Grinstead-Snell-Probability · Problem 16

Law of Large Numbers

Mathematics · weak-vs-strong-law-large-numbers

Problem

In this exercise, we shall construct an example of a sequence of random variables that satisfies the weak law of large numbers, but not the strong law. The distribution of XiX_i will have to depend on ii, because otherwise both laws would be satisfied. Suppose we have an infinite sequence of mutually independent events A1,A2,…A_1, A_2, \ldots. Let ai=P(Ai)a_i = P(A_i), and let rr be a positive integer. (a) Find an expression of the probability that none of the AiA_i with i>ri > r occur. (b) Use the fact that x−1≤e−xx - 1 \leq e^{-x} to show that P(No Ai with i>r occurs)≤e−∑i=r∞aiP(\text{No } A_i \text{ with } i > r \text{ occurs}) \leq e^{-\sum_{i=r}^{\infty} a_i} (c) (The first Borel-Cantelli lemma) Prove that if ∑i=1∞ai\sum_{i=1}^{\infty} a_i diverges, then P(infinitely many Ai occur)=1.P(\text{infinitely many } A_i \text{ occur}) = 1. Now, let XiX_i be a sequence of mutually independent random variables such that for each positive integer i≥2i \geq 2, P(Xi=i)=12ilog⁡i,P(Xi=−i)=12ilog⁡i,P(Xi=0)=1−1ilog⁡i.P(X_i = i) = \frac{1}{2i\log i}, \quad P(X_i = -i) = \frac{1}{2i\log i}, \quad P(X_i = 0) = 1 - \frac{1}{i\log i}. When i=1i = 1 we let Xi=0X_i = 0 with probability 1. As usual we let Sn=X1+⋯+XnS_n = X_1 + \cdots + X_n. Note that the mean of each XiX_i is 0. (d) Find the variance of SnS_n. (e) Show that the sequence ⟨Xi⟩\langle X_i \rangle satisfies the Weak Law of Large Numbers, i.e. prove that for any ϵ>0\epsilon > 0 P(∣Snn∣≥ϵ)→0,P\left(\left|\frac{S_n}{n}\right| \geq \epsilon\right) \to 0, as nn tends to infinity. (f) Let AiA_i be the event ∣Xi∣≥12i|X_i| \geq \frac{1}{2}i. Find P(Ai)P(A_i). Show that ∑i=1∞P(Ai)\sum_{i=1}^{\infty} P(A_i) diverges (use the Integral Test). (g) Prove that AiA_i occurs for infinitely many ii. (h) Prove that P(Snn→0)=0,P\left(\frac{S_n}{n} \to 0\right) = 0, and hence that the Strong Law of Large Numbers fails for the sequence {Xi}\{X_i\}.

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