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                                                    3.2 Basic Properties                      75

                        Then h is of bounded variation and for any a < b
                                           b            b              b

                                            |h(y)| dy =  F(x + y) dy −   F(y) dy
                                          a            a              a
                                                         b+x          a+x
                                                    =      F(z) dz −    F(z) dz
                                                       b            a
                                                    ≤ [F(b + x) − F(a)]x ≤ x.
                        Hence,

                                                      ∞
                                                        |h(y)| dy < ∞.
                                                     −∞
                          It follows from the Theorem 3.2 that
                                           1         T            ∞
                                    h(y) =    lim    exp{−ity}     h(z)exp(itz) dz dy
                                          2π T →∞  −T           −∞
                        provided that h is continuous at y; note that, in this expression, the integral with respect to
                        dz is simply the Fourier transform of h. Consider the integral
                                           ∞                    ∞

                                                             1
                                             h(z)exp{itz} dz =    h(z) d exp{itz}.
                                                            it
                                          −∞                    −∞
                        Using integration-by-parts, this integral is equal to
                                           1     ∞             exp{itz}     ∞

                                         −       exp{itz} dh(z) +     h(z)    .
                                           it  −∞                 it       −∞
                          Note that exp{itz} is bounded,
                                                 lim h(z) = lim h(z) = 0,
                                                 z→∞      z→−∞
                        and

                                  ∞                 ∞                     ∞
                                    exp{itz} dh(z) =  exp{itz} dF(x + z) −  exp{itz} dF(z)
                                 −∞                −∞                    −∞
                                                = ϕ(t)[exp{−itx}− 1].
                        Hence,
                                            ∞                    1 − exp(−itx)

                                              h(z)exp(itz) dz = ϕ(t)         ,
                                                                      it
                                           −∞
                        so that
                                           1         T  exp{−ity}− exp{−it(x + y)}
                                    h(y) =    lim                             ϕ(t) dt,
                                          2π T →∞  −T            it
                        provided that h is continuous at y, which holds provided that F is continuous at x and
                        x + y. Choosing y = x 0 and x = x 1 − x 0 yields the result.


                          Thus, given the characteristic function of a random variable X we may determine differ-
                        ences of F, the distribution function of X,of the form F(x 1 ) − F(x 0 ) for continuity points
                        x 0 , x 1 .However, since set of points at which F X is discontinuous is countable, and F is
                        right-continuous, the characteristic function determines the entire distribution of X. The
                        details are given in the following corollary to Theorem 3.3.
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