Page 94 - Elements of Distribution Theory
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            052184472Xc03  CUNY148/Severini  May 24, 2005  2:34





                            80                         Characteristic Functions

                              The remainder of the proof is essentially the same as the m = 1 case considered above.
                            Since
                                                              [i sin(hX/2)]
                                                                         2m
                                               ϕ  (2m) (0) = lim E          X  2m
                                                        h→0      (hX/2) 2m
                            and
                                                                        2m
                                                                sin(hX/2)   2m
                                                 (2m)
                                               |ϕ   (0)|= lim E           X    ,
                                                          h→0    (hX/2) 2m
                            and by Fatou’s Lemma (see Appendix 1),
                                                       sin(hX/2)   2m       2m
                                                               2m
                                                 lim E           X    ≥ E(X   ),
                                                h→0     (hX/2) 2m
                            it follows that

                                                        E(X 2m ) ≤|ϕ (2m) (0)|
                            so that E(X 2m ) < ∞.


                            Example 3.9 (Normal distribution). Let X denote a random variable with a normal dis-
                            tribution with parameters µ and σ; see Example 3.6. This distribution has characteristic
                            function
                                                                σ  2
                                                                 2
                                                    ϕ(t) = exp −  t + iµt .
                                                                 2
                                                                                       r
                            Clearly, ϕ(t)is m-times differentiable for any m = 1, 2,... so that E(X )exists for all
                            r = 1, 2,.... It is straightforward to show that
                                                                              2
                                                                          2


                                                ϕ (0) = µ  and  ϕ (0) =−(µ + σ )
                            so that
                                                                             2
                                                                         2
                                                                   2
                                                 E(X) = µ  and  E(X ) = µ + σ .
                            Example 3.10 (Gamma distribution). Consider a gamma distribution with parameters α
                            and β,asin Example 3.4; recall that this distribution has characteristic function
                                                              β α
                                                     ϕ(t) =        ,  t ∈ R.
                                                           (β − it) α
                            Since the density function p(x) decreases exponentially fast as x →∞,itis straightforward
                                          m
                            to show that E(X )exists for any m and, hence, the moments may be determined by
                            differentiating ϕ.
                              Note that
                                                                                 β α
                                          (m)         m
                                         ϕ  (t) =−(−i) [α(α + 1) ··· (α + m − 1)]
                                                                             (β − it) α+m
                            so that
                                                                                 1
                                             (m)
                                                          m
                                            ϕ  (0) =−(−i) [α(α + 1) ··· (α + m − 1)]
                                                                                 β m
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