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Given a markov chain $(X_n)_{n\ge0}$ with transition matrix $P$ I just proved that $(Z_n)_{n\ge0}:=(X_{kn})_{n\ge0}$ is a Markov chain with transition matrix $P^k$, $k\ge1$.

Does this also work for $(Z_n)_{n\ge0}:=(X_{2n+1})_{n\ge0}$?

In case $(Z_n)_{n\ge0}:=(X_{2n+1})_{n\ge0}$ isn't a Markov chain: why not?

And in case it is, how can I express its transition matrix using $P$?

Bernard
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user657391
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    If ${X_n}$ is a homogeneous Markov chain the ${X_{2n}}$ and ${X_{2n-1}}$ are both Markov chains with transition matrix $P^{2}$. – Kavi Rama Murthy Mar 29 '19 at 08:17
  • Okay, thanks! Is the proof for ${X_{2n-1}}$ analogous to the proof for ${X_{2n}}$? (using the definition of two step transition probabilities as in example 3 here: http://www2.ece.rochester.edu/~gmateosb/ECE440/Midterm/practice_midterm_solutions.pdf ).

    And lets say the state space of $X_n$ is $I={1,2,3,4}$ and its initial disstribution is $\lambda$. Does it change for ${X_{2n-1}}$, ${X_{2n}}$?

     – user657391 Mar 29 '19 at 08:28
  • The two cases are similar. – Kavi Rama Murthy Mar 29 '19 at 08:35
  • Do $\lambda$ and/or $I$ change? – user657391 Mar 29 '19 at 08:40

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