### 统计代写|蒙特卡洛方法代写monte carlo method代考| MARKOV PROCESSES

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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

## 统计代写|蒙特卡洛方法代写monte carlo method代考|MARKOV PROCESSES

Markov processes are stochastic processes whose futures are conditionally independent of their pasts given their present values. More formally, a stochastic process $\left{X_{t}, t \in \mathscr{T}\right}$, with $\mathscr{T} \subseteq \mathbb{R}$, is called a Markov process if, for every $s>0$ and $t$,
$$\left(X_{t+s} \mid X_{u}, u \leqslant t\right) \sim\left(X_{t+s} \mid X_{t}\right)$$
In other words, the conditional distribution of the future variable $X_{t+s}$, given the entire past of the process $\left{X_{u}, u \leqslant t\right}$, is the same as the conditional distribution of $X_{t+s}$ given only the present $X_{t}$. That is, in order to predict future states, we only need to know the present one. Property (1.30) is called the Markov property.
Depending on the index set $\mathscr{T}$ and state space $\mathscr{E}$ (the set of all values the $\left{X_{t}\right}$ can take), Markov processes come in many different forms. A Markov process with a discrete index set is called a Markov chain. A Markov process with a discrete state space and a continuous index set (such as $\mathbb{R}$ or $\mathbb{R}_{+}$) is called a Markov jump process.

## 统计代写|蒙特卡洛方法代写monte carlo method代考|Markov Chains

Consider a Markov chain $X=\left{X_{t}, t \in \mathbb{N}\right}$ with a discrete (i.e., countable) state space $\mathscr{E}$. In this case the Markov property (1.30) is
$$\mathbb{P}\left(X_{t+1}=x_{t+1} \mid X_{0}=x_{0}, \ldots, X_{t}=x_{t}\right)=\mathbb{P}\left(X_{t+1}=x_{t+1} \mid X_{t}=x_{t}\right)$$
for all $x_{0}, \ldots, x_{t+1}, \in \mathscr{E}$ and $t \in \mathbb{N}$. We restrict ourselves to Markov chains for which the conditional probabilities
$$\mathbb{P}\left(X_{t+1}=j \mid X_{t}=i\right), i, j \in \mathscr{E}$$
are independent of the time $t$. Such chains are called time-homogeneous. The probabilities in (1.32) are called the (one-step) transition probabilities of $X$. The distribution of $X_{0}$ is called the initial distribution of the Markov chain. The one-step transition probabilities and the initial distribution completely specify the distribution of $X$. Namely, we have by the product rule (1.4) and the Markov property (1.30),
\begin{aligned} &\mathbb{P}\left(X_{U}=x_{U}, \ldots, X_{t}=x_{t}\right) \ &\quad=\mathbb{P}\left(X_{0}=x_{0}\right) \mathbb{P}\left(X_{1}=x_{1} \mid X_{0}=x_{0}\right) \cdots \mathbb{P}\left(X_{t}=x_{t} \mid X_{0}=x_{0}, \ldots X_{t-1}=x_{t-1}\right) \ &\quad=\mathbb{P}\left(X_{0}=x_{0}\right) \mathbb{P}\left(X_{1}=x_{1} \mid X_{0}=x_{0}\right) \cdots \mathbb{P}\left(X_{t}=x_{t} \mid X_{t-1}=x_{t-1}\right) . \end{aligned}
Since $\mathscr{E}$ is countable, we can arrange the one-step transition probabilities in an array. This array is called the (one-step) transition matrix of $X$. We usually denote it by $P$. For example, when $\mathscr{E}={0,1,2, \ldots}$, the transition matrix $P$ has the form
$$P=\left(\begin{array}{cccc} p_{00} & p_{01} & p_{02} & \cdots \ p_{10} & p_{11} & p_{12} & \cdots \ p_{20} & p_{21} & p_{22} & \cdots \ \vdots & \vdots & \vdots & \ddots \end{array}\right) \text {. }$$
Note that the elements in every row are positive and sum up to unity.
Another convenient way to describe a Markov chain $X$ is through its transition graph. States are indicated by the nodes of the graph, and a strictly positive $(>0)$ transition probability $p_{i j}$ from state $i$ to $j$ is indicated by an arrow from $i$ to $j$ with weight $p_{i j}$.

