Give another proof of Exercise 38 by computing the moment generating function of X, and then differentiating to obtain its moments. Hint: Let Now,

(f) =E[exp(x)] =E[E[exp(x)N]] Exp(X) N = n] = [exp(x)] - 6 = (x(!))" since N is independent of the X's where ox() E[e] is the moment generating function for the X's. Therefore, (t) = E[(ox (!))"]

Differentiation yields '(t) =E[N(x(t)) (f)], "(t) = E[N(N-1)(x(t))-2(x(t)) + N(x(t))-x (1)] Evaluate at = 0 to get the desired result.