Mathematical Methods for Physics and Engineering - Matematica.NET

30.7 GENERATING FUNCTIONSand differentiating once more we obtaind 2 Φ X (t)∞∑dt 2 = n(n − 1)f n t n−2 ⇒ Φ ′′ X (1) = ∑∞ n(n − 1)f n = E[X(X − 1)].n=0n=0(30.75)Equation (30.74) shows that Φ ′ X (1) gives the mean of X. Using both (30.75) and(30.51) allows us to writeΦ ′′ X(1) + Φ ′ X(1) − [ Φ ′ X(1) ] 2= E[X(X − 1)] + E[X] − (E[X])2= E [ X 2] − E[X]+E[X] − (E[X]) 2= E [ X 2] − (E[X]) 2= V [X], (30.76)and so express the variance of X in terms of the derivatives of its probabilitygenerating function.◮A random variable X is given by the number of trials needed to obtain a first successwhen the chance of success at each trial is constant and equal to p. Find the probabilitygenerating function for X and use it to determine the mean and variance of X.Clearly, at least one trial is needed, and so f 0 =0.Ifn (≥ 1) trials are needed for the firstsuccess, the first n − 1 trials must have resulted in failure. ThusPr(X = n) =q n−1 p, n ≥ 1, (30.77)where q =1− p is the probability of failure in each individual trial.The corresponding probability generating function is thus∞∑∞∑Φ X (t) = f n t n = (q n−1 p)t nn=0n=1= p ∞∑(qt) n = p qq × qt1 − qt = pt1 − qt , (30.78)n=1where we have used the result for the sum of a geometric series, given in chapter 4, toobtain a closed-form expression for Φ X (t). Again, as must be the case, Φ X (1) = 1.To find the mean and variance of X we need to evaluate Φ ′ X (1) and Φ′′ X (1). Differentiating(30.78) givesΦ ′ pX(t) =⇒ Φ ′(1 − qt)X(1) = p 2 p = 1 2 p ,Φ ′′ X(t) =2pq ⇒ Φ ′′(1 − qt)X(1) = 2pq = 2q3 p 3 p . 2Thus, using (30.74) and (30.76),E[X] =Φ ′ X(1) = 1 p ,V [X] =Φ ′′ X(1) + Φ ′ X(1) − [Φ ′ X(1)] 2= 2qp 2 + 1 p − 1 p 2 = q p 2 .A distribution with probabilities of the general form (30.77) is known as a geometricdistribution and is discussed in subsection 30.8.2. This form of distribution is common in‘waiting time’ problems (subsection 30.9.3). ◭1159