## Statistical Identification of Reversible Markov Chain on Cyclic Graph

Xiang Xuyan,1, Fu Haiqin2, Zhou Jieming3, Deng Yingchun3, Yang Xiangqun3

 基金资助: 国家自然科学基金.  11671132湖南省教育厅科学研究重点项目.  19A342应用经济学湖南省应用特色学科

 Fund supported: the NSFC.  11671132the Key Scientific Research Project of Hunan Provincial Education Department.  19A342the Applied Economics of Hunan Province

Abstract

The statistical identification of Markov chain explores how to identify the transition rate matrix of the underlying Markov chain by partially observable data. SIMC on reversibly cyclic graphs (containing one cycle at least), as the most important and crucial class, is investigated then in this letter. As the differentials of hitting time distribution for a reversible Markov chain are expressed by taboo rates, the necessary condition is developed to identify the transition rate matrix and a general conclusion about sufficiency is provided. The proposed algorithms to exactly calculate all transition rates are developed. A numerical example is included to demonstrate the correctness of the proposed algorithms.

Keywords： Reversible Markov chain ; Transition rate matrix ; Statistical identification ; Cyclic graph ; Hitting time

Xiang Xuyan, Fu Haiqin, Zhou Jieming, Deng Yingchun, Yang Xiangqun. Statistical Identification of Reversible Markov Chain on Cyclic Graph. Acta Mathematica Scientia[J], 2020, 40(6): 1682-1698 doi:

## 2 主要结果

(Ⅰ)所有叶子状态的观测是确认无环子图的充分条件, 也是首选的有效方法.

(Ⅱ)两相邻状态的观测是确认一个环或子环的必要条件, 也是非常关键的打开环的方法; 如果它有$k$个子环, 那么$k$对来自每个子环的两相邻状态的观测是确认它的必要条件.

### 3 背景和预备知识

$\{X_t:t\geq0\}$是有限状态空间$S = \{0, 1, \cdots, M\}$上的连续时间不可约马尔可夫链, 其保守的$Q$ -矩阵为$Q = {(q_{ij})}_{S\times S}$, 使得$Q{\bf 1} = 0\ (q_{ij}\geq 0, j\neq i, q_{i} = -q_{ii}>0), $$\widetilde{\pi} = [\pi_0, \pi_{1}, \cdots, \pi_{M}]^{\top}$$ Q$的平稳分布, 满足

$$$\begin{array}{lll} &\widetilde{\pi}^{\top}{\bf 1} = 1, \\ &\widetilde{\pi}^{\top}Q = {\bf 0}. \end{array}$$$

$$$\pi_{i}q_{ij} = \pi_{j}q_{ji}, \forall i, j\in S.$$$

$\Pi\equiv {\rm diag}(\pi_0, \pi_{1}, \cdots, \pi_{M})$, 则它具有如下的矩阵形式

$$$\Pi^{1/2}Q\Pi^{-1/2} = (\Pi^{1/2}Q\Pi^{-1/2})^{\top}.$$$

$O$是状态空间中某些被观测的状态集合, 且满足可区分性条件, 见文献[32], 其他状态表示为$C = S-O$, 那么可以将Q矩阵写成如下的分块矩阵

$$$Q = \left(\begin{array}{ccc} Q_{oo} {\quad}& Q_{oc}\\ Q_{co} {\quad}& Q_{cc} \end{array}\right).$$$

$O$的击中时(相应地, 逗留时)可定义为$\tau = \inf\{t>0, X_t\in O\}$ (相应地, $\sigma = \inf\{t>0, $$X_t\not\in O\} ), 也称为 C 的逗留时(相应地, 击中时). 因为需要, 先回顾一下有关聚合马尔可夫链的基本结果. 为了记号简便, 令 C\equiv\{1, 2, \cdots, N\} , \Pi_c\equiv {\rm diag}(\pi_{1}, \pi_{2}, \cdots, \pi_{N}) .因为 Q_{cc} 是非奇异的, 所以它有 N 个负的特征值, 记为 -\alpha_{1}, -\alpha_{2}, \cdots, -\alpha_{N}, {(\alpha_{i}>0)} , 通过在 (\cdot, \cdot)_{\Pi_c} 下的对称参数化, 可将 Q_{cc} 进行对角化处理, 相应的正交矩阵为 E , 令 W = \Pi^{-1/2}_c E , D = W^{-1} , A = {\rm diag}(\alpha_1, \alpha_2, \cdots, \alpha_N) , 可得关于其击中时分布的如下引理[28, 32]. 引理3.1 O 的击中时 \tau (相应地, C 的逗留时)的概率密度函数服从 N -混合指数密度 其中 \gamma_i = \alpha_i(D{\bf 1})^{\top}_i(D{\bf 1})_i(\widetilde{\pi}^{\top}_{c}{\bf 1})^{-1} . 推论3.1 单个状态 i 的逗留时 \sigma 服从参数为 q_{i} 的指数分布 ### 3.2 击中分布与转移速率矩阵的微分关系 n\geq 1 , 令 $$d_n = \sum\limits_{i = 1}^{N}{\gamma_i}{\alpha_i^{n-1}}, \quad c_n = (\pi^{\top}_{c}{\bf 1})d_n = (1-\pi^{\top}_{o}{\bf 1})d_n,$$ (-1)^{n}d_n$$ O$的击中时分布函数在零时刻的$n$阶微分.由系3.1和(2.5)式, 可得如下引理.

