从Chapman–Kolmogorov 方程推导出n步转移矩阵

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In this article, I will derive the $n$-step transition matrix from Chapman–Kolmogorov equations.

Before the derivation, we define $P_{ij}^n=P(X_{n+k}=j|X_k=i)$ as the $n$-step transition probability from state $i$ to state $j$, ${\text{P}}^{(n)}$ as the $n$-step transition matrix and $\text{P}$ as the one-step transition matrix. Here, we note that ${\text{P}}^{(n)}$ does not denote the multiplication of $n$ matrix $\text{P}$, which will be derived in the following. To continue the derivation, the Chapman–Kolmogorov equations are given as
$$P_{ij}^{n+m}=\sum _{k=0}^{\infty }{P}_{ik}^n{P}_{kj}^m,(1)$$ for all $n,m\ge 0$ and all $i,j$.
My explanation of equation (1): equation (1) can be interpreted as the multiplication of two infinite matrix $i.e.$ $P_{ij}^{n+m}$ is the dot product of the $i_{th}$ row of the $n$-step transition matrix and the $j_{th}$ column of the $m$-step transition matrix. The preliminary sketch is as follows
Problem: Derive ${\text{P}}^{(n)}={\text{P}}^n$ based on equation (1).
Solution: Combining Fig. 1 with equation (1), we can obtain the relationship between ${\text{P}}^{(n+m)}$, ${\text{P}}^{(n)}$ and ${\text{P}}^{(m)}$, which is
$${\text{P}}^{(n+m)}={\text{P}}^{(n)}\cdot {\text{P}}^{(m)}.(2)$$
Based on equation (2), we can using induction to solve the problem. Induction involves two steps. The first step is to prove ${\text{P}}^{(n)}={\text{P}}^n$ when $n=1$. The second steo is to suppose when $n=k$, k$\ge 1$, ${\text{P}}^{(k)}={\text{P}}^k$. Then ${\text{P}}^{(k+1)}={\text{P}}^{k+1}$. For this problem, we have ${\text{P}}^{(1)}={\text{P}}^1$ by definition. For the second step, we suppose that when ($n=k-1,k-1\ge 1$) is given, ${\text{P}}^{(k-1)}={\text{P}}^{k-1}$. Then when ($n=(k-1)+1=k$), we have
$${\text{P}}^{((k-1)+1)}={\text{P}}^{(k-1)}\cdot {\text{P}}^{(1)}={\text{P}}^{k-1}\cdot \text{P}={\text{P}}^k,k\ge 2.(3)$$
The first equation in (3) is guaranteed by equation (1), the second equation is guaranteed by the fist step of induction and the supposition in the second step of induction and the third equation is guaranteed by Matrix multiplication. Now ${\text{P}}^{(n)}={\text{P}}^n$ if proved for $n\ge 1$ based on Chapman–Kolmogorov equations.

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