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Section 10.6 Recall from calculus: x 1 lim 1 + — = e x x kx 1 = ek lim 1 + — x x y k = ek lim 1 + — y y (Let y = kx in the previous limit.) If derivatives of f(x) up to order k are all continuous on an interval about 0 (zero), then for all x on this interval, we have (x – 0)2 f [2](0) (x – 0)3 f [3](0) + —————– + … f(x) = f(0) + (x – 0) f [1](0) + —————– 3! 2! (x – 0)k f [k](h) + —————– . k! for 0 < h < x . 1. Let X1 , X2 , … , Xn be a random sample from a Bernoulli distribution with success probability p. The following random variables are defined: n n Xi Xi – np n i=1 i=1 X V = i=1 i X= W= n np(1 – p) (a) Find the m.g.f. for each of V and X. From Corollary 5.4-1, we have that n (1) the m.g.f. of the random variable V = Xi is n i=1 MV (t) = (1 – p + pet) = (1 – p + pet)n . i=1 (We recognize that V has a b(n,p) distribution.) V (2) the m.g.f. of the random variable X = — is n t MX (t) = MV( — ) = (1 – p + pet / n)n . n (b) Find the limiting distribution for V with np equal to a given constant λ as n tends to infinity, forcing p to go to 0 (zero). Since np = is fixed, then lim MV(t) = lim (1 – p + n n et n pet)n npet n np = lim 1 – — + —— = n n n (et – 1) lim 1 – — + —— = lim 1 + ———— n n n n n n = t – 1) (e e The limiting distribution of V is a Poisson() distribution. Consequently, for small (or large!) values of p, a binomial distribution can be approximated by a Poisson distribution with mean = np. This should not be surprising, since the Poisson distribution was derived as the limit of a sequence of binomial distributions where p tended to zero. 1.-continued (c) Find the limiting distribution for V as n tends to infinity, with p a fixed constant. lim MV(t) = lim (1 – p + pet)n = n n We cannot find a limiting distribution for V. (d) Find the limiting distribution for X as n tends to infinity, with p a fixed constant. lim MX(t) = lim (1 – p + pet/n)n = n n lim (1 – p + p[1 + t/n + (t/n)2/2! + (t/n)3/3! + …])n = n lim (1 + p[t/n + (t/n)2/2! + (t/n)3/3! + …])n = n n n pt + pt2/(2!n) + pt3/(3!n2) + … pt ———————————— lim 1 + = lim 1 + — = n n n n It is intuitively obvious that all terms in the numerator except the first go to 0 as n , and (from advanced calculus) the terms going to 0 can be ignored. ept This is the moment generating function corresponding to a distribution where the value p has probability 1 (one). Suppose X1 , X2 , … , Xn is a random sample from any distribution with finite mean and finite variance 2. Let M(t) be the common moment generating function of Xi , that is, for each i = 1, 2, …, n, we have tX M(t) = E(e i ) From Corollary 5.4-1(b), we have that the moment generating function n Xi i=1 of the random variable X = is n n t t n MX (t) = M( — ) = M( — ) . n n i=1 With M(t) and M /(t) both continuous on an interval about 0 (zero), we have that for all t on this interval, M(t) = M(0) + t M /(h) = 1 + t M /(h) for 0 < h < t . Consequently, we have that for all t on this interval, n t t n / M (h) = MX (t) = M( — ) = 1 + — n n n t t / / + — [M (h) – M (0)] 1+ — n n t for 0 < h < — . n To investigate the limiting distribution of X as n, we consider n / / t + t [M (h) – M (0)] lim MX (t) = lim 1 + ————————— n n n It is intuitively obvious that the second t = et term in the numerator goes to 0 as n , = lim 1 + — n n and (from advanced calculus) this term can be ignored. This is the moment generating function corresponding to a distribution where the value has probability 1 (one). n Xi – For i = 1, 2, …, n, suppose Yi = ——— , and let n n Yi W= i=1 n Xi – n = i=1 (n) X– = / n . Let m(t) be the common m.g.f. for each Yi . Then for each i = 1, 2, …, n, we have E(Yi) = m /(0) = 0 , and Var(Yi) = E(Yi2) = m //(0) = 1 . From Theorem 5.4-1, we have that the moment generating function of the random variable W is n n t t MW (t) = m( — ) = m( — ) . i = 1 n n With m(t) and m /(t) both continuous on an interval about 0 (zero), we have that for all t on this interval, m(t) = m(0) + t m /(0) 1 2 m //(h) +— t 2 =1 1 2 m //(h) +— t 2 for 0 < h < t . Consequently, we have that for all t on this interval, t n MW (t) = m( — ) = n n t2 // 1+ — 2n m (h) = n t2 t2 // // (1) + — [m (h) – m (0)] 1+ — 2n 2n t for 0 < h < — . n To investigate the limiting distribution of W as n, we consider n 2 2 // // t / 2 + (t / 2) [m (h) – m (0)] lim MW (t) = lim 1 + ————————————– n n n It is intuitively obvious that the second t2 / 2 term in the numerator goes to 0 as n , and (from advanced calculus) this term can be ignored. This is the moment generating function corresponding to a standard normal (N(0,1)) distribution. /2 n = lim 1 + —— = e n n t2 This proves the following important Theorem in the text: Central Limit Theorem Theorem 5.6-1 1.-continued (e) Find the limiting distribution for W as n tends to infinity, with p a fixed constant. For each i, = E(Xi) = p , and 2 = Var(Xi) = p(1 p) . From the Central Limit Theorem, we have that limiting n n Xi – n Xi – np X–p i=1 i=1 distribution for W = = = —————– p(1 – p) / n n np(1 – p) is a N(0,1) (standard normal) distribution. 2. Suppose Y has a b(400, p) distribution, and we want to approximate P(Y 3). (a) If p = 0.001, explain why a Poisson distribution can be expected to give a good approximation of P(Y ≥ 3), and use the Poisson approximation to find this probability. In Class Exercise #1(b), we found that the limiting distribution of a sequence of b(n, p) distributions as n tends to infinity is Poisson when np remains fixed, which forces p to tend to 0 (zero). This suggests that the Poisson approximation to a binomial distribution is better when p is close to zero (or one). = np = (400)(0.001) = 0.4 P(Y 3) = 1 – 0.992 = 0.008 (b) What other distribution may potentially be used to approximate a binomial probability when p is not sufficiently close to zero (or one)? The Central Limit Theorem tells us that with a sufficiently large sample size n, the normal distribution can be used.