Eitxo and since leitxn i 1 we have by dominated

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eitXo and since leitXn I :;: 1, we have by dominated convergence that t/>n(t) = EeitXn -+ EeitXo = t/>o(t) . 0 The proof of the harder part (ii) is given in the next section. We close this section with some simple but illustrative examples of the use of the continuity theorem where we make use of the harder half to prove conver- gence . Example 9.5.1 (WLLN) Suppose {Xn, n :::: 1} is iid with common chf t/>(t) and assume E(IXID < oo and E(XI) = f.J. . Then p Snfn-+ f.J.. Since convergence in probability to a constant is equivalent to weak conver- gence to the constant, it suffices to show the chf of Sn In converges to the chf of f.J., namely eiJI.t . We have (9 . 16) The last equality needs justification, but suppose for the moment it is true; this would lead to as desired. To justify the representation in (9.16), note from (9.6) with n = 1 that t t (t 2 IX1I 2 t ) It/>(;;) -1- i;;JJ.I:;: E 2;;z- A 2;;1XII , so it suffices to show that (9.17) Bring the factor n inside the expectation. On the one hand
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306 9. Characteristic Functions and the Central Limit Theorem and on the other as n --+ oo. So by dominated convergence, (9.17) follows as desired. 0 Example 9.5.2 (Poisson approximation to the binomial) Suppose the random variable Sn has binomial mass function so that P[Sn = k] = G)lo- p)n-k, k = 0, .. . , n. If p = p(n) --+ 0 as n --+ oo in such a way that np--+ ). > 0, then Sn => PO(A.) where the limit is a Poisson random variable with parameter A.. To verify this, we first calculate the chf of PO(A.). We have 00 -'A).k Eeit(PO('A)) = Leitk_e __ k=O k! 00 = e-'A L().eitl/k! = e-'Ae'Ae;, k=O _ 'A(e;'-1) -e . Recall we can represent a binomial random variable as a sum of iid Bernoulli random variables ;1. .. . , ;n where P[;l = 1] = p = 1- P[;l = 0]. So Eei tSn = ( 1 - p+eitp)n ( np(eit 1))n =(1 + p(eit -1)r = 1 + n- --+e'A<ei' -I> . The limit is the chf of PO().) just computed. 0 The final example is a more sophisticated version of Example 9.5.2. In queue- ing theory, this example is often used to justify an assumption about traffic inputs being a Poisson process. Example 9.5.3 Suppose we have a doubly indexed array of random variables such that for each n = 1, 2, ... , {;n , k 2: 1} is a sequence of independent (but not necessarily identically distributed) Bernoulli random variables satisfying P[;n,k = 1] = Pk(n) = 1- P[;n,k = 0], (9.18)
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9.6 The Selection Theorem, Tightness, and Prohorov's theorem 307 V Pk(n) = : 8(n) -+ 0, n -+ oo, n n (9.19) LPk(n) = A. E (0, oo), n-+ oo. (9.20) k=l k=l Then n L => PO(A.). k=l The proof is left to Exercise 13 . 9.6 The Selection Theorem, Tightness, and Prohorov'stheorem This section collects several important results on subsequential convergence of probability distributions and culminates with the rest of the proof of the continuity theorem. 9.6.1 The Selection Theorem We seek to show that every infinite family of distributions contains a weakly con- vergent subseqence. We begin with a lemma . Lemma 9.6.1 (Diagonalization) Given a sequence {a j , j ::: 1} of distinct real numbers and a family {unO.
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