### 数学代写|随机过程统计代写Stochastic process statistics代考|STAT3021

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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|随机过程统计代写Stochastic process statistics代考|Discrete Time Markov Chain

1. Suppose $P$ is a stochastic matrix, then show that $P^{n}$ is also a stochastic matrix for all $n>1$.
2. If $P^{n}$ is stochastic, is $P$ stochastic?
3. Show that 1 is an eigenvalue if $A$ is a stochastic matrix, i.e.
$$|\lambda I-A|=0 \Rightarrow \lambda=1 \text {. }$$
Consider a sequence of trials with possible outcomes $E_{1}, E_{2}, \ldots, E_{k} \ldots$ To the pairs of outcomes $\left(E_{j}, E_{k}\right)$ we can associate some numbers (i.e. conditional probabilities) $P_{j k}$. The $\left{E_{k}\right}$ are referred to as the possible states of the system. Instead of saying that the $n$th trial results in $E_{k}$ one says that the $n$th step leads to $E_{k}$ or that $E_{k}$ is entered at the $n$th step.

We shall denote by $P_{j k}^{(n)}$ the probability of transition from $E_{j}$ to $E_{k}$ in exactly n steps i.e. the conditional probability of entering $E_{k}$ at the $n$th step from $E_{j}$. This is the sum of all the probabilities of all possible paths $E_{j} \rightarrow E_{j_{1}} \rightarrow \ldots E_{j_{n-1}} \rightarrow E_{k}$ of length $n$ starting at $E_{j}$. and ending at $E_{k}$
In particular,
$$p_{j k}^{(1)}=p_{j k} .$$
Theorem 2.2 (Chapman-Kolmogorov equation)
$$p_{i j}^{(n)}=\sum_{k \varepsilon s} p_{i k}^{(n-1)} p_{k j}$$
Proof
\begin{aligned} P_{i j}^{(n)} &=P\left[X_{n}=j \mid X_{0} \equiv i\right] \ &=\sum_{k E s} P\left[X_{n}=j\left|X_{n-1}=k, X_{0}=i\right| P\left[X_{n-1}=k \mid X_{0}=i\right]\right.\ &=\sum_{k \in s} P\left[X_{n}=j \mid X_{n-1}=k\right] P\left[X_{n-1}=k \mid X_{0}=i\right] \text { (by Markov property) } \ &=\sum_{k E s} P_{k j}^{(1)} P_{i k}^{(n-1)}=\sum_{k E s} P_{k j} P_{i k}^{(n-1)} \end{aligned}
Corollary For
$$0 \leq m \leq n, p_{i j}^{(n)}=\sum_{k \in s} p_{i k}^{(n-m)} p_{k j}^{(m)}$$

## 数学代写|随机过程统计代写Stochastic process statistics代考|A Few More Examples

(a) Independent trials
$P^{n}=P$ for all $n \geq 1$, where $p_{i j}=p_{j}$ i.e. all the rows are same.
(b) Success runs
Consider an infinite sequence of Bernoulli trials and at the $n$th trial the system is in the state $E_{j}$ if the last failure occurred at the trial number $n-j, j=0,1$, $2, \ldots$ and zero-th trial counts as failure. In other words, the index $j$ equals the length of uninterrupted run of successes ending at $n$th trial.
Here
$$p_{i j}^{(n)}=\left{\begin{array}{l} q p^{j} \text { for } j=0,1,2, \ldots, i+n-1 \ p^{j} \text { for } j=j+n \ 0 \text { otherwise } \end{array}\right.$$
This follows either directly or from Chapman-Kolmogorov’s equation. It can be shown that $P^{n}$ converges to a matrix whose all elements in the column $j$ equals $q p^{j}$, where the transition matrix $P$ is given by
$$P_{i j}=P\left(X_{n}=j \mid X_{n-1}=i\right)=\left{\begin{array}{l} p \text { if } j=i+1 \ q \text { if } j=0 \ 0 \text { otherwise } \end{array}\right.$$
(c) Two state M.C.
There are two possible states $E_{1}$ and $E_{2}$ in which the matrix of transition probability is of the form
$$P=\left(\begin{array}{cc} 1-p & p \ a & 1-a \end{array}\right), 0<p<1 \text { and } 0<a<1 .$$
The system is said to be in state $E_{1}$ if a particle moves in the positive direction and in $E_{2}$ if the direction is negative.
(d) Random walk with absorbing barriers
Let the possible states be $E_{0}, E_{1}, E_{2}, \ldots, E_{k}$. Consider the matrix of transition probabilities

