Math Problems Vault

Problem 11.4.1

In Claim 3, show that the function $g$ as defined there is an element of $L^{1}(\Omega , \mathcal{F}, \mu )$ , so that we may meaningfully discuss the integral $\int_{\Omega } g \, d\mu$ .

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Remark.

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Claim 3: Let $F \subseteq \mathbb {N}$ be nonempty and finite, and for each $i \in F$ let $f_{i}$ denote a function in $L^{1}(\Omega_{i}, \mathcal{F}_{i}, \mu_{i})$ . Let $g$ denote the mapping on $\Omega$ given for every $\omega \in \Omega$ by the rule $\omega \mapsto \prod_{i \in F} f_{i}(\omega_{i})$ . Then

\int _{\Omega } g(\omega ) \, d\mu (\omega )=\prod _{i \in F} \int _{\Omega _{i}} f_{i}(\omega _{i}) \, d\mu _{i}(\omega _{i}).

Problem 11.4.2

In the statement of the $L^{2}$ Strong Law, replace the assumption that $\sum_{k=1}^{\infty } \frac{\left\| g_{k}\right\|_{2}^{2}}{k^{2}}$ converges with the assumption that $\sum_{k=1}^{\infty } c_{k} \left\| g_{k}\right\|_{2}^{2}$ converges, where $\left\{ c_{k}\right\}_{k=1}^{\infty }$ is a sequence of positive numbers such that $\sum_{k=1}^{\infty } c_{k}<+\infty$ . Then $\sum_{k=1}^{\infty } c_{k} f_{k}(\omega )$ converges for $\mu$ -a.e. $\omega \in \Omega$ .

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Remark.

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$L^{2}$ Strong Law of Large Numbers: For each $k \in \mathbb {N}$ , let $g_{k}$ denote a function in $L^{2}(\Omega_{k}, \mathcal{F}_{k}, \mu_{k})$ such that $\int_{\Omega_{k}} g_{k} \, d\mu_{k}=0$ . Let $f_{k}$ denote the function on $\Omega$ that assigns each $\omega \in \Omega$ to $g_{k}(\omega_{k})$ . Assume that $\sum_{k=1}^{\infty } \frac{\left\| g_{k}\right\|_{2}^{2}}{k^{2}}$ converges. Then $\lim_{n} \frac{1}{n} \sum_{k=1}^{n} f_{k}(\omega )=0$ for $\mu$ -a.e. $\omega \in \Omega$ .

Problem 11.4.3

In the statement of the $L^{2}$ Strong Law, replace the assumption that $\sum_{k=1}^{\infty } \frac{\left\| g_{k}\right\|_{2}^{2}}{k^{2}}$ converges with the assumption that $\lim_{n} \frac{1}{n^{2}} \sum_{k=1}^{n} \left\| f_{k}\right\|_{2}^{2}=0$ . Deduce for every $\epsilon >0$ that

\lim _{n} \mu \left(\left\{ \omega \in \Omega : \left|\frac{1}{n} \sum _{k=1}^{n} f_{k}(\omega )\right|>\epsilon \right\} \right)=0,

which shows that $\left\{ \frac{1}{n} \sum_{k=1}^{n} f_{k}\right\}_{n=1}^{\infty }$ converges in $\mu$ -measure to zero. This result is known as the Weak Law of Large Numbers.

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Remark.

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$L^{2}$ Strong Law of Large Numbers: For each $k \in \mathbb {N}$ , let $g_{k}$ denote a function in $L^{2}(\Omega_{k}, \mathcal{F}_{k}, \mu_{k})$ such that $\int_{\Omega_{k}} g_{k} \, d\mu_{k}=0$ . Let $f_{k}$ denote the function on $\Omega$ that assigns each $\omega \in \Omega$ to $g_{k}(\omega_{k})$ . Assume that $\sum_{k=1}^{\infty } \frac{\left\| g_{k}\right\|_{2}^{2}}{k^{2}}$ converges. Then $\lim_{n} \frac{1}{n} \sum_{k=1}^{n} f_{k}(\omega )=0$ for $\mu$ -a.e. $\omega \in \Omega$ .

Problem 11.4.4

(a)(b)(c)

Reconsider the framework of Exercise 11.2.5, and recall the definitions of $f_{n}$ and $h_{n}$ for each $n \in \mathbb {N}$ .

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(a)

The hypotheses of the Strong Law are satisfied since $\left\| f_{j}\right\|_{2}^{2}=\frac{1}{4}$ for each $j \in \mathbb {N}$ , and hence $\lim_{n} h_{n}=\lim_{n} \frac{1}{n}(f_{1}+\cdots +f_{n})=0$ $\mu$ -a.e. on $\Omega$ , that is $h_{n}$ converges to zero $\mu$ -a.e. on $\Omega$ .

(b)

Show that the result of (a) implies the result of part (d) of Exercise 11.2.5.

(c)

As with part (e) of Exercise 11.2.5, discuss the result of (a) in terms of tossing a fair coin.

Remark.

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Framework of Exercise 11.2.5: let $I=\mathbb {N}$ , let $\Omega_{i}=\left\{ 0,1\right\}$ for each $i \in \mathbb {N}$ , and let $\mathcal{F}_{i}=2^{\Omega_{i}}$ for each $i \in \mathbb {N}$ . For each $i \in \mathbb {N}$ , let $\mu_{i}$ be defined on $\mathcal{F}_{i}$ by writing $\mu_{i}(\left\{ 0\right\} )=\mu_{i}(\left\{ 1\right\} )=\frac{1}{2}$ . Let $\Omega =\underset {i=1}{\overset {\infty }{\times }} \Omega_{i}$ , let $\mathcal{F}$ denote the usual $\sigma$ -field on $\Omega$ , and let $\mu$ denote the corresponding product measure. For each $i \in \mathbb {N}$ , define $f_{i}$ on $\Omega$ by writing $f_{i}(\omega )=\omega_{i}-\frac{1}{2}$ . Next, for each $n \in \mathbb {N}$ , define $h_{n}=\frac{1}{n}(f_{1}+\cdots +f_{n})$ . Part (d) of Exercise 11.2.5 shows that $h_{n}$ converges in measure to $0$ ; part (e) interprets that result by letting $0$ and $1$ correspond to obtaining heads or tails, respectively, upon flipping a fair coin, so that $\frac{1}{n}(\omega_{1}+\cdots +\omega_{n})$ may in a certain sense be thought of as the proportion of tails obtained in the first $n$ flips.

Problem 11.4.5

For each $n \in \mathbb {N}$ , make the following definitions: $\Omega_{n}=\left\{ 0,1, \cdots , k-1\right\}$ , $\mathcal{F}_{n}=2^{\Omega_{n}}$ , and $\mu_{n}$ is the counting measure. Consider the usual product space $(\Omega , \mathcal{F}, \mu )$ as constructed in Sections 11.1 and 11.2. Fix $m \in \left\{ 0,1, \cdots , k-1\right\}$ , and for each $n \in \mathbb {N}$ define the mapping $g_{n}$ on $\Omega$ by writing $g_{n}(\omega )=1$ if $\omega_{n}=m$ , and $g_{n}(\omega )=0$ otherwise. Show that $\frac{1}{n} \sum_{i=1}^{n} g_{i} \rightarrow \frac{1}{k}$ in $\mu$ -measure. That is, show for every $\epsilon >0$ that

\lim _{n} \mu \left(\left\{ \omega \in \Omega : \left|\frac{1}{n} \sum _{i=1}^{n} g_{i}(\omega )-\frac{1}{k}\right|>\epsilon \right\} \right)=0.

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