### 统计代写|随机过程作业代写stochastic process代考|Special Chains and Foster Type Theorems

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## 统计代写|随机过程作业代写stochastic process代考|Special chains

If the Markov Chain is infinite, the number of equations given by $\pi(P-I)=0$ will be infinite involving an infinite number of unknowns. In some particular cases we can solve these equations. The following examples will illustrate this point.
Example $2.5 \quad$ Birth and-Death Chain (Non-Homogeneous Random Walk) Consider a birth and death chain on ${0,1,2, \ldots, d}$ or a set of non-negative integers i.e. where $d=\infty$. Assume that the chain is irreducible i.e. $p_{j}>0$ and $q_{j}>0$ in case $0 \leq j \leq d$ (i.e. when $d$ is finite) $p_{j}>0$ for $0 \leq j<\infty$ and $q_{j}>0$ for $0<j<\infty$ if $d$ is infinite. Consider the transition matrix

$$X=\left(x_{0}, x_{1}, x_{2}, \ldots\right)=\left(x_{0}, x_{1}, x_{2}, \ldots\right)\left(\begin{array}{ccccc} 0 & 1 & 0 & 0 & \cdots \ q_{1} & 0 & p_{1} & 0 & \cdots \ 0 & q_{2} & 0 & p_{2} & \ldots \ & \cdots & \cdots & & \end{array}\right)$$
or $X=X P$. Let $x_{0} \neq 0$. Then
\begin{aligned} &x_{0}=x_{1} q_{1}, \ &x_{1}=x_{0}+x_{2} q_{2}, \ &x_{3}=x_{2} p_{2}+x_{4} q_{4}, \ &x_{4}=\ldots \ &\ldots \ &y_{i}=\frac{x_{i}}{x_{0}}, y_{0}=1, i=1,2,3, \ldots \end{aligned}
Define
Then
\begin{aligned} &y_{1}=1 / q_{1}, y_{1}=1+y_{2} q_{2} \text { or } y_{2}=\frac{y_{1}-1}{q_{2}}=\frac{1-q_{1}}{q_{1} q_{2}}=\frac{p_{1}}{q_{1} q_{2}} \ &y_{3}=\frac{p_{1} p_{2}}{q_{1} q_{2} q_{3}}, \ldots, y_{n}=\frac{p_{1} p_{2} \ldots p_{n-1}}{q_{1} q_{2} \ldots q_{n}}>0 \quad \text { for all } n=1,2, \ldots \end{aligned}
(by assumption that all $p, q$ ‘s are $>0$ ).
By Theorems $2.9$ and $2.11$, the non-homogeneous random walk is positive recurrent if $0<\sum_{0}^{\infty} x_{i}<\infty$ i.e. $\sum_{1}^{\infty} y_{i}<\infty$ i.e. iff $$\sum_{n=1}^{\infty} \frac{p_{1} p_{2} \cdots p_{n-1}}{q_{1} q_{2} \cdots q_{n}}<\infty$$ Note that $\sum_{0}^{\infty} y_{i}=\sum_{0}^{\infty} x_{i} / x_{0}=1 / x_{0}$, since $\sum_{0}^{\infty} y_{i}=1$. Therefore $x_{n}=y_{r} x_{0}=y_{n} / \sum_{0}^{\infty} y_{i}$ gives the stationary distribution provided $\sum_{0}^{\infty} y_{\mathrm{i}}<\infty \cdot x_{0}$ still has to be determined. Now $$1=x_{0}+\sum_{i=1}^{\infty} x_{i}=x_{0}+\sum_{i=1}^{\infty} x_{0} y_{i} \Rightarrow x_{0}=\frac{1}{1+\sum_{i=1}^{\infty} y_{i}}$$ and so $x_{0}>0$, iff $\sum_{i=1}^{\infty} y_{i}<\infty$ i.e. iff $\sum_{i=1}^{\infty} y_{i}<\infty\left(\right.$ since $\left.y_{0}=1\right)$.
In fact if $\sum_{i} y_{i}=\infty$, the solution to $(2.18)$ is either identically zero or has infinite sum $\left(\sum_{i} x_{i}=\infty\right)$ and hence has no stationary distribution.

## 统计代写|随机过程作业代写stochastic process代考|Foster type theorems

The following theorems, associated with Foster, give criteria for transient and recurrent chains in terms of solution of certain equations. Assume that the M.C. is irreducible.