$$$c_n = (-1)^{n}{\bf 1}^{\top}\Pi_c Q^n_{cc}{\bf 1}.$$$

$$$_H {\bar{q}_{ij}}^{(n)} = \sum\limits_{k_1, \cdots, k_{n-1}\notin H}q_{i, k_{1}}q_{k_{1}k_{2}}\cdots q_{k_{n-1}, j}, i, j\in S, n\geq1$$$

$$$_{H}\overline{q}^{(n)}_{ij} = \sum\limits_{k\notin H}\ _{H}\overline{q}^{(m)}_{ik}\ _{H}\overline{q}^{(n-m)}_{kj},$$$

$L(i, j;n) = (i, j_{1}, \cdots, j_{n-1}, j)$为从状态$i$到达状态$j$的一个路径, 这里, $n$表示路径长度.进一步, 称$i $$j 是互通的, 如果 q_{ij}>0$$ q_{ji}>0$, 记为$i\leftrightarrow j$.因此, 对于可逆马尔可夫链来说, $i\rightarrow j$意味着$i\leftrightarrow j$.与状态$i$直接可达的状态的数目称为状态$i$的节点度数, 记为$\deg(i)$; 如果一个状态$i$的度数$\deg(i) = 1$, 则称为叶子状态.

$_ {H}\overline{q}^{(n)}_{jj} $$j 出发, 经过 n 步转移后再要回到 j , 一般情况下, 它最多只能跑到距离它 n/2 步的状态, 所以, 对于给出的 j\in S , n\in N^{+} , 互不相同的 j, j_1, \cdots, j_m\notin H ( 1\leq m\leq [\frac{n}{2}] ), 我们用禁忌速率 _ {H}\overline{q}^{(n)}_{jj}(j\leftrightarrow j_1\leftrightarrow \cdots\leftrightarrow j_{m}) 表示从状态 j 出发, 沿着路径 L(j, j_m;m) = (j, j_1, \cdots, j_m) , 最远到达 j_m , 然后回到状态 j 的速率. 性质3.1 令 \overline{q}(j_1, j_2, \cdots j_k) = q_{j_1, j_2}q_{j_2, j_3}\cdots q_{j_{k-1}, j_k}q_{j_{k}, j_{k-1}}\cdots q_{j_3, j_2}q_{j_2, j_1}, \!\! 下列等式成立[29] \begin{eqnarray} _{\{i\}}\overline{q}^{(2k+1)}_{jj} (j\leftrightarrow j_1\leftrightarrow \cdots\leftrightarrow j_{k})& = &\bigg({\bf q}_{j_{k}}+2\sum\limits_{s\in \{j, j_1, \cdots, j_{k-1}\}}q_{s}\bigg)\ \overline{q}(j, j_1, \cdots, j_k){}\\ &&+ _{\{i\}}\overline{q}^{(2k+1)}_{jj}(j\leftrightarrow j_1\leftrightarrow \cdots\leftrightarrow j_{k-1}), \\ _{\{i\}}\overline{q}^{(2k+2)}_{jj}(j\leftrightarrow j_1\leftrightarrow \cdots\leftrightarrow j_{k+1})& = & q_{j_k, j_{k+1}}{{\bf q}_{j_{k+1}, j_k}}\ \overline{q}(j, j_1, \cdots, j_k){}\\ &&+_{\{i\}}\overline{q}^{(2k+2)}_{jj}(j\leftrightarrow j_1\leftrightarrow \cdots\leftrightarrow j_{k}).{} \end{eqnarray} 在后面的求解过程中, 每个公式中粗体字的量可能是其中的一个新的未知量. 推论3.2 O 的微分关系可用禁忌速率表达成[29] $$c_n = (-1)^{n+1}\sum\limits_{j\in O}\sum\limits_{i\in C}\pi_i\ _{O}\overline{q}^{(n)}_{ij} = (-1)^{n+1}\sum\limits_{j\in O}\pi_j\sum\limits_{i\in C}\ _{O}\overline{q}^{(n)}_{ji}.