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                                                 9.7 Approximation of Sums                   291

                        with error

                                                        r
                                                        P(x) f (x) dx.                      (9.9)

                                                      m
                        Example 9.15 (Discrete uniform distribution). Let X denote a discrete random variable
                        with a uniform distribution on the set {0, 1,..., m} for some positive integer m; hence,

                                                          1
                                             Pr(X = j) =     ,  j = 0,..., m.
                                                        m + 1
                        Let f denote a bounded, real-valued function on [0, 1] and consider the expected value
                        E[ f (X/m)]. Let U denote an absolutely continuous random variable with a uniform distri-
                        bution on (0, 1). Here we consider the approximation of E[ f (X/m)] by E[ f (U)] for large

                        m.We assume that f is differentiable and that f satisfies the Lipschitz condition
                                           | f (s) − f (t)|≤ K|s − t|, s, t ∈ [0, 1]


                        for some constant K.
                          Using Theorem 9.17,
                                                     m
                                                1
                                  E[ f (X/m)] =        f ( j/m)
                                              m + 1
                                                    j=0
                                                1      m              1
                                            =           f (x/m) dx +      [ f (0) + f (1)]
                                              m + 1  0             2(m + 1)
                                                    1       m

                                              +             (x −	x − 1/2) f (x/m) dx.
                                                m(m + 1)  0
                        Changing the variable of integration,

                                    1      m             m      1          m
                                            f (x/m) dx =        f (u)du =     E[ f (U)].
                                  m + 1  0             m + 1  0          m + 1

                          Note that, for j ≤ x < j + 1,
                                                x −	x − 1/2 = x − j − 1/2;

                        hence,

                                   m                       m−1    j+1


                                   (x −	x − 1/2) f (x/m) dx =      (x − j − 1/2) f (x/m) dx
                                 0                          j=0  j
                                                                 1
                                                           m−1
                                                                 2     u + j + 1/2
                                                         =        uf  
            du.
                                                                 1         m
                                                            j=0  −  2
                        Since
                                                        1

                                                        2
                                                         cu du = 0
                                                        1
                                                       −
                                                        2
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