Let $(A_n)$ be a discrete Markov chain with transition matrix $M$ and $B_n = A_{(mn)}$ where $m\in \mathbb N$, I want to show that $(B_n)$ is a Markov chain with transition matrix $M^m$.
Please help me refine my argument. I have a picture of what is going on, but I don't know how to argue in symbols/more rigorously.
$B_n = A_{(mn)}$ means that you need to jump $m$ steps for each 1 jump in $A_n$ hence $(B_n)$ has the $m$-th transition matrix $M^m$. (How do I show that I can use $M$ as the "base transition matrix"?) Then to show that $(B_n) = Markov(\lambda, M^m)$, would I have to show that $\mathbb P(B_{n+1}=i_{n+1}|B_0=i_0,...,B_n=i_n)=(M^m)_{i_ni_{n+1}}$, or does that follow from the fact that $(A_n)$ is Markov?
Thanks.