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# 数学代写|随机过程Stochastic Porcesses代考|AMATH562 Theorems Regarding Finite Markov Chain

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## 数学代写|随机过程Stochastic Porcesses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
$$p_{i j}^{(n)} \rightarrow \pi_j>0 \text { (by Theorem 2.9). }$$
Hence $1=\sum_{i=1}^k p_{i j}^{(n)}$ for all $j$ and $n \geq 1$,
Therefore $k \pi_j=1 \Rightarrow \pi_j=\frac{1}{k}$.

## 数学代写|随机过程Stochastic Porcesses代考|Methods of Evaluation of the n-Step Transition Probability

(a) Method of Spectral Decomposition
Let $P$ be a NXN matrix with latent roots $\lambda_1, \ldots, \lambda_N$ all distinct and simple. Then $\left(P-\lambda_i I\right) U_i=0$ for the column latent vector $U_i$ and
$V_i^{\prime}\left(P-\lambda_i I\right)=0$ for the row latent vector $V_i$.
$A_i=U_i V_i^{\prime}$ are called latent or spectral matrix associated with $\lambda_i, i=1, \ldots, N$. The following properties of $A_i$ ‘s are well known:

(i) $A_i$ ‘s are idempotent, i.e. $A_i^2=A_i$,
(ii) they are orthogonal, i.e. $A_i A_j=0(i \neq j)$,
(iii) they give spectral decomposition $P=\sum_{i=1}^N \lambda_i A_i$. It follows from (i) to (iii), that
$$P^k=\left(\sum_{i=1}^N \lambda_i A_i\right)^k=\sum_{i=1}^N \lambda_i^k A_i=\sum_{i=1}^N \lambda_i^k U_i V_i^{\prime}$$
Also we know that $P^k=U D^k U^{-1}$ (by Diagonalisation Theorem) where
and
\begin{aligned} U &=\left(U_1, U_2, \ldots, U_N\right) \ D &=\left[\begin{array}{ccc} \lambda_1 & 0 & 0 \ 0 & \lambda_2 & \vdots \ 0 & \ldots & \lambda_N \end{array}\right] \end{aligned}

## 数学代写|随机过程Stochastic Porcesses代考|Theorems Regarding Finite Markov Chain

$$p_{i j}^{(n)} \rightarrow \pi_j>0(\text { by Theorem } 2.9)$$

## 数学代写|随机过程Stochastic Porcesses代考|Methods of Evaluation of the nStep Transition Probability

(a) 谱分解法

$A_i=U_i V_i^{\prime}$ 被称为与相关联的渚在或光谱矩阵 $\lambda_i, i=1, \ldots, N$. 的以下属性 $A_i$ 众所周知:
（一世) $A_i$ 是帛等的，即 $A_i^2=A_i$ ，
(ii) 它们是正交的，即 $A_i A_j=0(i \neq j)$ ，
(iii) 他们给出谱分解 $P=\sum_{i=1}^N \lambda_i A_i$. 从 (i) 到 (iii) 可知，
$$P^k=\left(\sum_{i=1}^N \lambda_i A_i\right)^k=\sum_{i=1}^N \lambda_i^k A_i=\sum_{i=1}^N \lambda_i^k U_i V_i^{\prime}$$

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