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

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


$P(Y_n) = P(X_{kn}) = P^k \lambda_{i_0}$ 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!


1 Answer 1


I think you have indeed over-confused yourself!

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 $X$.


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