Homogeneous Markov Chain $(X_{2n})_{n\in\mathbb{N_0}}$

Question

Let $(X_n)_{n\in\mathbb{N_0}}$ be a homogeneous markov chain with initial distribution $\mu$ and transition matrix $P$.

Is $(X_{2n})_{n\in\mathbb{N}_0}$ a homogeneous markov chain? What are the initial distribution and transition matrix of it?

How can I prove it?

I already know it is a homogeneous markov chain. Should I firstly prove if the markov property holds? I haven't done an example. — Tino, May 15 '20 at 15:49
If you figure that out, the fact that it's a homogeneous Markov chain is obvious. If the chain is in state $i$ what is the probability that it is in state $j$ after two steps? — saulspatz, May 15 '20 at 15:57

score 1 · Accepted Answer · answered May 15 '20 at 18:14

1

Let $p_{ij}$ be the probability that the chain will be in state $j$ at time $t+1$ given that it is in state $i$ at time $t$, so that the transition matrix is $$P=(p_{ij})_{n\times n}$$ Then then probability that the chain will be in state $j$ at time $t+2$ given that it is in state $i$ at time $t$ is $$\sum_{k=1}^np_{ik}p_{kj}\tag1$$ by the law of total probability, because the chain must first transition from state $i$ to some state $k$, and then from state $k$ to state $j$.

I'm sure you recognize the expression in $(1)$. It's just $P^2_{ij}$. That is, the transition matrix is $P^2$. Since this depends only on the state, and not the time, $X_{2n}$ is homogeneous.

answered May 15 '20 at 18:14

saulspatz

53,131

Thanks for your answer. What is the initial distribution of $(X_{2n})_{n\in\mathbb{N}_0}$? – Tino May 15 '20 at 19:29
The same as the initial distribution of $X_n$ How could it be any different? It's the distribution of $X_0$ in both cases. – saulspatz May 15 '20 at 19:43
That means, the initial distribution of $X_{2n}$ is also $\mu$? – Tino May 15 '20 at 20:06
Yes, of course. – saulspatz May 15 '20 at 20:07

Homogeneous Markov Chain $(X_{2n})_{n\in\mathbb{N_0}}$

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