            Example 5.34 (Cauchy distribution). Consider X = R and for x ∈ R , x = (x 1 , x 2 ,
                            x 3 , x 4 ), let
                                                                x 1 − x 2
                                                         T (x) =      .
                                                                x 3 − x 4
                                                                               (4)
                            In Example 5.32, it is shown that T is invariant with respect to G .
                                                                               ls
                              Consider two elements of X, x = (1, 0, 1, 0) and ˜ x = (1, 0, 2, 1). Note that since for
                            a > 0,
                                             (1, 0, 2, 1) − a(1, 0, 1, 0) = (1 − a, 0, 2 − a, 1)

                            there does not exist an a such that the elements of x − a ˜ x are all equal. Hence, x and ˜ x lie
                            on different orbits; see Example 5.30. Since T (x) = T (˜ x) = 1, it follows from part (ii) of
                            Theorem 5.6 that T is not a maximal invariant.

                            Equivariance
                            Consider a group G acting on a set X and let T denote a statistic, T : X → Y for some set
                            Y. Suppose that G also acts on Y. The statistic T is said to be equivariant if for each g ∈ G,

                                                     T (gx) = gT (x), x ∈ X.

                            Note that two different applications of the transformation g are being used in this expression:
                            gx refers to the action of g on X, while gT (x) refers to the action of g on Y.
                              Equivariance is an important concept in statistics. For instance, consider a transformation
                            model for a random variable X, with respect to a group of transformations G; let X denote
                            the range of X and let   denote the parameter space. Let T denote an estimator of θ,a
                            function T : X →  . Hence, if X = x is observed, we estimate the value of θ to be T (x).
                            The estimator is equivariant if the estimate corresponding to gx, g ∈ G,is gT (x).


                            Example 5.35 (Estimation of a location parameter). Let X denote a real-valued random
                            variable with an absolutely continuous distribution with density p 0 satisfying p 0 (x) > 0,
                            x ∈ R. Consider the location model based on p 0 consisting of the class of distributions of
                                    (1)      (1)
                            gX, g ∈ G  where G  denotes the group of location transformations on R.
                                    l        l
                                                                                             (1)
                              Suppose X is distributed according to p 0 and let g = θ denote an element of G . Then
                                                                                             l
                            the distribution of gX is absolutely continuous with density function
                                                p(x; θ) = p 0 (x − θ), −∞ < x < ∞.
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