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Let $X_n,n \geq0$ be an irreducible, positive recurrent, aperiodic markov chain with state space $S$ and transition matrix $P$. $Y_n$ has the same $S$ and $P$ but independent with $X_n$. Then $\epsilon=(X_n,Y_n)$ becomes a new markov chian with state spacde $S \times S$. My question is $\epsilon$ recurrent or positive recurrent?

It's obvious $\epsilon$ is irreducible. I do some research on it and can not figure this out... Only know this kind is called "coupled markov chain"... Any hint? Thank you!

abc1m2x3c
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    Are you assuming a finite or countably infinite state space $S$? If so, then "yes" because you can find a probability vector that satisfies the global balance equations. That vector is just a multiplication of the individual steady state vectors. That is $\pi(i,j) = \pi(i)\pi(j)$. – Michael Mar 15 '16 at 19:17
  • Another way to argue is that you know, for all states $i$ and $j$ and all initial conditions $x_0,y_0$, we have $\lim_{n\rightarrow\infty}P[X_n=i|X_0=x_0] = \pi(i)>0$ and $\lim_{n\rightarrow\infty}P[Y_n=j|Y_0=y_0]=\pi(j)>0$, and by independence $P[X_n=i,Y_n=j|X_0=x_0,Y_0=y_0]=P[X_n=i|X_0=x_0]P[Y_n=j|Y_0=y_0]\rightarrow \pi(i)\pi(j)>0$. – Michael Mar 15 '16 at 19:29
  • Yes, I do assume it's a coutably infinite state. but..sorry, I am a beginner on stochastic process. I have trouble getting you. Your first comment shows we have a stationary state, How can we know it's reccurent from this? The second shows we have positve elements for transition matrix. I cannot get why it's recurrent either.... Thanks a lot! – abc1m2x3c Mar 15 '16 at 19:35
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    What tools do you have that you are comfortable using? Specifically, can you assume you have the steady state theorem available? Or, are you trying to prove the steady state theorem? You are using language of "irreducible" and "aperiodic" so I assumed you knew the steady state theorem. – Michael Mar 15 '16 at 22:09
  • Sure! I will learn the steady state theorem. Thanks a lot! – abc1m2x3c Mar 16 '16 at 02:03

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