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            052184472Xc07  CUNY148/Severini  May 24, 2005  3:59





                            222             Distribution Theory for Functions of Random Variables

                            Note that, for X ∈ X(τ), X (·) = τ X. Hence,

                                               E{h(X (·) )I {X∈X(τ)} }= E{h(τ X)I {X∈X(τ)} }.
                            Let τ 0 denote the identity permutation. Then the event that X ∈ X(τ)is equivalent to the
                            event that τ X ∈ X(τ 0 ). Hence,

                                                 E[h(X (·) )] =  E{h(τ X)I {τ X∈X(τ 0 )} }.
                                                            τ
                              Since X 1 ,..., X n are independent and identically distributed, for any permutation τ,
                            τ X has the same distribution as X.It follows that

                                         E[h(X (·) )] =  E{h(X)I {X∈X(τ 0 )} }= n!E[h(X)I {X∈X(τ 0 )} ];
                                                    τ
                            the factor n!is due to the fact that there are n! possible permutations. The result follows.

                              As noted above, the density function of (X (1) ,..., X (n) ) may be used to determine the
                            density function of some smaller set of order statistics. The following lemma is useful in
                            carrying out that approach.

                            Lemma 7.1. Let p denote the density function of an absolutely continuous distribution on
                            R and let F denote the corresponding distribution function. Then, for any n = 1, 2,...,
                            and any a < b,
                                    ∞      ∞

                                                                                              n
                                n!    ···    p(x 1 ) ··· p(x n )I {a<x 1 <x 2 <···<x n <b} dx 1 ··· dx n = [F(b) − F(a)] .
                                   −∞     −∞
                            Proof. Let X 1 , X 2 ,..., X n denote independent, identically distributed random variables,
                            each distributed according to the distribution with distribution function F and density
                            function p. Then, according to Theorem 7.9, the density function of (X (1) ,..., X (n) )is
                            given by

                                          n!p(x 1 ) ··· p(x n ), −∞ < x 1 < x 2 < ··· < x n < ∞.
                            It follows that
                                      Pr(a < X (1) < ··· < X (n) < b)
                                              ∞     ∞

                                       = n!     ···    p(x 1 ) ··· p(x n )I {a<x 1 <x 2 <···<x n <b} dx 1 ··· dx n .
                                             −∞    −∞
                            Note that the event a < X (1) < ··· < X (n) < b is simply the event that all observations fall
                            in the interval (a, b). Hence,

                               Pr(a < X (1) < ··· < X (n) < b) = Pr(a < X 1 < b, a < X 2 < b,..., a < X n < b)
                                                        = Pr(a < X 1 < b) ··· Pr(a < X n < b)
                                                                      n
                                                        = [F(b) − F(a)] ,
                            proving the result.

                              Using Lemma 7.1 together with Theorem 7.9 yields the density function of any pair of
                            order statistics; note that the same approach may be used to determine the density function
                            of any subset of the set of all order statistics.
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