Math 711 Homework 1 Austin Mohr January 27, 2011 Problem 1 Proposition 1. Convergence in probability of a sequence of random variables does not imply almost sure convergence of the sequence. Proof. Let ([0; 1]; B([0; 1]); λ) be the ambient probability space. Consider a process wherein, at the nth stage, we choose (independently at random) a closed 1 subinteral Fn of [0; 1] of length n . pr Define now the random variable Xn = I(Fn). Evidently, X −! 0, since, for 1 each n, Xn is nonzero only on the interval of length n . Thus, for any > 0, 1 P (jX j > ) = ! 0: n n To see that Xn does not enjoy almost sure convergence to zero, observe that, for each ! 2 [0; 1], 1 1 X X 1 P (X (!) 6= 0) = n n n=1 n=1 = 1: Thus, by the Borel Zero-One Law, P (Xn(!) 6= 0 i.o.) = 1, and so it cannot be that Xn converges almost surely to zero. Problem 2 Proposition 2. For any sequence fXng of random variables, there exists a Xn sequence of constants fang such that converges almost surely to zero. an 1 Proof. Let > 0 be given. Choose an such that P (jXnj > ) ≤ 2n . Observe that 1 1 X X 1 P (jX j > ) ≤ n 2n n=1 n=1 < 1; 1 and so the Borel-Cantelli Lemma gives that P ([jXnj > ] i.o.) = 0. That is, a.s. Xn −! 0. Problem 3 Proposition 3. For a monotone sequence of random variables, convergence in probability implies almost sure convergence. pr Proof. Let Xn be a monotone sequence of random variables with Xn −! X. It follows that, for all > 0, 0 = lim P (jXn − Xj > ) n!1 0 1 [ = lim P @ [jXn − Xj > ]A (by monotonicity of the Xn) N!1 n≥N = P lim sup[jXn − Xj > ] n!1 = P ([jXn − Xj > ] i.o.): a.s. Therefore, Xn −! X. Problem 4 Proposition 4. Let fXng be a sequence of random variables and define, for each n, Yn = XnIfjXnj ≤ ang. There exists a sequence fang of positive real numbers such that P fXn 6= Yn i.o.g = 0. 1 Proof. For each n, choose an such that P (jXnj > an) ≤ 2n . Thus, 1 1 X X P (Xn 6= Yn) = P (jXnj > an) n=1 n=1 1 X 1 ≤ 2n n=1 < 1: By the Borel-Cantelli Lemma, we conclude that P ([Xn 6= Yn] i.o.) = 0. Problem 5 Proposition 5. Suppose n points are chosen randomly on the unit circle. De- fine the random variable Xn to be the arc length of the largest arc not containing any of the chosen points. In this case, Xn ! 0 almost surely. 2 Proof. Let > 0 be given. We have that P ([Xn > ] i.o.) = P lim sup[Xn > ] n!1 0 1 [ = lim P @ [Xn > ]A N!1 n≥N = lim P (Xn > ) (by monotonicity of the Xn): n!1 4π We bound this last probability by breaking the unit circle into disjoint in- tervals of length 2 . Thus, P (Xn ≤ ) is no larger than the probability of having a point contained in a every interval. Thus, P ([Xn > ] i.o.) = lim P (Xn > ) n!1 4π 2π − n ≤ lim 2 n!1 2π = 0; a.s. and so Xn −! 0. Problem 6 Proposition 6. If, for all a < b, P f[fXn < ag i.o.] and [fXn > bg i.o.]g = 0; then limn!1Xn exists almost surely. Proof. By taking complements, we have P f[Xn ≥ a] or [Xn ≤ b]g = 1; for all sufficiently large n. If it is the case that the former holds for all a, then limn!1 Xn = 1 almost surely. Similarly, if the latter holds for all b, then limn!1 Xn = −∞ almost surely. Otherwise, there is some c so that P (Xn ≤ b) = 1 for all b ≥ c and P (Xn ≤ b) = 0 for all b < c. By keeping b = c fixed and letting a " b, we see that limn!1 Xn = c almost surely. Problem 7 Proposition 7. If Xn ! 0 in probability, then, for any α > 0, α jXnj pr α −! 0: 1 + jXnj 3 Proof. Let > 0 be given. Choose N such that 1 P (jXnj > α ) < , for all n ≥ N. It follows that α P (jXnj > ) < for all n ≥ N. Now, α jXnj α ≤ jXnj: 1 + jXnj Thus, for all n ≥ N, α jXnj α P α > ≤ P (jXnj > ) 1 + jXnj < , α jXnj pr and so α −! 0. 1+jXnj Problem 8 Proposition 8. If, for some α > 0, α jXnj pr α −! 0; 1 + jXnj pr then Xn −! 0. Proof. Let > 0 be given. Choose N such that α jXnj P α > < 1 + jXnj for all n ≥ N. Now, α α jXnj jXnj ≤ α : 1 + jXnj 1 + jXnj Thus, for all n ≥ N, α α jXnj jXnj P > ≤ P α 1 + jXnj 1 + jXnj < , and so jXnj 1 P > α < , 1 + jXnj pr from which it follows that Xn −! 0. 4 Problem 9 Proposition 9. Let fXng be a collection of independent random variables with 1 1 P fX = n2g = and P fX = −1g = 1 − n n2 n n2 Pn for all n. In this case, i=1 Xi converges almost surely to −∞ as n ! 1. 2 Proof. Observe first that, by definition of the Xn, P (Xn 2 fn ; −1g) = 1. That 2 is, except for a null set, Xn takes on only the values n or −1. Now, we have 1 1 X X 1 P (X = n2) = n n2 n=1 n=1 < 1; 2 and so P ([Xn = n ] i.o.) = 0 by the Borel-Cantelli Lemma. Similarly, 1 1 X X 1 P (X = −1) = 1 − n n2 n=1 n=1 = 1; and so P ([Xn = −1] i.o.) = 1 by the Borel Zero-One Law (note here that we require the independence of the Xn). Thus, for almost all ! 2 Ω, there exists N! such that Xn(!) = −1 for all n ≥ N!. It follows that n X lim Sn(!) = lim Xi(!) n!1 n!1 i=1 N X! X = Xi(!) + Xi(!) i=1 i>N! N X! X ≤ n2 + −1 i=1 i>N! = −∞; a.s. and so Sn −! −∞. 5.