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                                      4.3 Laplace Transforms and Moment-Generating Functions  105

                          For fixed t, define
                                                           n
                                                                 j
                                                    f n (x) =  (tx) /j!.
                                                           j=0
                        Note that for each fixed x,
                                                f n (x) → exp{tx} as n →∞.

                        Also,

                                                 n            n

                                                       j           j

                                                   (tx) /j!  ≤  |tx| /j! ≤ exp{|tx|}
                                       | f n (x)|≤
                                                 j=0         j=0

                        and, for |t| <δ,
                                                    E[exp(|tX|)] < ∞.
                        Hence, by the Dominated Convergence Theorem (see Appendix 1),
                                                     ∞
                                                         j   j
                                       lim E[ f n (X)] =  t E(X )/j! = M X (t), |t| <δ.
                                       n→∞
                                                     j=0
                        That is,
                                                      ∞
                                                         n    n
                                             M X (t) =  t E(X )/n!, |t| <δ.
                                                     n=0
                          The remaining part of the theorem now follows from the fact that a power series may be
                        differentiated term-by-term within its radius of convergence (see Appendix 3).

                        Example 4.13 (Poisson distribution). Let X denote a random variable with a Poisson
                        distributionwithparameterλ;seeExample4.10.Recallthatthemoment-generatingfunction
                        of this distribution is given by

                                          M(t) = exp{[exp(t) − 1]λ}, −∞ < t < ∞.
                          Note that


                                                   M (t) = M(t)exp(t)λ,
                                                              2

                                            M (t) = M(t)exp(2t)λ + M(t)exp(t)λ
                        and
                                                                     2
                                                                                3

                                      M (t) = M(t)[λ exp(t) + 3(λ exp(t)) + (λ exp(t)) ].
                        It follows that
                                                                            2
                                                               2


                                       E(X) = M (0) = λ,    E(X ) = M (0) = λ + λ
                        and
                                                      3
                                                                2
                                                                    3
                                                  E(X ) = λ + 3λ + λ .
                        In particular, X has mean λ and variance λ.
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