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                            114                        Moments and Cumulants

                            Hence, by Leibnitz’s rule,
                                                          r
                                                d r              d  j   d r− j
                                                              r


                                                   α (t) =         β (t)    α(t).
                                                dt r          j dt  j  dt r− j
                                                         j=0
                            The result now follows by evaluating both sides of this expression at t = 0.
                              Hence, applying this result to the moment-generating and cumulant-generating functions
                            yields a general formula relating moments to cumulants. In this context, α 1 ,α 2 ,... are the
                            moments and β 1 ,β 2 ,... are the cumulants. It follows that
                                                      r
                                                          r        r− j
                                               r+1
                                            E(X   ) =       κ j+1 E(X  ), r = 0, 1,....
                                                          j
                                                      j=0
                                                                                            r
                            An important consequence of this result is that κ r is a function of E(X),..., E(X ).
                              Lemma 4.1 can also be used to derive an expression for central moments in terms of
                            cumulants by interpreting α 1 ,α 2 ,... as the central moments and β 1 ,β 2 ,..., under the
                            assumption that α 1 = β 1 = 0. Hence,
                                                       2                   3
                                               E[(X − µ) ] = κ 2 ,  E[(X − µ) ] = κ 3
                            and
                                                                         2
                                                              4
                                                     E[(X − µ) ] = κ 4 + 3κ .
                                                                        2
                              The approach to cumulants taken thus far in this section requires the existence of the
                            moment-generating function of X.A more general approach may be based on the charac-
                                                                                          m
                            teristic function. Suppose that X has characteristic function ϕ X (t) and that E(X )exists and
                                                       (m)                                     (m)
                            is finite. Then, by Theorem 3.5, ϕ  (0) exists and, hence, the mth derivative of log ϕ  (t)
                                                       X                                       X
                            at t = 0exists. We may define the jth cumulant of X,1 ≤ j ≤ m,by
                                                          1 d  j

                                                    κ j =       log ϕ X (t)   .
                                                           j
                                                         (i) dt  j      t=0
                              Of course, if the cumulant-generating function X exists, it is important to confirm that
                            the definition of cumulants based on the characteristic function agrees with the definition
                            based on the cumulant-generating function. This fact is established by the following lemma.
                            Lemma 4.2. Let X denote a random variable with moment-generating function M X and
                            characteristic function ϕ X . Then, for any m = 1, 2,...,
                                              d m                1 d m

                                                             =         log ϕ X (t)   .
                                                                  m
                                                  log M X (t)
                                              dt m        t=0  (i) dt m        t=0
                            Proof. Fix m. Since the mth moment of X exists we may write
                                                        m
                                                           j    j       m
                                            M X (t) = 1 +  t E(X )/j! + o(t )as t → 0
                                                        j=1
                            and
                                                       m
                                                            j   j        m
                                           ϕ X (t) = 1 +  (it) E(X )/j! + o(t )as t → 0.
                                                      j=1
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