Theorem 2.11 (Foster, 1953) Let the Markov chain be irreducible. Assume that there exists $x_{k}, k \in S$ such that $x_{k}=\sum_{k \in S} x_{i} p_{i k}$ and $0<\sum_{k \in S}\left|x_{k}\right|<\infty$. Then the Markov Chain is positive recurrent (this is a sort of converse of Theorem $2.9$ ). Proof Since $y_{k}=\frac{1}{\sum_{k \in S}\left|x_{k}\right|}>0, \sum_{k \in S} y_{k}=1$.

Without loss of generality $\left{x_{k}, k \in S\right}$ is a stationary distribution of a M.C. Then

$$x_{k}=\sum_{k \in S} x_{i} p_{i k}^{(n)} \text { for all } n=1,2, \ldots$$
Suppose that there is no positive state.
Since the M.C. is irreducible, then all the states are either transient or null. In that case $p_{i k}^{(n)} \rightarrow 0$ as $n \rightarrow \infty$ for all $i, k \in S$. By Lebesgue Dominated Convergence Theorem, taking $n \rightarrow \infty$ in (2.19)
$$x_{k}=\sum_{i \in S}\left(x_{i}\right) .0=0 \text { for all } k \in S$$
But $0<\sum_{k \in S} x_{k}<\infty$ is a contradiction to $(2.20)$.
Hence, there is at least one positive recurrent state. Since M.C. is irreducible, by Solidarity Theorem the M.C. must be positive recurrent. Conclusion An ireducible aperiodic M.C. has a stationary distribution iff all states are positive recurrent.

Theorem 2.11(a) If the M.C. is positive recurrent the system of equations $x_{i}=\sum_{j=0}^{\infty} x_{j} p_{j i}$ has a solution such that $0<\sum_{j=0}^{\infty} x_{j}<\infty$.
(Proof may be found in Karlin and Taylor’s book.)
Theorem 2.12 The M.C. is transient iff $x_{i}=\sum_{j=0}^{\infty} p_{i j} x_{j}$ has a solution for $i \neq 0$, which is bounded and non-constant i.e. all $x_{i}$ ‘s are not equal.

Theorem $2.13$ The M.C. is positive recurrent if $x_{i} \geq \sum_{j=0}^{\infty} p_{i j} x_{j}$ has a solution such that $x_{i} \rightarrow \infty$ as $i \rightarrow \infty$ (see Chung’s book on Markov Chains with Stationary Transition Probabilities).

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

Theorem 2(a). In a M.C. with a finite number of states, there is no null state and not all states can be transient.

Proof Suppose the chain has $N<\infty$ states. If all states are transient, then letting $n \rightarrow \infty$ in the relation $\sum_{j=0}^{N} p_{i j}^{(n)}=1$ we get $0=1$ (since by Theorem $2.8$, $\lim {n \rightarrow \infty} p{i j}^{(n)}=0$ for each $j$, which is absured and hence not all states in a finite M.C. are transient. Consider the subchain $C_{1}$ formed by a closed set of null recurrent states. Then $\sum_{j \in C_{1}} p_{i j}^{(n)}=\alpha$ (say) $>0$. Letting $n \rightarrow \infty, 0=\alpha>0$ which is also absurd. So there cannot be any null recurrent state in a finite M.C.
Theorem 2(b). An irreducible M.C. having a finite number of states is positive recurrent.

Proof By previous theorem, there is no null recurrent state and not all states are

transient. Suppose there is one transient state. Then all states are transient by Solidarity Theorem. Hence, all states are positive recurrent.

Exercise $2.6$ If a finite M.C. is irreducible, aperiodic and has doubly stochastic transition matrix, then show that $\lim {n \rightarrow \infty} p{i j}^{(n)}=1 / k$, where $k$ is the number of states in the chain.

Solution If $j$ is a positive recurrent state in an aperiodic irreducible chain then $p_{i j}^{(n)} \rightarrow \pi_{j}>0($ by Theorem $2.9)$
Hence $1=\sum_{i=1}^{k} p_{i j}^{(n)}$ for all $j$ and $n \geq 1$,
Therefore $k \pi_{j}=1 \Rightarrow \pi_{j}=\frac{1}{k}$.

## 统计代写|随机过程作业代写stochastic process代考|Special chains

X0=X1q1, X1=X0+X2q2, X3=X2p2+X4q4, X4=… … 是一世=X一世X0,是0=1,一世=1,2,3,…

（假设所有p,q是>0）。

## 统计代写|随机过程作业代写stochastic process代考|Foster type theorems

Xķ=∑一世∈小号(X一世).0=0 对全部 ķ∈小号

（证明可以在 Karlin 和 Taylor 的书中找到。）

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