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• Foundations of Data Science 数据科学基础

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

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 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
$$\left(\begin{array}{cccccc} r_0 & p_0 & 0 & \ldots & . & \ q_1 & r_1 & p_1 & 0 . & \ldots & \ 0 & q_2 & r_2 & p_2 & 0 \ldots & . \ 0 & 0 & q_3 & r_3 & p_3 & 0 \ & . & . & . & \cdots & \end{array}\right)$$
when $d<\infty$ we assume that $r_i=0$ for $i \geq 0$ and $p_0=1$.
Particular Case: First consider that $d$ is still infinite and $r_1=0$ for $i \geq 0$, $p_0=1$. The stationary distribution is given by $$X=\left(x_0, x_1, x_2, \ldots\right)=\left(x_0, x_1, x_2, \ldots\right)\left(\begin{array}{ccccc} 0 & 1 & 0 & 0 & \cdots \ q_1 & 0 & p_1 & 0 & \cdots \ 0 & q_2 & 0 & p_2 & \cdots \ & \cdots & \ldots & & \end{array}\right)$$
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 \end{aligned}

## 统计代写|随机过程代写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)
$$x_k=\sum_{i \in S}\left(x_i\right) \cdot 0=0 \text { for all } k \in S$$
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.

# 随机过程代考

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

$\left(\begin{array}{lllllllllllllllllllllllll}r_0 & p_0 & 0 & \ldots & q_1 & r_1 & p_1 & 0 & \ldots & 0 & q_2 & r_2 & p_2 & 0 & \ldots & .0 & 0 & q_3 & r_3 & p_3 & 0 & . & . & \ldots & \end{array}\right)$

$$x_0=x_1 q_1, \quad x_1=x_0+x_2 q_2, x_3=x_2 p_2+x_4 q_4, \quad x_4=\ldots$$

## 统计代写|随机过程代写stochastic process代考|Foster type theorems

$$x_k=\sum_{k \in S} x_i p_{i k}^{(n)} \text { for all } n=1,2, \ldots$$

(2.19)
$x_k=\sum_{i \in S}\left(x_i\right) \cdot 0=0$ for all $k \in S$

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