# Markov Chain Question/Notation Confusion

Show that if $$(X_n)_{n \geq 0}$$ is a discrete-time Markov chain with transition matrix $$P$$ and $$Y_n = X_{kn}$$, then $$(Y_n)_{n \geq 0}$$ is a Markov chain with transition matrix $$P^k$$.

I am a little confused as to what is meant by $$Y_n = X_{kn}$$. I read $$Y_n$$ as meaning the row vector of probabilities for the $$Y$$ chain at time $$n$$ but am unsure what is meant by $$X_{kn}$$? This makes it a matrix and I suppose the row index $$k$$ corresponds to the number of transitions i.e. number of applications of $$P$$ (although this is not clear to me from the question - just the fact that it is the only possibility that makes sense). Can somebody clear this up for me please?

Assuming this is the case $$X_{kn}$$ as a matrix that is constructed by appending row vectors representing the probabilities of the $$X$$ chain after $$k$$ transitions and this matrix will grow in row size according to the value of $$k$$. Therefore it makes sense that $$k$$ is a constant picked by the user before generating the $$Y$$ chain.

In that case for an initial probability distribution of $$\lambda_{i_0} = P(X_0 = i_0)$$ for $$i_0 \in I$$ where $$I$$ is the state space, we have

$$P(X_{kn}) = P(X_k = i_n) = P_{i_k i_{k-1}} P_{i_{k-1}i_{k-2}} \dots P_{i_2 i_1} P_{i_1 i_0} \lambda_{i_0} = P^k \lambda_{i_0}$$

Thus,

$$P(Y_n) = P(X_{kn}) = P^k \lambda_{i_0}$$ and so the transition matrix of $$Y_n$$ is $$P^k$$.

On an interpretation point, the state space of $$Y$$ would still be $$I$$, right? In other words $$P(Y_n)$$ really means $$P(Y_n = i_n)$$ but then this should equal $$P(X_k = i_n)$$ and this confuses me in terms of indices?

Does that look correct?

Is my notation ok?

Thanks for any help. I think I've over-confused myself on this one!

You have one Markov chain $$(X_n)_{n \geq 0}$$ with a transition matrix $$P$$. That means, this generates the process $$X_1, X_2, X_3, \dots$$.
Define $$Y_n = X_{kn}$$. The notation $$X_{kn}$$ means $$X$$ at times $$kn$$, for $$n \geq 0$$. So, this generates the process, $$X_{k}, X_{2k}, X_{3k}, \dots$$.
This has now been defined as $$Y_n$$, so that $$Y_1 = X_k, Y_2 = X_{2k}, Y_3 = X_{3k} \dots$$.
Thus, $$Y$$ just picks out every $$k$$th time point of $$X$$.