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Question: Let $P$ be the stationary transition probability matrix of the Markov Chain $\{X_n:n\ge0\}$ which is irreducible and every state has period $2$. Further, suppose that Markov Chain $\{Y_n:n\ge 0\}$ on the same state space has transition probability matrix $P^2$. Both the chains are assumed to have the same initial distribution. Then

$1.$ $\text{Pr}\{X_0=Y_0, X_2=Y_1\}=1$

$2.$ all states of the chain $\{Y_n:n\ge 0\}$ are aperiodic

$3.$ the chain $\{Y_n:n\ge 0\}$ is irreducible

$4.$ If a state is recurrent for the chain $\{X_n:n\ge 0\}$, then it is also recurrent for the chain $\{Y_n:n\ge 0\}$.

My Attempt:

From the given information, we can say that the chain $\{X_n:n \ge 0\}$ is periodic and irreducible and hence, it has infinitely many stationary distributions. However, we don't know if the state space is finite.

First option can be discarded easily as initial distribution is same but we can't say anything when the processes are in states $1$ and $2$.

For second option, I took $P=\left [ \begin{matrix} 0 & 1 \\ 1 & 0 \\ \end{matrix} \right ] $ and found $P^2$ to be aperiodic. But I was unable to prove this.

Options third and fourth completely stumped me.

Kindly help me to prove these last three options as these are correct options(given in the answer key). Is there any theorem related to $P$ and $P^2$, i.e. if we know the properties of $P$, then we can tell about $P^2$ or in general $P^n$ ?

Thank you!

P.S. The correct options are $2$ and $4$.

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1 Answer 1

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Your answer key is wrong if it says that the third option is true. For your example with $\ P=\pmatrix{0&1\\1&0}\ $, for instance, $\ P^2=\pmatrix{1&0\\0&1}\ $. The two states of any Markov chain with the latter transition Matrix belong to two separate communication classes, and it is therefore reducible.

Hints for the remaining options:

  • The period of a state $\ i\ $ in a chain with transition matrix $\ P\ $ is $$ \gcd\big\{n\,\big|\,P^{(n)}_{ii}>0\big\}\ . $$ If this is $2$, then what is the value of $$ \gcd\big\{n\,\big|\,P^{(2n)}_{ii}>0\big\}\ ? $$
  • For all $\ i\ $ and $\ n\ $, $\ P\big(Y_n=i\,\big|\,Y_0=i\big)=P\big(X_{2n}=i\,\big|\,X_0=i\big)\ $. A state $\ i\ $ of $\ X_n\ $ is recurrent if $$ P\Big(\bigcup_{m=1}^\infty \big\{X_m=i\big\}\,\big|\,X_0=i\ \Big)=1\ . $$ But because $\ i\ $ has period $2$, then \begin{align} P\Big(\bigcup_{m=1}^\infty \big\{X_m=i\big\}\,\big|\,X_0=i\ \Big)&=P\Big(\bigcup_{n=1}^\infty \big\{X_{2n}=i\big\}\,\big|\,X_0=i\ \Big)\\ &=P\Big(\bigcup_{n=1}^\infty \big\{Y_n=i\big\}\,\big|\,Y_0=i\ \Big)\ . \end{align}
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  • $\begingroup$ $\gcd\big\{n\,\big|\,P^{(2n)}_{ii}>0\big\}=\gcd\{1,2,3, \cdots\}=1$. Therefore, each state in second chain will be aperiodic. $\endgroup$
    – Learning
    Aug 3, 2021 at 5:34
  • $\begingroup$ Yes, that's right. $\endgroup$ Aug 3, 2021 at 5:43
  • $\begingroup$ please have a look here...math.stackexchange.com/q/4216360/516174 $\endgroup$
    – Learning
    Aug 4, 2021 at 10:53

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