### 数学代写|随机过程统计代写Stochastic process statistics代考|MXB334

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (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|Theorems Regarding Finite Markov Chain

Theorem 2(a). In a M.C. with a finite number of states, there is no null state and not all states can be transient.

Proof Suppose the chain has $N<\infty$ states. If all states are transient, then letting $n \rightarrow \infty$ in the relation $\sum_{j=0}^{N} p_{i j}^{(n)}=1$ we get $0=1$ (since by Theorem $2.8$, $\lim {n \rightarrow \infty} p{i j}^{(n)}=0$ for each $j$ ), which is absured and hence not all states in a finite M.C. are transient. Consider the subchain $C_{1}$ formed by a closed set of null recurrent states. Then $\sum_{j \in C_{1}} p_{i j}^{(n)}=\alpha$ (say) $>0$. Letting $n \rightarrow \infty, 0=\alpha>0$ which is also absurd. So there cannot be any null recurrent state in a finite M.C.
Theorem 2(b). An irreducible M.C. having a finite number of states is positive recurrent.

Proof By previous theorem, there is no null recurrent state and not all states are transient. Suppose there is one transient state. Then all states are transient by Solidarity Theorem. Hence, all states are positive recurrent.

Exercise 2.6 If a finite M.C. is irreducible, aperiodic and has doubly stochastic transition matrix, then show that $\lim {n \rightarrow \infty} p{i j}^{(n)}=1 / k$, where $k$ is the number of states in the chain.

Solution If $j$ is a positive recurrent state in an aperiodic irreducible chain then
$$p_{i j}^{(n)} \rightarrow \pi_{j}>0(\text { by Theorem 2.9) }$$
Hence $1=\sum_{i=1}^{k} p_{i j}^{(n)}$ for all $j$ and $n \geq 1$,
$$\begin{array}{cc} \left(\begin{array}{cc} p_{11} & p_{12} \ldots p_{1 k} \ p_{21} & p_{22} \ldots p_{2 k} \ \ldots & \ p_{k 1} & p_{k 2} \ldots p_{k k} \end{array}\right)=1 \ 1 & 1 \ldots 1 \end{array}$$
Therefore $k \pi_{j}=1 \Rightarrow \pi_{j}=\frac{1}{k}$.

## 数学代写|随机过程统计代写Stochastic process statistics代考|Methods of Evaluation of the n-Step Transition Probability

(a) Method of Spectral Decomposition
Let $P$ be a NXN matrix with latent roots $\lambda_{1}, \ldots, \lambda_{N}$ all distinct and simple. Then $\left(P-\lambda_{j} I\right) U_{j}=0$ for the column latent vector $U_{j}$ and
$V_{i}^{\prime}\left(P-\lambda_{i} I\right)=0$ for the row latent vector $V_{i}$.
$A_{i}=U_{i} V_{i}^{\prime}$ are called latent or spectral matrix associated with $\lambda_{i}, i=1, \ldots, N$.
The following properties of $A_{i}$ ‘s are well known:

(i) $A_{i}$ ‘s are idempotent, i.e. $A_{i}^{2}=A_{i}$,
(ii) they are orthogonal, i.e. $A_{i} A_{j}=0(i \neq j)$,
(iii) they give spectral decomposition $P=\sum_{i=1}^{N} \lambda_{i} A_{i}$. It follows from (i) to (iii), that
$$P^{k}=\left(\sum_{i=1}^{N} \lambda_{i} A_{i}\right)^{k}=\sum_{i=1}^{N} \lambda_{i}^{k} A_{i}=\sum_{i=1}^{N} \lambda_{i}^{k} U_{i} V_{i}^{\prime} .$$
Also we know that $P^{k}=U D^{k} U^{-1}$ (by Diagonalisation Theorem) where
\begin{aligned} &U=\left(U_{1}, U_{2}, \ldots, U_{N}\right) \ &D=\left[\begin{array}{ccc} \lambda_{1} & 0 \ldots & 0 \ 0 & \lambda_{2} & \vdots \ 0 & \cdots & \lambda_{N} \end{array}\right] \end{aligned}
Since the latent vectors are determined uniquely only upto a multiplicative constant, we have chosen them such that $U_{i}^{\prime} V_{i}=1$. From $(2.21)$ one can get any power of $P$ knowing $\lambda_{i}$ ‘s and $A_{i}{ }^{\circ}$ ‘s.

Example $2.7$ We shall illustrate the last method with the help of Exercise $2.8$ of Section 2.7.

In our problem, $P=\left[\begin{array}{ccc}1 & 0 & 0 \ \frac{1}{4} & \frac{1}{2} & \frac{1}{4} \ \frac{1}{18} & \frac{8}{18} & \frac{9}{18}\end{array}\right]$ with characteristic equation
$$\left[\begin{array}{ccc} 1-\lambda & 0 & 0 \ \frac{1}{4} & \frac{1}{2} \lambda & \frac{1}{4} \ \frac{1}{18} & \frac{8}{18} & \frac{9}{18} \lambda \end{array}\right]=0 \text { or }(1-\lambda)\left(\lambda^{2}-\lambda+\frac{5}{36}\right)=0$$
So the eigenvalues are $\lambda_{1}=1, \lambda_{2}=\frac{1}{6}, \lambda_{3}=\frac{5}{6}$.

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

Let $\left{X_{n}, n=0,1,2,3, \ldots\right}$ be a sequence of independent discrete random variables taking integral values only and $S_{n}=X_{1}+X_{2} \ldots+X_{n}(n=0,1,2, \ldots)$. Then the sequence $\left{S_{n}\right}$ is a M.C. whose transition probabilities are given by,
$${ }^{(m)} p_{i j}=P\left(S_{m+1}=j \mid S_{m}=i\right)=P\left(X_{m+1}=j-i\right), i, j=\ldots,-2,-1,0,1,2, \ldots$$
(non-homogeneous random walk).
The chain represents a Random walk of a particle along a straight line, the magnitude of ‘jump’ at time $n$ being given by the random variable $X_{n}$. If $X_{0}$ is denotes the initial position of a particle then its position after $n$ jumps (at time $n$ ) is given by $S_{n}$. When $X_{n}$ ‘s are also indentically distributed, ${ }^{(n)} p_{i j}=p_{j-i}$ where $p_{j}$ $=P\left(X_{n}=j\right.$ ). We have then a homogeneous Random walk (RW). Such Random walks occur in fluctuation theory (sums of discrete or continuous random variables). In classical RW, $P\left(X_{n}=+1\right)=p, p\left(X_{n}=-1\right)=q=1-p$.

## 数学代写|随机过程统计代写Stochastic process statistics代考|Theorems Regarding Finite Markov Chain

p一世j(n)→圆周率j>0( 由定理 2.9)

(p11p12…p1ķ p21p22…p2ķ … pķ1pķ2…pķķ)=1 11…1

## 数学代写|随机过程统计代写Stochastic process statistics代考|Methods of Evaluation of the n-Step Transition Probability

(a) 谱分解

（一世）一个一世是幂等的，即一个一世2=一个一世,
(ii) 它们是正交的，即一个一世一个j=0(一世≠j),
(iii) 他们给出谱分解磷=∑一世=1ñλ一世一个一世. 从 (i) 到 (iii) 得出，

[1−λ00 1412λ14 118818918λ]=0 或者 (1−λ)(λ2−λ+536)=0

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

(米)p一世j=磷(小号米+1=j∣小号米=一世)=磷(X米+1=j−一世),一世,j=…,−2,−1,0,1,2,…
（非均匀随机游走）。

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