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                            218             Distribution Theory for Functions of Random Variables

                              When the distribution given by F is either absolutely continuous or discrete, it is possible
                            to derive the density function or frequency function, respectively, of the distribution.

                            Theorem 7.7. Let X 1 , X 2 ,..., X n denote independent, identically distributed, real-valued
                            random variables each with distribution function F and range X.
                               (i) If X 1 has a discrete distribution with frequency function p, then X (m) has a discrete
                                  distribution with frequency function
                                            m                 n+k−m
                                                  n       m−k       n − m + k     j        n−m+k− j
                                   p (m) (t) =        F(t 0 )                 p(t) [1 − F(t)]    ,
                                                m − k                   j
                                           k=1                 j=k
                                  for t ∈ X; here t 0 is the largest element of X less than t.
                              (ii) If X 1 has an absolutely continuous distribution with density function p, then X (m)
                                  has an absolutely continuous distribution with density function
                                                   n − 1     m−1        n−m

                                        p (m) (t) = n    F(t)   [1 − F(t)]  p(t), −∞ < t < ∞.
                                                   m − 1
                            Proof. First consider the case in which X 1 has a discrete distribution. Let t denote a fixed
                            element of X. Each observation X 1 ,..., X n falls into one of three sets: (−∞, t 0 ], {t},or
                            [t 1 , ∞). Here t 0 denotes the largest element of X less than t, and t 1 denotes the smallest
                            element of X greater than t. Let N 1 , N 2 , N 3 denote the number of observations falling into
                            these three sets, respectively. Then

                                           m                         m n+k−m

                             Pr(X (m) = t) =  Pr(N 1 = m − k, N 2 ≥ k) =     Pr(N 1 = m − k, N 2 = j).
                                          k=1                        k=1  j=k
                            Note that (N 1 , N 2 ) has a multinomial distribution with
                                                               n

                                                                              n 2
                                                                                        n 3
                                                                         n 1
                                      Pr(N 1 = n 1 , N 2 = n 2 ) =   F(t 0 ) p(t) (1 − F(t)) ,
                                                            n 1 , n 2 , n 3
                            n 1 + n 2 + n 3 = n, where F and p denote the distribution function and frequency function,
                            respectively, of the distribution of X 1 . Hence,
                                Pr(X (m) = t)
                                      m n+k−m
                                                         n               m−k   j        n−m+k− j
                                   =                                 F(t 0 )  p(t) (1 − F(t))
                                               m − k, j, n − m + k − j
                                      k=1  j=k
                                      m                  n+k−m
                                             m−k   n          n − m + k     j        n−m+k− j
                                   =     F(t 0 )                        p(t) (1 − F(t))    ,
                                                 m − k            j
                                      k=1                j=k
                            the result in part (i).
                              Now suppose that X 1 has an absolutely continuous distribution. Recall from Theorem
                            7.6 that X (m) has distribution function

                                                         n
                                                             n
                                                                   i        n−i
                                                F (m) (t) =    F(t) (1 − F(t))  ,
                                                             i
                                                        i=m
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