$$ 特别地, 若 O = \{i\} , 则推论3.2为 推论3.3 如果单个状态 i 在一个树(或 m 个状态的环)中, 那么关于 i 的微分关系能够解码为如下形式, 对所有的 n (或 n<m ), 有 \begin{eqnarray} c_1& = &\pi_i\sum\limits_{j}\ _{\{i\}}\overline{q}^{(1)}_{ij} = \sum\limits_{j\neq i}\pi_iq_{ij} = \sum\limits_{j\neq i}\pi_jq_{ji}, {}\\ c_2& = &\pi_i\ _{\{i\}}\overline{q}^{(2)}_{ii} = \sum\limits_{j\neq i}\pi_iq_{ij}q_{ji} = \sum\limits_{j\neq i}\pi_j (q_{ji})^2, {}\\ c_3& = &\pi_i\ _{\{i\}}\overline{q}^{(3)}_{ii} = \sum\limits_{j\neq i}\pi_iq_{ij}\ _{\{i\}}\overline{q}^{(1)}_{jj}q_{ji} = \sum\limits_{j\neq i}\pi_iq_{ij}q_jq_{ji} = \sum\limits_{j\neq i}\pi_j (q_{ji})^2q_j, \\ c_4& = &\pi_i\ _{\{i\}}\overline{q}^{(4)}_{ii} = \sum\limits_{j\neq i}\pi_iq_{ij}\ _{\{i\}}\overline{q}^{(2)}_{jj}q_{ji} = \sum\limits_{j\neq i}\pi_iq_{ij}\bigg[\sum\limits_{k\neq i}q_{jk}q_{kj}\bigg]q_{ji}{}\\ & = &\sum\limits_{j\neq i}\pi_j(q_{ji})^2\bigg[\sum\limits_{k\neq i}q_{jk}q_{kj}\bigg], {}\\ c_{n}& = &\pi_i\ _{\{i\}}\overline{q}^{(n)}_{ii} = \sum\limits_{j\neq i}\pi_iq_{ij}\ _{\{i\}} \overline{q}^{(n-2)}_{jj}q_{ji} = \sum\limits_{j\neq i}\pi_j(q_{ji})^2\ _{\{i\}}\overline{q}^{(n-2)}_{jj}, n\geq 5.{} \end{eqnarray} 右边第一个方程就是状态 i 的所有流出(或流入)速率的和.这里为了记号简洁, 让 q_{kk} = q_k , 即可去掉 c_n 右边前面的系数 (-1)^n . 上述可能的转移解码了 Q_{cc}$$ c_n$之间的信息, 使得大部分的转移都可由$c_n$求解出.为了求解过程的简单和计算的精确性, 通常叶子状态观测是首选.

## 5 环形链:推论2.1的证明

$$$c^{(m, m+1)}_1 = \pi_mq_{m, m-1}+\pi_{m+1}q_{m+1, m+2},$$$

$$$c^{(m, m+1)}_2 = \pi_mq^2_{m, m-1}+\pi_{m+1}q^2_{m+1, m+2}.$$$

$$$\frac{1}{E\sigma^{(i)}} = q_i = q_{i, i+1}+q_{i, i-1},$$$

$$$c^{(i)}_1 = \pi_{i-1}q_{i-1, i}+\pi_{i+1}q_{i+1, i},$$$

$$$c^{(i)}_2 = \pi_{i-1}q^2_{i-1, i}+\pi_{i+1}q^2_{i+1, i},$$$

$$$c^{(i)}_3 = \pi_{i-1}q^2_{i-1, i}\ {{\bf q}_{i-1}}+\pi_{i+1}q^2_{i+1, i}\ q_{i+1},$$$

$$$c^{(i)}_4 = \pi_{i-1}q^2_{i-1, i}(q^2_{i-1}+q_{i-1, i-2}\ {{\bf q}_{i-2, i-1}})+\pi_{i+1}q^2_{i+1, i}(q^2_{i+1}+q_{i+1, i+2}\ q_{i+2, i+1}),$$$

