### 数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Math 632

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Proof of the Strong Law of Large Numbers

In this section we give a proof of the strong law of large numbers. Our proof of the strong law makes use of the Borel-Cantelli lemma.

Borel-Cantelli Lemma. For a sequence of events $A_{i}, i \geq 1$, let $N$ denote the number of these events that occur. If $\sum_{i=1}^{\infty} P\left(A_{i}\right)<\infty$, then $P(N=\infty)=0$.

Proof. Suppose that $\sum_{i=1}^{\infty} P\left(A_{i}\right)<\infty$. Now, if $N=\infty$, then for every $n<\infty$ at least one of the events $A_{n}, A_{n+1}, \ldots$ will occur. That is, $N=\infty$ implies that $\cup_{i=n}^{\infty} A_{i}$ occurs for every $n$. Thus, for every $n$
\begin{aligned} P(N=\infty) & \leq P\left(\cup_{i=n}^{\infty} A_{i}\right) \ & \leq \sum_{i=n}^{\infty} P\left(A_{i}\right) \end{aligned}

where the final inequality follows from Boole’s inequality. Because $\sum_{i=1}^{\infty} P\left(A_{i}\right)<\infty$ implies that $\sum_{i=n}^{\infty} P\left(A_{i}\right) \rightarrow 0$ as $n \rightarrow \infty$, we obtain from the preceding upon letting $n \rightarrow \infty$ that $P(N=\infty)=0$, which proves the result.

Remark. The Borel-Cantelli lemma is actually quite intuitive, for if we define the indicator variable $I_{i}$ to equal 1 if $A_{i}$ occurs and to equal 0 otherwise, then $N=\sum_{i=1}^{\infty} I_{i}$, implying that
$$E[N]=\sum_{i=1}^{\infty} E\left[I_{i}\right]=\sum_{i=1}^{\infty} P\left(A_{i}\right)$$
Consequently, the Borel-Cantelli theorem states that if the expected number of events that occur is finite then the probability that an infinite number of them occur is 0 , which is intuitive because if there were a positive probability that an infinite number of events could occur then $E[N]$ would be infinite.

## 数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Stochastic Processes

A stochastic process ${X(t), t \in T}$ is a collection of random variables. That is, for each $t \in T, X(t)$ is a random variable. The index $t$ is often interpreted as time and, as a result, we refer to $X(t)$ as the state of the process at time $t$. For example, $X(t)$ might equal the total number of customers that have entered a supermarket by time $t$; or the number of customers in the supermarket at time $t$; or the total amount of sales that have been recorded in the market by time $t$; etc.

The set $T$ is called the index set of the process. When $T$ is a countable set the stochastic process is said to be a discrete-time process. If $T$ is an interval of the real line, the stochastic process is said to be a continuous-time process. For instance, $\left{X_{n}, n=0,1, \ldots\right}$ is a discrete-time stochastic process indexed by the nonnegative integers; while ${X(t), t \geq 0}$ is a continuous-time stochastic process indexed by the nonnegative real numbers.

The state space of a stochastic process is defined as the set of all possible values that the random variables $X(t)$ can assume.

Thus, a stochastic process is a family of random variables that describes the evolution through time of some (physical) process. We shall see much of stochastic processes in the following chapters of this text.

Example 2.54. Consider a particle that moves along a set of $m+1$ nodes, labeled $0,1, \ldots, m$, that are arranged around a circle (see Fig. 2.3). At each step the particle is equally likely to move one position in either the clockwise or counterclockwise direction. That is, if $X_{n}$ is the position of the particle after its $n$th step then
$$P\left{X_{n+1}=i+1 \mid X_{n}=i\right}=P\left{X_{n+1}=i-1 \mid X_{n}=i\right}=\frac{1}{2}$$ where $i+1 \equiv 0$ when $i=m$, and $i-1 \equiv m$ when $i=0$. Suppose now that the particle starts at 0 and continues to move around according to the preceding rules until all the nodes $1,2, \ldots, m$ have been visited. What is the probability that node $i, i=1, \ldots, m$, is the last one visited?

## 数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|The Discrete Case

Recall that for any two events $E$ and $F$, the conditional probability of $E$ given $F$ is defined, as long as $P(F)>0$, by
$$P(E \mid F)=\frac{P(E F)}{P(F)}$$
Hence, if $X$ and $Y$ are discrete random variables, then it is natural to define the conditional probability mass function of $X$ given that $Y=y$, by
\begin{aligned} p_{X \mid Y}(x \mid y) &=P{X=x \mid Y=y} \ &=\frac{P{X=x, Y=y}}{P{Y=y}} \ &=\frac{p(x, y)}{p_{Y}(y)} \end{aligned}
for all values of $y$ such that $P{Y=y}>0$. Similarly, the conditional probability distribution function of $X$ given that $Y=y$ is defined, for all $y$ such that $P{Y=y}>0$, by
\begin{aligned} F_{X \mid Y}(x \mid y) &=P{X \leq x \mid Y=y} \ &=\sum_{a \leq x} p_{X \mid Y}(a \mid y) \end{aligned}
Finally, the conditional expectation of $X$ given that $Y=y$ is defined by
$$E[X \mid Y=y]=\sum_{x} x P{X=x \mid Y=y}$$ $$=\sum_{x} x p_{X \mid Y}(x \mid y)$$
In other words, the definitions are exactly as before with the exception that everything is now conditional on the event that $Y=y$. If $X$ is independent of $Y$, then the conditional mass function, distribution, and expectation are the same as the unconditional ones. This follows, since if $X$ is independent of $Y$, then
\begin{aligned} p_{X \mid Y}(x \mid y) &=P{X=x \mid Y=y} \ &=P{X=x} \end{aligned}

## 数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Proof of the Strong Law of Large Numbers

Borel-Cantelli引理。对于一系列事件一个一世,一世≥1， 让ñ表示发生的这些事件的数量。如果∑一世=1∞磷(一个一世)<∞， 然后磷(ñ=∞)=0.

## 数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|Stochastic Processes

P\left{X_{n+1}=i+1 \mid X_{n}=i\right}=P\left{X_{n+1}=i-1 \mid X_{n}=i\right }=\frac{1}{2}P\left{X_{n+1}=i+1 \mid X_{n}=i\right}=P\left{X_{n+1}=i-1 \mid X_{n}=i\right }=\frac{1}{2}在哪里一世+1≡0什么时候一世=米， 和一世−1≡米什么时候一世=0. 现在假设粒子从0开始，并继续根据前面的规则继续四处移动，直到所有节点1,2,…,米被访问过。节点的概率是多少一世,一世=1,…,米, 是最后一个访问吗？

## 数学代写|概率模型和随机过程代写Probability Models and Stochastic Processes代考|The Discrete Case

pX∣是(X∣是)=磷X=X∣是=是 =磷X=X,是=是磷是=是 =p(X,是)p是(是)

FX∣是(X∣是)=磷X≤X∣是=是 =∑一个≤XpX∣是(一个∣是)

=∑XXpX∣是(X∣是)

pX∣是(X∣是)=磷X=X∣是=是 =磷X=X

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## MATLAB代写

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