## 数学代写|随机过程统计代写Stochastic process statistics代考|Decomposition of state space

It may be possible that $p_{i j}=0, p_{i j}^{(2)}=0$ but $p_{i j}^{(3)}>0$. We say that the state $j$ is accessible from state $i$ if $p_{i j}^{(n)}>0$ for some $n>0$. In notation $i \rightarrow j$, i.e. $i$ leads to $j$. If $i \rightarrow j$ and $j \rightarrow i$, then $i$ and $j$ communicate and we denote this by $i \leftrightarrow j$.
Definition 2.4 The state $i$ is essential if $i \rightarrow j$ implies $i \leftarrow j$, i.e. if any state $j$ is accessible from $i$, then $i$ is accessible from that state. We shall let $\mathcal{S}$ denote the set of all essential states. States that are not essential are called inessential.
Lemma $2.1 \quad i \leftrightarrow j$ defines an equivalence relation on $\mathfrak{S}$, the class of essential states.
Proof $i \leftrightarrow i$ (reflexivity)
(i) Since for each $i, \sum_{j E s} p_{i j}=1$ there exists at least one $j$ for which $p_{i j}>0$. But if $i$ is essential then there exists $m \geq 1$ such that $p_{j i}^{(m)}>0$. So by ChapmenKolmogrov equation $p_{i i}^{(m+1)} \geq p_{i j} p_{j i}^{(m)}>0$.
(ii) $i \leftrightarrow j \Leftrightarrow j \leftrightarrow i$ (symmetry)
(iii) $i \leftrightarrow j$ and $j \leftrightarrow k \Rightarrow i \leftrightarrow k$ (transitivity)
Proof of (iii)
To prove $i \rightarrow k$, since $i \rightarrow j p_{j i}^{(n)}>0$ for some $n \geq 1$ and $j \rightarrow k, p_{j k}^{(m)}>0$ for some $m \geq 1$.
Claim: $p_{i k}^{(l)}>0$ for some $l \geq 1$
$$0<p_{i j}^{(n)} p_{j k}^{(m)} \leq \sum_{j e s} p_{i j}^{(n)} p_{j k}^{(m)}=p_{i k}^{(n+m)} \text { (Chapman-Kolmogorov) }$$
Taking $l=m+n$,
$$i \rightarrow k \text { and similarly } k \rightarrow i \Rightarrow i \leftrightarrow k \text {. }$$
By Lemma 2.1, i.e. $\mathfrak{J}=\cup{C(i)$, where $C(i)={j \in \mathcal{S} \mid i \leftrightarrow j}$ is called a communicating class, i.e. the class of essential states is partitioned into disjoint equivalent classes (communicating classes).
Definition $2.5$
A Markov chain is called irreducible (or ergodic) if there is only one communicating class, i.e. if all states communicate with each other or every state can be reached from every other state.
Definition $2.6$
A subset $C$ of $S$ is called closed (or transient) if it is impossible to leave $C$ in one step i.e. $p_{i j}=0$ for all $i \in C$ and all $j \notin C$.

## 数学代写|随机过程统计代写Stochastic process statistics代考|Discrete Time Markov Chain

1. 认为磷是一个随机矩阵，那么证明磷n也是一个随机矩阵n>1.
2. 如果磷n是随机的，是磷随机？
3. 证明 1 是一个特征值，如果一个是一个随机矩阵，即
|λ我−一个|=0⇒λ=1.
考虑一系列具有可能结果的试验和1,和2,…,和ķ…到成对的结果(和j,和ķ)我们可以关联一些数字（即条件概率）磷jķ. 这\left{E_{k}\right}\left{E_{k}\right}称为系统的可能状态。而不是说n试验结果和ķ有人说n这一步导致和ķ或者那个和ķ被输入在n第一步。

pjķ(1)=pjķ.

p一世j(n)=∑ķesp一世ķ(n−1)pķj

0≤米≤n,p一世j(n)=∑ķ∈sp一世ķ(n−米)pķj(米)

## 数学代写|随机过程统计代写Stochastic process statistics代考|A Few More Examples

(a) 独立审判

(b) 成功运行

$$p_{ij}^{(n)}=\left{ qpj 为了 j=0,1,2,…,一世+n−1 pj 为了 j=j+n 0 否则 \正确的。 吨H一世sF○ll○在s和一世吨H和rd一世r和C吨l是○rFr○米CH一个p米一个n−ķ○l米○G○r○在′s和q在一个吨一世○n.我吨C一个nb和sH○在n吨H一个吨磷nC○n在和rG和s吨○一个米一个吨r一世X在H○s和一个ll和l和米和n吨s一世n吨H和C○l在米nj和q在一个lsqpj,在H和r和吨H和吨r一个ns一世吨一世○n米一个吨r一世X磷一世sG一世在和nb是 P_{ij}=P\left(X_{n}=j \mid X_{n-1}=i\right)=\left{ p 如果 j=一世+1 q 如果 j=0 0 否则 \正确的。 (C)吨在○s吨一个吨和米.C.吨H和r和一个r和吨在○p○ss一世bl和s吨一个吨和s和1一个nd和2一世n在H一世CH吨H和米一个吨r一世X○F吨r一个ns一世吨一世○npr○b一个b一世l一世吨是一世s○F吨H和F○r米 P=\左( 1−pp 一个1−一个\right), 0<p<1 \text { 和 } 0<a<1 。$$

(d) 具有吸收障碍的随机游走

## 数学代写|随机过程统计代写Stochastic process statistics代考|Decomposition of state space

(i) 因为对于每个一世,∑j和sp一世j=1至少存在一个j为此p一世j>0. 但如果一世是必不可少的，那么存在米≥1这样pj一世(米)>0. 所以由 ChapmenKolmogrov 方程p一世一世(米+1)≥p一世jpj一世(米)>0.
(二)一世↔j⇔j↔一世（对称）
（iii）一世↔j和j↔ķ⇒一世↔ķ（及物性）

0<p一世j(n)pjķ(米)≤∑j和sp一世j(n)pjķ(米)=p一世ķ(n+米) （查普曼-科尔莫哥洛夫）

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