Chapter 4: E4.9 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

Let $Y_{0},Y_{1},Y_{2},\dotsc$ be independent random variables with $\mathbb{P}(Y_{n}=+1)=\mathbb{P}(Y_{n}=-1)=\tfrac{1}{2},\forall n.$ For $n\in\mathbb{N}$ , define $X_{n}:=Y_{0}Y_{1}\dotsc Y_{n}.$ Prove that the variables $X_{1},X_{2},\dotsc$ are independent. Define $\begin{align*} \mathcal{Y}:=\sigma(Y_{1},Y_{2},\dotsc), & \mathcal{T}_{n}:=\sigma(X_{r}:r>n). \end{align*}$ Prove that $\mathcal{L}:=\bigcap_{n}\sigma(\mathcal{Y},\mathcal{T}_{n})\neq\sigma\left(\mathcal{Y},\bigcap_{n}\mathcal{T}_{n}\right)=:\mathcal{R}.$ Hint. Prove that $Y_{0}\in\text{m}\mathcal{L}$ and that $Y_{0}$ is independent of $\mathcal{R}$ .

Let $n_{1}\le\dotsb\le n_{k}$ . The probability space can be divided into $2^{k}$ parts, where in each part $X_{n_{1}},\dotsc,X_{n_{k}}$ are fixed. There is a bijective correspondence between every pair of parts (by inverting particular $Y_{n}$ ’s we can invert specific $X_{n}$ ’s). Therefore, for every $a_{1},\dotsc,a_{k}\in\{0,1\}$ , $\begin{align*} \mathbb{P}(X_{n_{i}}=a_{i},i=1,\dotsc,k) & =\tfrac{1}{2^{k}}\\ & =\prod_{i=1}^{k}\mathbb{P}(X_{n_{i}}=a_{i}). \end{align*}$

Now, $Y_{0}=\tfrac{X_{n}}{Y_{1}\dotsm Y_{n}},$ so that $\{Y_{0}\le x\}\in\sigma(\mathcal{Y},\mathcal{T}_{n})$ for each $n\in\mathbb{N}$ , and $Y_{0}\in\text{m}\mathcal{L}$ ; however, $Y_{0}$ is independent of both $\mathcal{Y}$ (clearly) and $\bigcap_{n}\mathcal{T}_{n}$ (a tail $\sigma$ -algebra), and, hence, $\text{m}\mathcal{R}$ (consider, for example, the $\pi$ -system consisting of intersections of $Y\in\mathcal{Y}$ and $T\in\bigcap_{n}\mathcal{T}_{n}$ ). Since $Y_{0}$ is not a.s. constant, we conclude that $\mathcal{L}\neq\mathcal{R}$ .

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Chapter 4: E4.9 Solution

Chapter 4: E4.9 Solution