### 统计代写|随机过程作业代写stochastic process代考|Homogeneous Random Walk

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## 统计代写|随机过程作业代写stochastic process代考|Homogeneous Random Walk

Here state space is given by
\begin{aligned} &S={\ldots,-2,-1,0,1,2 \ldots} \ &p_{i}=p, q_{i}=q \text { for all } i \geq 1 . \end{aligned}
This is an irreducible M.C. Hence by solidarity theorem it is enough to consider the state ${0}$ only. The $n$-step recurrence probability is

${0}$ is transient iff $\sum_{n=0}^{\infty} P_{00}^{(n)}<\infty$ and recurrent iff $\sum_{n=0}^{\infty} P_{00}^{(n)}=\infty$ (by Theorem 2.5) \begin{aligned} \sum_{m=1}^{\infty}\left(\begin{array}{c} 2 m \ m \end{array}\right) p^{m} q^{m} & \cong \sum_{m=1}^{\infty} \frac{(4 p q)^{m}}{(\pi m)^{1 / 2}}<\infty \text { if } 4 p q<1 \\ &=\infty \text { if } 4 p q \geq 1 . \end{aligned} (using Stirling’s approximation for $\left.m ! \cong \sqrt{2 \pi} e^{-m} m^{m+1 / 2}\right) 4 p q>1$ is impossible for if $4 p q>(p+q)^{2}$ then $0>(p-q)^{2}$.
Hence
$4 p q<1$ if $p \neq q$
$$=1 \text { if } p=q=\frac{1}{2} \text {. }$$
Therefore $\sum_{m=0}^{\infty} p_{00}^{(2 m)}$ converges faster than the geometric series $\sum_{m}(4 p q)^{m}$ if $p \neq 1 / 2$.
Hence Random walk is recurrent iff $p=\frac{1}{2}$ and transient iff $p \neq \frac{1}{2}$.
We have shown in Exercise $2.1$ that a symmetric Random walk in one dimension is recurrent. Similarly it can be proved that in 2 -dimensions a symmetric Random walk is recurrent. But Polya proved that in $k \geq 3$ dimensions a symmetric Random walk is transient.

## 统计代写|随机过程作业代写stochastic process代考|Limit Theorems for Markov Chain

Definition 2.10 Let $d$ (i) be the greatest common divisor of those $n \geq 1$ for which $p_{i i}^{(n)}>0$. Then $d(i)$ is called the period of the state $i$. If $d(i)=1$, then the state $i$ is called aperiodic.
Note $i \leftrightarrow j$, then $d(i)=d(j)$.
There exists $n_{1}$ and $n_{2}$ such that $p_{i j}^{\left(n_{1}\right)}>0$ and $p_{j i}^{\left(n_{2}\right)}>0$.
Now $p_{i i}^{\left(n_{1}+n_{2}\right)} \geq p_{i j}^{\left(n_{1}\right)} p_{j i}^{\left(n_{2}\right)}>0$ and hence $d(i)$ is a divisor of $n_{1}+n_{2}$.
If $p_{j j}^{(n)}>0$, then $p_{i i}^{\left(n_{1}+n+n_{2}\right)} \geq p_{i j}^{\left(n_{1}\right)} p_{j j}^{(n)} p_{j i}^{\left(n_{2}\right)}>0$ (by Chapman Kolmogorov equation).

Hence, $d(i)$ is a divisor of $n_{1}+n+n_{2}$. So $d(i)$ must be a divisor of $n$ if $p_{j i}^{(n)}>0$.

Thus $d(i)$ is a divisor of $\left{n \geq 1: p_{j j}^{(n)}>0\right}$. Since $d(j)$ is the largest of such divisors, $d(i) \leq d(j)$. Hence, by symmetry $d(j) \leq d(i)$.
Hence $d(i)=d(j)$. Therefore having a period $d$ is a class property.
Note If $p_{i i}>0$, then $d(i)=1$ and this implies that a sufficient condition for an irreducible M.C. to be aperiodic is that $p_{i i}>0$ for some $i \in S$. Hence a queueing chain is aperiodic.
Theorem $2.7$ Limit Theorem (for diagonal elements)
Let $j$ be any state in a M.C. As $n \rightarrow \infty$.
(i) if $j$ is transient, then $p_{j j}^{(n)} \rightarrow 0$
(ii) if $j$ is null recurrent, then $p_{j j}^{(n)} \rightarrow 0$
(iii) if $j$ is positive (recurrent) and
(a) aperiodic, then $p_{j j}^{(n)} \rightarrow \frac{1}{\sum_{n=1}^{\infty} n f_{j j}^{(n)}}=\frac{1}{\mu_{j}}$ (mean recurrence time of $j$ )(b) periodic with period $d(j)$ then $p_{j j}^{(n d(j))} \rightarrow \frac{d(j)}{\mu_{j}}$. Write $d(j) / \mu_{j}=\pi_{j}$.

