# 统计代写|随机过程代写stochastic process代考|STATS217

#### Doug I. Jones

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## 统计代写|随机过程代写stochastic process代考|Proof of Borel’s Strong Law

Let $S_n:=\sum_{k=1}^n X_k$ and $Z_n:=\frac{1}{n} S_n$.
Lemma 1.5.7 If
$$\sum_{n \geq 1} P\left(\left|Z_n-p\right| \geq \varepsilon_n\right)<\infty$$
for some sequence of positive numbers $\left{\varepsilon_n\right}_{n \geq 1}$ converging to 0 , then the sequence $\left{Z_n\right}_{n \geq 1}$ converges $P$-a.s. to $p$.

Proof. If, for a given $\omega,\left|Z_n(\omega)-p\right| \geq \varepsilon_n$ finitely often (or f.o.; that is, for all but a finite number of indices $n$ ), then $\lim {n \uparrow \infty}\left|Z_n(\omega)-p\right| \leq \lim {n \uparrow \infty} \varepsilon_n=0$. Therefore
$$P\left(\lim {n \uparrow \infty} Z_n=p\right) \geq P\left(\left|Z_n-p\right| \geq \varepsilon_n \quad \text { f.o. }\right) .$$ On the other hand, $$\left{\left|Z_n-p\right| \geq \varepsilon_n \quad \text { f.o. }\right}=\overline{\left{\left|Z_n-p\right| \geq \varepsilon_n \quad \text { i.o. }\right}} .$$ Therefore $$P\left(\left|Z_n-p\right| \geq \varepsilon_n \quad \text { f.o. }\right)=1-P\left(\left|Z_n-p\right| \geq \varepsilon_n \quad \text { i.o. }\right) .$$ Hypothesis (1.30) implies (Borel-Cantelli lemma) that $$P\left(\left|Z_n-p\right| \geq \varepsilon_n \quad \text { i.o. }\right)=0 .$$ By linking the above facts, we obtain $$P\left(\lim {n \uparrow \infty} Z_n=p\right) \geq 1$$
and of course, the only possibility is $=1$.
In order to prove almost-sure convergence using Lemma 1.5.7, we must find some adequate upper bound for the general term of the series occurring in the left-hand side of (1.30).

## 统计代写|随机过程代写stochastic process代考|Measurable Functions

Remember the definition of a $\sigma$-field:
Definition 2.1.1 Denote by $\mathcal{P}(X)$ the collection of all subsets of a given set $X . A$ collection of subsets $\mathcal{X} \subseteq \mathcal{P}(X)$ is called a $\sigma$-field on $X$ if:
( $\alpha) X \in \mathcal{X}$
(B) $A \in \mathcal{X} \Longrightarrow \bar{A} \in \mathcal{X}$ and
One then says that $(X, \mathcal{X})$ is a measurable space. A set $A \in \mathcal{X}$ is called a measurable set.
Observe that (See Exercise 1.6.5):
$\left(\gamma^{\prime}\right) A_n \in \mathcal{X}$ for all $n \in \mathbb{N} \Longrightarrow \cap_{n=0}^{\infty} A_n \in \mathcal{X}$.
In fact, given the properties $(\alpha)$ and $(\beta)$, properties $(\gamma)$ and $\left(\gamma^{\prime}\right)$ are equivalent. Note also that $\varnothing \in \mathcal{X}$, being the complement of $X$. Therefore, a $\sigma$-field on $X$ is a collection of subsets of $X$ that contains $X$ and $\varnothing$, and is closed under countable unions, countable intersections and complementation.

The two simplest examples of $\sigma$-fields on $X$ are the gross $\sigma$-field $\mathcal{X}={\varnothing, X}$ and the trivial $\sigma$-field $\mathcal{X}=\mathcal{P}(X)$

Definition 2.1.2 The $\sigma$-field generated by a non-empty collection of subsets $\mathcal{C} \subseteq \mathcal{P}(X)$ is, by definition, the smallest $\sigma$-field on $X$ containing all the sets in $\mathcal{C}$ (see Exercise 2.4.1). It is denoted by $\sigma(\mathcal{C})$.
Of course, if $\mathcal{G}$ is a $\sigma$-field, $\sigma(\mathcal{G})=\mathcal{G}$, a fact that will often be used.
We now review some basic notions of topology.

# 随机过程代考

## 统计代写|随机过程代写stochastic process代考|Proof of Borel’s Strong Law

$$\sum_{n \geq 1} P\left(\left|Z_n-p\right| \geq \varepsilon_n\right)<\infty$$

$\lim n \uparrow \infty\left|Z_n(\omega)-p\right| \leq \lim n \uparrow \infty \varepsilon_n=0$. 所以
$P\left(\lim n \uparrow \infty Z_n=p\right) \geq P\left(\left|Z_n-p\right| \geq \varepsilon_n \quad\right.$ f.o. $)$

$$P\left(\left|Z_n-p\right| \geq \varepsilon_n \quad \text { f.o. }\right)=1-P\left(\left|Z_n-p\right| \geq \varepsilon_n\right.$$

$$P\left(\left|Z_n-p\right| \geq \varepsilon_n \quad \text { i.o. }\right)=0 .$$

$$P\left(\lim n \uparrow \infty Z_n=p\right) \geq 1$$

## 统计代写|随机过程代写stochastic process代考|Measurable Functions

$:(\alpha) X \in \mathcal{X}$
(二) $A \in \mathcal{X} \Longrightarrow \bar{A} \in \mathcal{X}$ 然后

$\left(\gamma^{\prime}\right) A_n \in \mathcal{X}$ 对全部 $n \in \mathbb{N} \Longrightarrow \cap_{n=0}^{\infty} A_n \in \mathcal{X}$.

2.4.1）。它表示为 $\sigma(\mathcal{C})$.

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