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assignmentutor-lab™ 为您的留学生涯保驾护航 在代写应用随机过程Stochastic process方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写应用随机过程Stochastic process代写方面经验极为丰富，各种代写应用随机过程Stochastic process相关的作业也就用不着说。

• 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|应用随机过程代写Stochastic process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
\begin{aligned} p_{i j}^{(n)}=P\left(S_{n}=j-i\right) &=\left(\begin{array}{c} n \ (n-j+i) / 2 \end{array}\right) p^{\frac{n+j-i}{2}} q^{\frac{n-j+i}{2}} \text { if } n \text { is even } \ &=0 \text { if } n \text { is odd. } \ p_{00}^{(n)} &=\left(\begin{array}{c} n \ \frac{n}{2} \end{array}\right)(p q)^{n / 2} \end{aligned}
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
$$f_{i 0}=\left{\begin{array}{l} 1 \text { if } p \leq q \ (q / p)^{i} \text { if } p \geq q \end{array}\right.$$

## 统计代写|应用随机过程代写Stochastic process代考|Waiting time for a gain

Let $\left{X_{i}\right}$ be sequence of i.i.d r.v.s with common distribution
$$P\left(X_{k}=1\right)=p, p\left(X_{k}=-1\right)=q, p+q=1 \text { and } S_{n}=X_{1}+\ldots+X_{n}, S_{0}=0 .$$
In gambling terminology $S_{n}, n \geq 1$ is the Peter’s accumulated gain at the end of the $n^{\text {th }}$ trial if Peter and Paul play for unit stakes. Now consider the event $A_{n}=\left{S_{1} \leq 0, S_{2} \leq 0, \ldots, S_{n-1} \leq 0, S_{n}=1\right}$.

Thus, the $n^{\text {th }}$ trial is the first to render Peter’s accumulated gain positive. The event $A_{n}$ is called first visit to $+1$ or the index $n$ is the passage time through 1 in random walk terminology.
Let $\phi_{n}=P\left(A_{n}\right)$. Define $\phi_{0}=0, \phi_{1}=p$.
If the event holds for $n>1$, then $S_{1}=-1$ and there exists a smallest integer $v$ $<n$ such that $S_{v}=0$. The outcome of the first $n$ trials may be described as follows: (1) At the first trial Peter looses an unit amount. (2) It takes $v-1$ further trials for Peter to reestablish the initial situation. (3) It takes exactly $n-v$ further trials for Peter to attain a positive net gain. These events depend on non-overlapping blocks of trials and are therefore mutually independent, (2) and (3) have probabilities $\phi_{v-1}$ and $\varphi_{n-v} .$

## 统计代写|应用随机过程代写Stochastic process代考|Different Types of Random Walks

(a) 无限制随机游走

$$p_{i j}^{(n)}=P\left(S_{n}=j-i\right)=(n(n-j+i) / 2) p^{\frac{n+j i}{2}} q^{\frac{n j+i}{2}} \text { if } n \text { is even } \quad=0 \text { if } n \text { is odd. } p_{00}^{(n)}=\left(n \frac{n}{2}\right)(p q)^{n / 2}$$

(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。 $0,-1,-2,-3, \ldots$ 凝聚成单一的吸收态’ $0^{\prime}$ 。

$$=(i n)(n(n-1) / 2) p^{(n-i) / 2} q^{(n+i) / 2}$$

$$f_{i 0}=\sum_{n} f_{i 0}^{(n)} \text { satisfies difference equations } f_{i 0}=p f_{i+1,0}+q f_{i-1,0} \text { for } i>1, f_{10}=p f_{20}+q$$

## 统计代写|应用随机过程代写Stochastic process代考|Waiting time for a gain

$$P\left(X_{k}=1\right)=p, p\left(X_{k}=-1\right)=q, p+q=1 \text { and } S_{n}=X_{1}+\ldots+X_{n}, S_{0}=0 .$$

$\backslash 1$ eft 的分隔符缺失或无法识别

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assignmentutor™您的专属作业导师
assignmentutor™您的专属作业导师