Chapter 4: E4.5 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

Prove that if $G$ is a random variable with the normal $N(0,1)$ distribution, then, for $x>0$ , $\mathbb{P}(G>x)=\tfrac{1}{\sqrt{2\pi}}\int_{x}^{\infty}e^{-\tfrac{1}{2}y^{2}}dy\le\tfrac{1}{x\sqrt{2\pi}}e^{-\tfrac{1}{2}x^{2}}.$ Let $X_{1},X_{2},\ldots$ be a sequence of independent $N(0,1)$ variables. Prove that, with probability 1, $L\le1$ , where $L:=\limsup(X_{n}/\sqrt{2\log n}).$ (Harder. Prove that $\mathbb{P}(L=1)=1$ .) [Hint. See Section 14.8.]

Let $S_{n}:=X_{1}+X_{2}+\cdots+X_{n}$ . Recall that $S_{n}/\sqrt{n}$ has the $N(0,1)$ distribution. Prove that $\mathbb{P}(|S_{n}|<2\sqrt{n\log n},\mbox{ ev})=1.$ Note that this implies the Strong Law: $\mathbb{P}(S_{n}/n\to0)=1$ .

Remark. The Law of the Iterated Logarithm states that $\mathbb{P}\left(\limsup\tfrac{S_{n}}{\sqrt{2n\log\log n}}=1\right)=1.$ Do not attempt to prove this now! See Section 14.7.

We have $\begin{align*} \tfrac{1}{\sqrt{2\pi}}\int_{x}^{\infty}e^{-\tfrac{1}{2}y^{2}}dy & =\tfrac{1}{\sqrt{2\pi}}[\tfrac{1}{x}e^{-\tfrac{1}{2}x^{2}}-\int_{x}^{\infty}\tfrac{1}{y^{2}}e^{-\tfrac{1}{2}y^{2}}dy]\\ & \le\tfrac{1}{x\sqrt{2\pi}}e^{-\tfrac{1}{2}x^{2}}. \end{align*}$

Let $E_{n}^{\epsilon}=\{X_{n}/\sqrt{2\log n}>1+\epsilon\}$ . Then, for every $\epsilon>0$ , $\begin{eqnarray*} \sum_{n\in\mathbb{N}}\mathbb{P}(E_{n}^{\epsilon}) & \le & \sum_{n\in\mathbb{N}}\tfrac{1}{2(1+\epsilon)\sqrt{\pi\log n}}n^{-(1+\epsilon)^{2}}\\ & < & +\infty, \end{eqnarray*}$ $\mathbb{P}(E_{n}^{\epsilon},\mbox{ i.o.})=0$ , and, by taking a sequence of $\epsilon_{n}\searrow0$ , $n\in\mathbb{N}$ , we conclude that $\mathbb{P}(L\le1)=1$ .

Using Section 14.8, $\begin{eqnarray*} \sum_{n\in\mathbb{N}}\mathbb{P}(E_{n}^{0}) & \ge & \sum_{n\in\mathbb{N}}\tfrac{\sqrt{2\log n}}{(1+2\log n)\sqrt{2\pi}}n^{-1}\\ & = & +\infty, \end{eqnarray*}$ hence, $\mathbb{P}(E_{n}^{0},\mbox{ i.o.})=1$ , and $\mathbb{P}(L\ge1)=1$ .

Now, $\begin{eqnarray*} \mathbb{P}(|S_{n}|<2\sqrt{n\log n},\mbox{ ev}) & = & 1-\mathbb{P}(|S_{n}|\ge2\sqrt{n\log n},\mbox{ i.o.}). \end{eqnarray*}$ Let $E_{n}^{+}=\{S_{n}\ge2\sqrt{n\log n}\}$ and $E_{n}^{-}=\{S_{n}\le-2\sqrt{n\log n}\}$ . Then, $\begin{eqnarray*} \sum_{n\in\mathbb{N}}\mathbb{P}(E_{n}^{+}) & \le & \tfrac{1}{2\sqrt{2\pi\log n}}n^{-2}\\ & < & +\infty, \end{eqnarray*}$ hence, $\mathbb{P}(E_{n}^{+},\mbox{ i.o.})=0$ , and, similarly, $\mathbb{P}(E_{n}^{-},\mbox{ i.o.})=0$ . Therefore, $\mathbb{P}(|S_{n}|\ge2\sqrt{n\log n},\mbox{ i.o.})=0$ .

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Chapter 4: E4.5 Solution

Chapter 4: E4.5 Solution