$\begin{eqnarray} c^{(i)}_{2s+1}& = &\pi_{i-1}q^2_{i-1, i}\big[({{\bf q}_{i-s}}+2\sum\limits^{s-1}_{k = 1}q_{i-k})\ \overline{q}(i-1, i-2, \cdots, i-s){}\\ &&+ _{\{i\}}\overline{q}^{(2s-1)}_{i-1, i-1}(i-1\leftrightarrow i-2\leftrightarrow \cdots\leftrightarrow i-s+1)\big]{}\\ & &+\pi_{i+1}q^2_{i+1, i}\big[({{\bf q}_{i+s}}+2\sum\limits^{s-1}_{k = 1}q_{i+k})\ \overline{q}(i+1, i+2, \cdots, i+s){}\\ &&+ _{\{i\}}\overline{q}^{(2s-1)}_{i+1, i+1}(i+1\leftrightarrow i+2\leftrightarrow \cdots\leftrightarrow i+s-1)\big], \end{eqnarray}$

$\begin{eqnarray} c^{(i)}_{2s+2}& = &\pi_{i-1}q^2_{i-1, i}[q_{i-s, i-s-1}{{\bf q}_{i-s-1, i-s}}\ \overline{q}(i-1, i-2, \cdots, i-s){}\\ &&+_{\{i\}}\overline{q}^{(2s)}_{i-1, i-1}(i-1\leftrightarrow i-2\leftrightarrow \cdots\leftrightarrow i-s)]{}\\ & &+\pi_{i+1}q^2_{i+1, i}[q_{i+s, i+s+1}{{\bf q}_{i+s+1, i+s}}\ \overline{q}(i+1, i+2, \cdots, i+s){}\\ &&+_{\{i\}}\overline{q}^{(2s)}_{i+1, i+1}(i+1\leftrightarrow i+2\leftrightarrow \cdots\leftrightarrow i+s)]. \end{eqnarray}$

$$$\label{eq-cycle-1} \begin{array}{lll} { } q_{m-1} = \frac{c^{(m)}_3-\pi_{m+1}q^2_{m+1, m}q_{m+1}}{\pi_{m-1}q^2_{m-1, m}}, \quad { } q_{m+2} = \frac{c^{(m+1)}_3-\pi_{m}q^2_{m, m+1}q_m}{\pi_{m+2}q^2_{m+2, m+1}}, {\nonumber}\\ [2mm] { } q_{m-1, m-2} = q_{m-1}-q_{m-1, m}, q_{m+2, m+3} = q_{m+2}-q_{m+2, m+1}. \end{array}$$$

$\begin{eqnarray} &&q_{m-2, m-1} = \frac{c^{(m)}_4-\pi_{m+1}q^2_{m+1, m}(q^2_{m+1}+q_{m+1, m+2}\ q_{m+2, m+1})-\pi_{m-1}q^2_{m-1, m}q^2_{m-1}}{\pi_{m-1}q^2_{m-1, m}q_{m-1, m-2}}, {}\\ &&q_{m+3, m+2} = \frac{c^{(m+1)}_4-\pi_{m}q^2_{m, m+1}(q^2_{m}+q_{m, m-1}q_{m-1, m})-\pi_{m+2}q^2_{m+2, m+1}q^2_{m+2}}{\pi_{m+2}q^2_{m+2, m+1}q_{m+2, m+3}}, \\ &&\pi_{m-2} = \frac{\pi_{m-1}q_{m-1, m-2}}{q_{m-2, m-1}}, \quad \pi_{m+3} = \frac{\pi_{m+2}q_{m+2, m+3}}{q_{m+3, m+2}}. {} \end{eqnarray}$

1) 打开环:根据证明中的第一部分可以计算出$q_m, q_{m+1}, \pi_m, \pi_{m+1}$, $q_{m, m-1} $$q_{m+1, m+2} , q_{m, m+1}$$ q_{m+1, m}$;

2) 根据(5.4)和(5.5)式, 向左计算出$\pi_{m-1}, q_{m-1, m}$, 向右计算出$\pi_{m+2}, q_{m+2, m+1}$;

3) 根据(5.10)式, 向左计算出$q_{m-1}, q_{m-1, m-2}$, 向右计算出$q_{m+2}, q_{m+2, m+3}$;

4) 根据(5.11)式, 向左计算出$q_{m-2, m-1}, \pi_{m-2}$, 向右计算出$q_{m+3, m+2}, \pi_{m+3}$;

5) 重复步骤3)和步骤4):对于$s = 2, 3, \cdots$, 根据(5.4)和(5.5)式分别向左和向右可以计算出剩余的转移速率.

