统计代写|随机过程作业代写stochastic process代考|Random Walks

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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|随机过程作业代写stochastic process代考|Different Types of Random Walks

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
$p_{i j}^{(n)}=P\left(S_{n}=j-i\right)=\left(\begin{array}{c}n \ (n-j+i) / 2\end{array}\right) p^{\frac{n+j-i}{2}} q^{\frac{n-j+i}{2}}$ if $n$ 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 $p=\frac{1}{2}$.

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|随机过程作业代写stochastic process代考|Random Walks with Absorbing Barriers

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
$$p(x)=p \cdot p(x+1)+q \cdot p(x-1) \text { if } 1p \ (q / p)^{x} \text { if } q<p \end{array}\right.$$
Let us investigate the effect of changing stakes,
If the amount of money held by two players are doubled, then
$$p_{2}(x)=\frac{(q / p)^{2 s}-(q / p)^{2 x}}{(q / p)^{2 s}-1}=p(x) \cdot \frac{(q / p)^{s}+(q / p)^{x}}{(q / p)^{s}+1}$$
depends only on the ratio $(q / p)$.
Let $p(s)$ be the Gambler’s ultimate winning probability.

统计代写|随机过程作业代写stochastic process代考|Random Walk with a Reflecting Barrier

$$P_{i, i+1}=p, P_{i, i-1}=q \text { for } i \geq 1, p_{00}=q, p_{01}=p$$
Here we imagine a barrier placed at $-1 / 2$ such that every time the particle moves to the left from 0 , it is reflected at the barrier and returns to ‘ 0 ‘. The chain is irreducible if $0<p<1$. To classify its states, consider the system of equations $y_{i}=\sum_{j=1}^{\infty} P_{i j} y_{j}$. Then we get
$$y_{i}=p y_{i+1}+q y_{i-1} \quad(i \geq 1)$$
i.e. $\quad p\left(y_{i+1}-y_{i}\right)=q\left(y_{i}-y_{i-1}\right)(i \geq 2), y_{1}=p y_{2}$.
Therefore by iteration we obtain
and
\begin{aligned} y_{i+1}-y_{i} &=y_{1}(q / p)^{i}, i \geq 1 \ y_{i}-y_{1} &=y_{1}\left{(q / p)+(q / p)^{2}+\ldots+(q / p)^{i-1}\right} \end{aligned}

Hence $y_{i}=\frac{1-(q / p)^{i}}{1-(q / p)} y_{1}, i \geq 1$, so that $y_{i}$ is bounded if $p>q$. Thus by Theorem $2.13$ (Foster-type theorem) The states are all transient if $p>q$ and recurrent if $p \leq q$, then the stationary distribution is given by
\begin{aligned} &\pi_{j}=\sum_{i=0}^{\infty} \pi_{i} p_{i j}=p \pi_{j-1}+q \pi_{j+1} \quad(j \geq 1) \ &\pi_{0}=q \pi_{0}+q \pi_{1} \Rightarrow \pi_{1}=\pi_{0} \frac{(1-q)}{q}=\pi_{0} \frac{p}{q} \end{aligned}
Proceeding successively $\pi_{j}=(p / q)^{j} \pi_{0}(j \geq 0)$,
where
$$\pi_{0}\left{1+(p / q)+(p / q)^{2}+\ldots\right}=1 .$$
If $p=q$, the series diverges and consequently $\pi_{0}=0$ and $\pi_{j}=0(j \geq 0)$ so that stationary distribution does not exist. Thus, if $p=q$, the states are null recurrent. If $p0$, and is the stationary distribution (the states are positive recurrent).

统计代写|随机过程作业代写stochastic process代考|Different Types of Random Walks

p一世j(n)=磷(小号n=j−一世)=(n (n−j+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

“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.

$$f_{i 0}=\left{1 如果 p≤q (q/p)一世 如果 p≥q\对。$$

统计代写|随机过程作业代写stochastic process代考|Random Walks with Absorbing Barriers

p(x)=p \cdot p(x+1)+q \cdot p(x-1) \text { if } 1p \ (q / p)^{x} \text { if } q<p \end {数组}\对。p(x)=p \cdot p(x+1)+q \cdot p(x-1) \text { if } 1p \ (q / p)^{x} \text { if } q<p \end {数组}\对。

p2(X)=(q/p)2s−(q/p)2X(q/p)2s−1=p(X)⋅(q/p)s+(q/p)X(q/p)s+1

统计代写|随机过程作业代写stochastic process代考|Random Walk with a Reflecting Barrier

IEp(是一世+1−是一世)=q(是一世−是一世−1)(一世≥2),是1=p是2.

\begin{对齐} y_{i+1}-y_{i} &=y_{1}(q / p)^{i}, i \geq 1 \ y_{i}-y_{1} &=y_{ 1}\left{(q / p)+(q / p)^{2}+\ldots+(q / p)^{i-1}\right} \end{对齐}\begin{对齐} y_{i+1}-y_{i} &=y_{1}(q / p)^{i}, i \geq 1 \ y_{i}-y_{1} &=y_{ 1}\left{(q / p)+(q / p)^{2}+\ldots+(q / p)^{i-1}\right} \end{对齐}

\pi_{0}\left{1+(p / q)+(p / q)^{2}+\ldots\right}=1 。\pi_{0}\left{1+(p / q)+(p / q)^{2}+\ldots\right}=1 。

广义线性模型代考

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MATLAB代写

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