Chapter 4: E4.6 Solution

Working problems is a crucial part of learning mathematics. No one can learn topology merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.

James R. Munkres

Converse to SLLN

Let $Z$ be a non-negative RV. Let $Y$ be the integer part of $Z$ . Show that $Y=\sum_{n\in\mathbb{N}}I_{\{Z\ge n\}},$ and deduce that $\begin{array}{ccc} (*) & & \sum_{n\in\mathbb{N}}\mathbb{P}[Z\ge n]\le\mathbb{E}(Z)\le1+\sum_{n\in\mathbb{N}}\mathbb{P}[Z\ge n].\end{array}$

Let $(X_{n})$ be a sequence of lID RVs (independent, identically distributed random variables) with $\mathbb{E}(|X_{n}|)=\infty$ , $\forall n$ . Prove that $\sum_{n}\mathbb{P}[|X_{n}|>kn]=\infty(k\in\mathbb{N})\mbox{ and }\limsup\tfrac{|X_{n}|}{n}=\infty,\mbox{ a.s.}$

Deduce that if $S_{n}=X_{1}+X_{2}+\cdots+X_{n}$ , then $\limsup\tfrac{|S_{n}|}{n}=\infty,\mbox{ a.s.}$

From the first (trivial) equality and $Y\le Z<Y+1$ , we obtain the two inequalities. Let $E_{n}^{k}=\left\{ \tfrac{|X_{n}|}{n}>k\right\}$ . Then, $\begin{align*} \sum_{n}\mathbb{P}(E_{n}^{k}) & \ge\sum_{n}\mathbb{P}(\tfrac{|X_{n}|}{n}\ge k+1)\\ & \ge\mathbb{E}(\tfrac{|X_{n}|}{k+1})-1\\ & =\infty, \end{align*}$ and, hence, by BC2, $\mathbb{P}(E_{n}^{k},\text{ i.o.})=1,$ and $\begin{align*} \mathbb{P}(\limsup\tfrac{|X_{n}|}{n}\ge k) & \ge\mathbb{P}(E_{n}^{k},\text{ i.o.})\\ & =1. \end{align*}$ Hence, $\mathbb{P}(\limsup\tfrac{|X_{n}|}{n}=\infty)=1.$

Finally, $\begin{align*} \mathbb{P}(\limsup\tfrac{|S_{n}|}{n}<k) & \le\mathbb{P}(\tfrac{|S_{n}|}{n}<k,\text{ ev}). \end{align*}$ Now, if $-nk<S_{n}<nk$ , and $-(n+1)k<S_{n+1}<(n+1)k$ , then $-2k<\tfrac{X_{n+1}}{n+1}<2k$ , therefore, $\begin{align*} \mathbb{P}(\tfrac{|S_{n}|}{n}<k,\text{ ev}) & \le\mathbb{P}(\tfrac{|X_{n}|}{n}<2k,\text{ ev})\\ & =1-\mathbb{P}(\tfrac{|X_{n}|}{n}\ge2k,\text{ i.o.})\\ & =0. \end{align*}$ Hence, $\mathbb{P}(\limsup\tfrac{|S_{n}|}{n}=\infty)=1.$

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Chapter 4: E4.6 Solution

Chapter 4: E4.6 Solution