### 图 1

$\begin{eqnarray} c^{(m+1)}_{2s+3}& = &\pi_{m}q^2_{m, m+1}\ _{\{m+1\}}\overline{q}^{(2s+1)}_{m, m} +\pi_{m+2}q^2_{m+2, m+1}\ _{\{m+1\}}\overline{q}^{(2s+1)}_{m+2, m+2}{}\\ & = &\pi_{m}q^2_{m, m+1}q_{m, m-1}q_{m-1, m}\big[A^{s}_1+\overline{A}^{s}_2+A^{s}_3+\overline{A}^{s}_4\big] +\pi_{m+2}q^2_{m+2, m+1}\big[\overline{B}^{s}_1+\overline{B}^{s}_2\big], {\qquad} \end{eqnarray}$

$\begin{eqnarray} c^{(m+1)}_{2s+4}& = &\pi_{m}q^2_{m, m+1}\ _{\{m+1\}}\overline{q}^{(2s+2)}_{m, m}+\pi_{m+2}q^2_{m+2, m+1}\ _{\{m+1\}}\overline{q}^{(2s+2)}_{m+2, m+2}{}\\ & = &\pi_{m}q^2_{m, m+1}q_{m, m-1}q_{m-1, m}\big[A^{s}_5+\overline{A}^{s}_6+A^{s}_7+ \overline{A}^{s}_8\big]+\pi_{m+2}q^2_{m+2, m+1}\big[\overline{B}^{s}_3+\overline{B}^{s}_4\big], {\qquad} \end{eqnarray}$

2) 对于$s = m-1$来说, 通过把(6.1)式中的$A^{m-1}_j$ ($j = 1, 2$)代入到(6.3)式中, 向右计算出$q_{2m+1}$; 于是有

3) 对于$s = m-1$来说, 通过把(6.2)式中的$A^{m-1}_j$ ($j = 5, 6$)代入到(6.4)式中, 向右计算出$q_{2m+2, 2m+1}$; 于是有

4) 重复步骤2)和步骤3):计算出状态$2m+2 $$N 之间的其余转移速率, 即 \pi_j, q_j, q_{j, j+1},$$ q_{j+1, j}$ ($j = 2m+2, \cdots, N$, 且$q_{N+1, N} = q_{1N}, q_{N, N+1} = q_{N1})$;

5) 对于$s = m-1$, 由(6.1)和(6.2)式计算出$q_{1' } $$q_{1' , 1} ; 于是有 6) 重复步骤5):在 s = m, m+1, \cdots 的情况下, 通过(6.1)与(6.2)式, 沿着直线的部分, 计算出从 1'$$ 0$的所有转移速率.

### 图 3

1) 由文献[32]中树形链的确认算法, 通过观测所有叶子可以确认出整个树形部分转移速率;

2) 类似于一条直线的单环链的确认算法, 通过观察环中的任何两相邻状态就可以计算子环中的所有转移速率.

### 图 4

关键的方法是将两个环分别打开.正如由一条直线的单环链的统计确认理论中所述, 两相邻状态距离节点度数为$3$的状态越远, 越容易确认出该子环的转移速率, 特别是当两个邻接状态是该子环的中心状态时.而且, 应该先确认状态数较少的环.不失一般性, 假设左环内的状态数少于右环内的状态数, 即$N<M'$.

### 图 6

$$$Q = \left(\begin{array}{cccccccccccc} -10 &0 &0 &0 &10 &0 &0 &0 &0 &0 &0 &0\\ 0 &-25 &0 &0 &25 &0 &0 &0 &0 &0 &0 &0\\ 0 &0 &-30 &0 &0 &30 &0 &0 &0 &0 &0 &0\\ 0 &0 &0 &-15 &0 &15 &0 &0 &0 &0 &0 &0\\ 5 &15 &0 &0 &-50 &0 &30 &0 &0 &0 &0 &0\\ 0 &0 &10 &15 &0 &-40 &15 &0 &0 &0 &0 &0\\ 0 &0 &0 &0 &5 &10 &-50 &10 &0 &0 &0 &50\\ 0 &0 &0 &0 &0 &0 &25 &-75 &50 &0 &0 &0\\ 0 &0 &0 &0 &0 &0 &0 &20 &-50 &30 &0 &0\\ 0 &0 &0 &0 &0 &0 &0 &0 &15 &-40 &25 &0\\ 0 &0 &0 &0 &0 &0 &0 &0 &0 &80 &-160 &80\\ 0 &0 &0 &0 &0 &0 &25 &0 &0 &0 &25 &-50 \end{array}\right).$$$

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