[math] \newcommand{\NA}{{\rm NA}} \newcommand{\mat}[1]{{\bf#1}} \newcommand{\exref}[1]{\ref{##1}} \newcommand{\secstoprocess}{\all} \newcommand{\NA}{{\rm NA}} \newcommand{\mathds}{\mathbb}[/math]

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 [math]X_i[/math] will have to depend on [math]i[/math], because otherwise both laws would be satisfied. (This problem was communicated to us by David Maslen.) \vskip .1in Suppose we have an infinite sequence of mutually independent events [math]A_1, A_2, \ldots[/math]. Let [math]a_i = P(A_i)[/math], and let [math]r[/math] be a positive integer.

• Find an expression of the probability that none of the [math]A_i[/math] with [math]i \gt r[/math] occur.
• Use the fact that [math]x-1 \leq e^{-x}[/math] to show that
  [[math]] P(\mbox{No\ $A_i$\ with\ \ltmath\gti \gt r[[/math]]
  \ occurs}) \leq e^{-\sum_{i=r}^{\infty} a_i} </math>
• (The first Borel-Cantelli lemma) Prove that if [math]\sum_{i=1}^{\infty} a_i[/math] diverges, then
  [[math]] P(\mbox{infinitely\ many\ $A_i$\ occur}) = 1. [[/math]]
  \vskip .1in Now, let [math]X_i[/math] be a sequence of mutually independent random variables such that for each positive integer [math]i \geq 2[/math],
  [[math]] 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}. [[/math]]
  When [math]i=1[/math] we let [math]X_i=0[/math] with probability [math]1[/math]. As usual we let [math]S_n = X_1 + \cdots + X_n[/math]. Note that the mean of each [math]X_i[/math] is [math]0[/math].
• Find the variance of [math]S_n[/math].
• Show that the sequence [math]\langle X_i \rangle[/math] satisfies the Weak Law of Large Numbers, i.e. prove that for any [math]\epsilon \gt 0[/math]
  [[math]] P\biggl(\biggl|{\frac{S_n}{n}}\biggr| \geq \epsilon\biggr) \rightarrow 0\ , [[/math]]
  as [math]n[/math] tends to infinity. \vskip .1in We now show that [math]\{ X_i \}[/math] does not satisfy the Strong Law of Large Numbers. Suppose that [math]S_n / n \rightarrow 0[/math]. Then because
  [[math]] \frac{X_n}{n} = \frac{S_n}{n} - \frac{n-1}{n} \frac{S_{n-1}}{n-1}\ , [[/math]]
  we know that [math]X_n / n \rightarrow 0[/math]. From the definition of limits, we conclude that the inequality [math]|X_i| \geq \frac{1}{2} i[/math] can only be true for finitely many [math]i[/math].
• Let [math]A_i[/math] be the event [math]|X_i| \geq \frac{1}{2} i[/math]. Find [math]P(A_i)[/math]. Show that [math]\sum_{i=1}^{\infty} P(A_i)[/math] diverges (use the Integral Test).
• Prove that [math]A_i[/math] occurs for infinitely many [math]i[/math].
• Prove that
  [[math]] P\biggl(\frac{S_n}{n} \rightarrow 0\biggr) = 0, [[/math]]
  and hence that the Strong Law of Large Numbers fails for the sequence [math]\{ X_i \}[/math].