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 (|X | > ) = → 0. n n

To see that Xn does not enjoy almost sure convergence to zero, observe that, for each ω ∈ [0, 1],

∞ ∞ X X 1 P (X (ω) 6= 0) = n n n=1 n=1 = ∞.

Thus, by the Borel Zero-One Law, P (Xn(ω) 6= 0 i.o.) = 1, and so it cannot be that Xn converges almost surely to zero.

Problem 2

Proposition 2. For any sequence {Xn} of random variables, there exists a Xn sequence of constants {an} such that converges almost surely to zero. an 1 Proof. Let > 0 be given. Choose an such that P (|Xn| > ) ≤ 2n . Observe that ∞ ∞ X X 1 P (|X | > ) ≤ n 2n n=1 n=1 < ∞,

1 and so the Borel-Cantelli Lemma gives that P ([|Xn| > ] 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 (|Xn − X| > ) n→∞   [ = lim P  [|Xn − X| > ] (by monotonicity of the Xn) N→∞ n≥N = P lim sup[|Xn − X| > ] n→∞

= P ([|Xn − X| > ] i.o.).

a.s. Therefore, Xn −→ X.

Problem 4

Proposition 4. Let {Xn} be a sequence of random variables and deﬁne, for each n, Yn = XnI{|Xn| ≤ an}. There exists a sequence {an} of positive real numbers such that P {Xn 6= Yn i.o.} = 0.

1 Proof. For each n, choose an such that P (|Xn| > an) ≤ 2n . Thus,

∞ ∞ X X P (Xn 6= Yn) = P (|Xn| > an) n=1 n=1 ∞ X 1 ≤ 2n n=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→∞   [ = lim P  [Xn > ] N→∞ n≥N

= lim P (Xn > ) (by monotonicity of the Xn). n→∞

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→∞ 4π 2π − n ≤ lim 2 n→∞ 2π = 0,

a.s. and so Xn −→ 0.

Problem 6

Proposition 6. If, for all a < b,

P {[{Xn < a} i.o.] and [{Xn > b} i.o.]} = 0, then limn→∞Xn exists almost surely. Proof. By taking complements, we have

P {[Xn ≥ a] or [Xn ≤ b]} = 1, for all suﬃciently large n. If it is the case that the former holds for all a, then limn→∞ Xn = ∞ almost surely. Similarly, if the latter holds for all b, then limn→∞ 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→∞ Xn = c almost surely.

Problem 7

Proposition 7. If Xn → 0 in probability, then, for any α > 0,

α |Xn| pr α −→ 0. 1 + |Xn|

3 Proof. Let > 0 be given. Choose N such that

1 P (|Xn| > α ) < , for all n ≥ N. It follows that

α P (|Xn| > ) < for all n ≥ N. Now, α |Xn| α ≤ |Xn|. 1 + |Xn| Thus, for all n ≥ N,

α |Xn| α P α > ≤ P (|Xn| > ) 1 + |Xn| < ,

α |Xn| pr and so α −→ 0. 1+|Xn|

Problem 8

Proposition 8. If, for some α > 0,

α |Xn| pr α −→ 0, 1 + |Xn|

pr then Xn −→ 0. Proof. Let > 0 be given. Choose N such that

α |Xn| P α > < 1 + |Xn| for all n ≥ N. Now, α α |Xn| |Xn| ≤ α . 1 + |Xn| 1 + |Xn| Thus, for all n ≥ N,

α α |Xn| |Xn| P > ≤ P α 1 + |Xn| 1 + |Xn| < , and so |Xn| 1 P > α < , 1 + |Xn| pr from which it follows that Xn −→ 0.

4 Problem 9

Proposition 9. Let {Xn} be a collection of independent random variables with 1 1 P {X = n2} = and P {X = −1} = 1 − n n2 n n2 Pn for all n. In this case, i=1 Xi converges almost surely to −∞ as n → ∞. 2 Proof. Observe ﬁrst that, by deﬁnition of the Xn, P (Xn ∈ {n , −1}) = 1. That 2 is, except for a null set, Xn takes on only the values n or −1. Now, we have

∞ ∞ X X 1 P (X = n2) = n n2 n=1 n=1 < ∞,

2 and so P ([Xn = n ] i.o.) = 0 by the Borel-Cantelli Lemma. Similarly,

∞ ∞ X X 1 P (X = −1) = 1 − n n2 n=1 n=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 ω ∈ Ω, there exists Nω such that Xn(ω) = −1 for all n ≥ Nω. It follows that

n X lim Sn(ω) = lim Xi(ω) n→∞ n→∞ i=1 N Xω X = Xi(ω) + Xi(ω)

i=1 i>Nω N Xω X ≤ n2 + −1

i=1 i>Nω = −∞,

a.s. and so Sn −→ −∞.