### 数学代写|随机过程统计代写Stochastic process statistics代考|MTH 3016

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• Statistical Inference 统计推断
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• 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|Different Types of Random Walks

(a) Unrestricted Random Walk
In this the elements of transition matrix is given by $p_{i, i+1}=p, p_{i, i-1}=q$, for all integer $i(\ldots,-1,0,1,2, \ldots)$.
If $0<p<1$, the chain is irreducible. Then we have
$$p_{i j}^{(n)}=P\left(S_{n}=j-i\right)=\left(\begin{array}{c} n \ (n-j+i) i 2 \end{array}\right) p^{\frac{n+j-i}{2}} q^{\frac{n-j+i}{2}} \text { if } n \text { is even }$$
$=0$ if $n$ is odd.
and
$$p_{00}^{(n)}=\left(\begin{array}{c} n \ \frac{n}{2} \end{array}\right)(p q)^{n / 2}$$
The period of the chain is 2 .
It is transient if $p \neq \frac{1}{2}$ and null recurrent if $p=\frac{1}{2}$.

(b) Random Walk with an Absorbing Barrier
In this walk the elements of transition matrix are given by $p_{i, i+1}=p, p_{i, i-1}=q$, $(p+q=1), p_{00}=1$ for all $i \geq 1$.
‘ 0 ‘ is an absorbing state and the remaining states are all transient. $0,-1,-2$, $-3, \ldots$ are condensed into a single absorbing state ‘ 0 ‘.
Let $f_{i 0}^{(n)}=$ Probability of visiting ‘ 0 ‘ from $i$, first time in $n$ steps
$$=\left(\begin{array}{l} i \ n \end{array}\right)\left(\begin{array}{c} n \ (n-1) / 2 \end{array}\right) p^{(n-i) / 2} q^{(n+i) / 2}$$
Probability of visiting ‘ 0 ‘ from $i$ ever,
$$\begin{gathered} f_{i 0}=\sum_{n} f_{i 0}^{(n)} \text { satisfies difference equations } \ f_{i 0}=p f_{i+1,0}+q f_{i-1,0} \text { for } i>1, f_{10}=p f_{20}+q . \end{gathered}$$
Hence solving we get
$$f_{i 0}=\left{\begin{array}{l} 1 \text { if } p \leq q \ (q / p)^{i} \text { if } p \geq q \end{array}\right.$$
(c) Random Walk with Two Absorbing Barries
Here the elements of transition matrix is given by
$$p_{i, i+1}=p, p_{i, i-1}=q \text { for } 1 \leq i \leq a-1, p_{00}=1, p_{a a}=1 .$$
$’ 0$ ‘ and ‘ $a$ ‘ are absorbing and remaining states are transient.

## 数学代写|随机过程统计代写Stochastic process statistics代考|Examples of Random Walks with Absorbing Barriers

Gambler’s Ultimate Ruin Problem
The fortune of a gambler forms a M.C. with transition matrix
$p_{i j}=\left{\begin{array}{l}p \text { if } j=i+1 \ q \text { if } j=i-1 \ 0, \text { otherwise }\end{array}\right.$ and $i=2,3, \ldots, s$
$p_{i j}=\left{\begin{array}{l}1 \text { if } j=1 \ 0 \text { if } j \neq 1\end{array} \quad\right.$ and $i=1$ and $s .$
More explicitly the transition matrix is given by
$$P=\left[\begin{array}{ccccc} 1 & 0 & 0 & & 0 \ q & 0 & p & 0 & 0 \ 0 & q & 0 & p & 0 \ \cdots & \cdots & \cdots & \cdots & \cdots \ 0 & 0 & 0 & 0 & 1 \end{array}\right]$$

## 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, $P\left(X_{n}=+1\right)=p, p\left(X_{n}=-1\right)=q=1-p$.

In terms of gambling this can be described as follows:
If two gamblers play a series of games in which the probability of a particular player winning is $p$ for each game ( $q=1-p$ is the probability of losing a game). If the player loses he gives one unit of money to his opponent and if he wins he receives one unit from his opponent. If this particular player starts with $x$ units of money and his opponent with $s-x$ units, what is the probability of the player losing all his money? The absorbing barriers are ‘ 0 ‘ and ‘ $s$ ‘. When barrier ‘ 0 ‘ is reached the gambler is ruined.