## 统计代写|随机过程作业代写stochastic process代考|Stationary Distribution

Definition $2.10$ A probability distribution is $\left{v_{j}\right}$ (i.e. $v_{j} \geq 0, \sum_{j} v_{j}=1$ ) is called a stationary distribution for a Markov chain with transition matrix $\left(p_{i j}\right)$ if
$v_{j}=\sum_{i} v_{i} p_{i j}$ for all $j=1,2, \ldots$
$=\sum_{i}\left(\sum_{k} v_{k} p_{k i}\right) p_{i j}$
$=\sum_{k} v_{k} \sum_{i} p_{k i} p_{i j} \quad$ (by Fubini’s Theorem)
$=\sum_{k} v_{k} p_{k j}^{(2)} \quad$ (by Chapman-Kolmogorov)
$\ldots=\sum_{k} v_{k} p_{k j}^{(n)} \quad$ (by induction)

Suppose a stationary distribution $\pi=\left(\pi_{1}, \pi_{2}, \ldots\right)$ exists. Also suppose
$$\lim {n \rightarrow \infty} p{i j}^{(n)}=\pi_{j} \geq 0 \text { for all } i \geq 1$$
Then $\pi$ is called the steady state distribution of the M.C. with transition matrix $\left(p_{i j}\right)$.

If the initial distribution $\left{a_{j}^{(0)}\right}\left(a_{j}^{(0)}=P\left(X_{0}=j\right)\right)$ is stationary, we have the marginal distribution of $X_{n}$ given by $a_{j}^{(n)}$ (i.e. $\left.a_{j}^{(n)}=P\left(X_{n}=j\right)\right)=\sum_{i} a_{j}^{(0)} a_{i j}^{(n)}=a_{j}^{(0)}$ (using (2.10)).

Thus, the unconditional (or marginal) distribution of $X_{n}$ is independent of $n$ and we may therefore say that the system (or the process) is in statistical equilibrium. Suppose conversely, that the distribution of $X_{n}$ is independent of $n$. Then the initial distribution $a_{0}=\left(a_{1}^{(0)}, a_{2}^{(0)}, \ldots\right)$ i.e. $a_{j}^{(0)}=P\left(X_{0}=j\right)=P\left(X_{1}=j\right)$ $=\sum_{i} a_{i}^{(0)} p_{i j}=a_{j}^{(1)}=\ldots$ and consequently, $a_{0}$ is a stationary distribution. Therefore the distribution of $X_{n}$ is independent of $n$ iff the initial distribution is a stationary distribution. Suppose (2.11) holds. Since $a_{j}^{(n)}=\sum_{i} a_{i}^{(0)} p_{i j}^{(n)} \rightarrow \pi_{j}$, we see that the limiting distribution of $X_{n}$ is given by $\pi$.

In other words, we can say that if (2.11) holds and a stationary distribution $\pi$ as $n \rightarrow \infty$. Denote $\frac{1}{\mu_{j}}=\pi_{j}$.

## 统计代写|随机过程作业代写stochastic process代考|Homogeneous Random Walk

0是瞬态的∑n=0∞磷00(n)<∞和复发当先∑n=0∞磷00(n)=∞（由定理 2.5）∑米=1∞(2米 米)p米q米≅∑米=1∞(4pq)米(圆周率米)1/2<∞ 如果 4pq<1=∞ 如果 4pq≥1.（使用斯特林的近似为米!≅2圆周率和−米米米+1/2)4pq>1如果是不可能的4pq>(p+q)2然后0>(p−q)2.

4pq<1如果p≠q
=1 如果 p=q=12.

## 统计代写|随机过程作业代写stochastic process代考|Limit Theorems for Markov Chain

(i) 如果j是瞬态的，那么pjj(n)→0
(ii) 如果j是零循环的，那么pjj(n)→0
(iii) 如果j是正的（经常性的）和
(a) 非周期性的，那么pjj(n)→1∑n=1∞nFjj(n)=1μj（平均复发时间j)(b) 有周期的周期性d(j)然后pjj(nd(j))→d(j)μj. 写d(j)/μj=圆周率j.

## 统计代写|随机过程作业代写stochastic process代考|Stationary Distribution

=∑一世(∑ķ在ķpķ一世)p一世j
=∑ķ在ķ∑一世pķ一世p一世j（由富比尼定理）
=∑ķ在ķpķj(2)（查普曼-科尔莫哥洛夫）
…=∑ķ在ķpķj(n)（通过感应）

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