## 统计代写|蒙特卡洛方法代写monte carlo method代考|Random Walk on the Integers

Let $p$ be a number between 0 and 1 . The Markov chain $X$ with state space $\mathbb{Z}$ and transition matrix $P$ defined by
$$P(i, i+1)=p, \quad P(i, i-1)=\bar{q}=1-\bar{p}, \quad \text { for all } i \in \mathbb{Z}$$
is called a random walk on the integers. Let $X$ start at $0 ;$ thus, $\mathbb{P}\left(X_{0}=0\right)=1$. The corresponding transition graph is given in Figure 1.4. Starting at 0 , the chain takes subsequent steps to the right with probability $p$ and to the left with probability $q$.

We show next how to calculate the probability that, starting from state $i$ at some (discrete) time $t$, we are in $j$ at (discrete) time $t+s$, that is, the probability $\mathbb{P}\left(X_{t+s}=j \mid X_{t}=i\right)$. For clarity, let us assume that $\mathscr{E}={1,2, \ldots, m}$ for some fixed $m$, so that $P$ is an $m \times m$ matrix. For $t=0,1,2, \ldots$, define the row vector
$$\boldsymbol{\pi}^{(t)}=\left(\mathbb{P}\left(X_{t}=1\right), \ldots, \mathbb{P}\left(X_{t}=m\right)\right)$$
We call $\pi^{(t)}$ the distribution vector, or simply the distribution, of $X$ at time $t$ and $\pi^{(0)}$ the initial distribution of $X$. The following result shows that the $t$-step probabilities can be found simply by matrix multiplication.
Theorem 1.13.1 The distribution of $X$ at time $t$ is given by
$$\pi^{(t)}=\pi^{(0)} P^{t}$$
for all $t=0,1, \ldots .$ (Here $P^{0}$ denotes the identity matrix.)
Proof: The proof is by induction. Equality (1.33) holds for $t=0$ by definition. Suppose that this equality is true for some $t=0,1, \ldots$. We have
$$\mathbb{P}\left(X_{t+1}=k\right)=\sum_{i=1}^{m} \mathbb{P}\left(X_{t+1}=k \mid X_{t}=i\right) \mathbb{P}\left(X_{t}=i\right)$$
But (1.33) is assumed to be true for $t$, so $\mathbb{P}\left(X_{t}=i\right)$ is the $i$-th element of $\pi^{(0)} P^{t}$. Moreover, $\mathbb{P}\left(X_{t+1}=k \mid X_{t}=i\right)$ is the $(i, k)$-th element of $P$. Therefore, for every $k$,
$$\sum_{i=1}^{m} \mathbb{P}\left(X_{t+1}=k \mid X_{t}=i\right) \mathbb{P}\left(X_{t}=i\right)=\sum_{i=1}^{m} P(i, k)\left(\boldsymbol{\pi}^{(0)} P^{t}\right)(i)$$
which is just the $k$-th element of $\pi^{(0)} P^{t+1}$. This completes the induction step, and thus the theorem is proved.

By taking $\pi^{(0)}$ as the $i$-th unit vector, $\mathbf{e}{i}$, the $t$-step transition probabilities can be found as $\mathbb{P}\left(X{t}=j \mid X_{0}=i\right)=\left(\mathbf{e}_{i} P^{t}\right)(j)=P^{t}(i, j)$, which is the $(i, j)$-th element of matrix $P^{t}$. Ihus, to find the $t$-step transition probabilities, we just have to compute the $t$-th power of $P$.

## 统计代写|蒙特卡洛方法代写monte carlo method代考|MARKOV PROCESSES

(X吨+s∣X在,在⩽吨)∼(X吨+s∣X吨)

## 统计代写|蒙特卡洛方法代写monte carlo method代考|Random Walk on the Integers

∑一世=1米磷(X吨+1=ķ∣X吨=一世)磷(X吨=一世)=∑一世=1米磷(一世,ķ)(圆周率(0)磷吨)(一世)

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## MATLAB代写

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