Solution Let $p(x)$ be the probability of the particular player losing all his money if he now has $x$ units. Then we have the difference equation
\begin{aligned} &p(x)=p \cdot p(x+1)+q \cdot p(x-1) \text { if } 1<x<s-1 \ &p(1)=p \cdot p(2)=q, p(s-1)=\dot{q} \cdot p(s-2) \end{aligned}
Boundary conditions are: $p(0)=1, p(s)=0$
Auxiliary equation is $p x^{2}-x+q=0$ or $(x-1)(x-q / p)=0$
Solutions are $x=1$ and $q / p$.
General solution is $p(x)=A+B(q / p)^{x}$
From the boundary conditions $1=p(0)=A+B$
Hence
\begin{aligned} &0=p(s)=A+B(q / p)^{s} \ &B=\frac{1}{1-(q / p)^{s}}, A=\frac{-(q / p)^{s}}{1-(q / p)^{s}} \end{aligned}
The last expression follows from the fact that if $\frac{q}{p}=r \rightarrow 1$, then $\lim p(x)=1-\frac{x}{\varepsilon}$ (by L’Hospital’s Rule).

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

(a) 无限制随机游走

p一世j(n)=磷(小号n=j−一世)=(n (n−j+一世)一世2)pn+j−一世2qn−j+一世2 如果 n 甚至
=0如果n很奇怪。

p00(n)=(n n2)(pq)n/2

(b) 带有吸收障碍的随机游走

“0”是吸收状态，其余状态都是瞬态的。0,−1,−2,−3,…凝聚成单一的吸收态‘0’。

=(一世 n)(n (n−1)/2)p(n−一世)/2q(n+一世)/2

F一世0=∑nF一世0(n) 满足差分方程  F一世0=pF一世+1,0+qF一世−1,0 为了 一世>1,F10=pF20+q.

$$f_{i 0}=\left{ 1 如果 p≤q (q/p)一世 如果 p≥q\正确的。 (C)R一个nd○米在一个lķ在一世吨H吨在○一个bs○rb一世nG乙一个rr一世和sH和r和吨H和和l和米和n吨s○F吨r一个ns一世吨一世○n米一个吨r一世X一世sG一世在和nb是 p_{i, i+1}=p, p_{i, i-1}=q \text { for } 1 \leq i \leq a-1, p_{00}=1, p_{aa}=1 。$$
′0′ 和 ‘一个’是吸收和剩余的状态是短暂的。

## 数学代写|随机过程统计代写Stochastic process statistics代考|Examples of Random Walks with Absorbing Barriers

$p_{ij}=\left{的 MC p 如果 j=一世+1 q 如果 j=一世−1 0, 否则 \正确的。一个ndi=2,3, \ldots, sp_{ij}=\左{ 1 如果 j=1 0 如果 j≠1\四\右。一个nd我=1一个nd小号米○r和和Xpl一世C一世吨l是吨H和吨r一个ns一世吨一世○n米一个吨r一世X一世sG一世在和nb是磷=[1000 q0p00 0q0p0 ⋯⋯⋯⋯⋯ 00001]$

## 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）。

p(X)=p⋅p(X+1)+q⋅p(X−1) 如果 1<X<s−1 p(1)=p⋅p(2)=q,p(s−1)=q˙⋅p(s−2)

0=p(s)=一个+乙(q/p)s 乙=11−(q/p)s,一个=−(q/p)s1−(q/p)s

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