A Probability Path.pdf

# D lemma 762 suppose xn n 1 are independent random

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D Lemma 7.6.2 Suppose {Xn, n :::: 1} are independent random variables which are uniformly bounded in the sense that there exists a > 0 such that IXn(w)l a for all n :::: 1 and w E n. Then Ln Xn convergent almost surely implies that Ln E (X n) converges. Proof. The proof uses a technique called symmetrization. Define an independent sequence {Yn, n :::: 1} which is independent of {Xn, n :::: 1} satisfying Yn 4 Xn. Let

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7.6 The Kolmogorov Three Series Theorem 233 Then {Zn. n 2:: 1} are independent random variables, E(Zn) = 0, and the distri- bution of each Zn is symmetric which amounts to the statement that d Zn = -Zn. Further, Var(Zn) = Var(Xn) + Var(Yn) = 2Var(Xn) and IZnl IXnl + IYnl 2a. Since {X n, n 2:: 1} 4: {Yn, n 2:: 1} as random elements of IR 00 , the convergence properties of the two sequences are identical and since Ln X n is assumed almost surely convergent, the same is true of Ln Yn. Hence also Ln Zn is almost surely convergent. Since {Zn} is also uniformly bounded, we conclude from Lemma 7.6.1 that I:var(Zn) = L2Var(Xn) < oo. n n From the Kolmogorov convergence criterion we get Ln(Xn - E(Xn)) almost surely convergent. Since we also assume Ln X n is almost surely convergent, it can only be the case that Ln E (X n) converges. 0 We now tum to the proof of necessity of the Kolmogorov three series theorem. Re-statement: Given independent random variables {Xn. n 2:: 1} such that Ln X n converges almost surely, it follows that the following three series converge for any c > 0: (i) Ln P(IXnl >c), (ii) Ln Var(Xn 1[1Xnl:::c]), (iii) Ln E(Xn1[1Xn!:5c]}. Proof of necessity. Since Ln X n converges almost surely, we have X n 0 and thus P([IXnl >c) i.o.) = 0. By the Borel zero-one law, it follows that LP[IXnl >c)< 00. (7.33) n If (7.33) holds, then {Xn} and {Xn 1[iXnl:::cJ} are tail equivalent and one converges iff the other does. So we get that the uniformly bounded sequence {Xn1[1Xnl:::cJ} satisfies Ln Xn1[1Xnl:::c] converges almost surely . By Lemma 7.6.2, Ln £(Xn1[1Xnl:::cJ) (the series in (iii)) is convergent. Thus the infinite series of uniformly bounded summands L (Xn1[1Xnl:5c]- £(Xn1[1Xnl:5c])) n
234 7. Laws of Large Numbers and Sums of Independent Random Variables is almost surely convergent and by Lemma 7.6.1 I:var(Xn1[1Xnl=:cJ) < oo n which is (ii). 7. 7 Exercises 0 1. Random signs. Does Ln 1/ n converge? Does Ln ( -1 )n 1/ n converge? Let {Xn} be iid with Does Ln Xn/n converge? 1 P[Xn = ±1] = 2. Let {Xn} be iid, EXn = J1. , Var(Xn) = a 2 Set X = E?=t X;/n. Show that - 2 P 2 - - X) --+ a . n i=I 3. Occupancy problems. Randomly distribute r balls in n boxes so that the sample space n consists of nr equally likely elements. Write n Nn = L 1[ith box is empty) i=l for the number of empty boxes . Check P[i th box is empty] = (1 - )r n so that E(Nn) = n(l - n- 1 )r. Check that as r /n --+ c E(Nn)/n--+ e-c N I p -c n n-+e . For the second result, compute Var(Nn) and show Var(Nn/n)--+ 0. (7.34) (7 .35) 4. Suppose g : [0, 1] lR is measurable and Lebesgue integrable. Let fUn, n 2: 1} be iid uniform random variables and define X; = g(U;). In what sense does E?=t X; /n approximate Jd g(x)dx?

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