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Let $X$ be an irreducible homogeneous Markov chain $(\mu,P)$ with state space $S$. The following statements should be proved or disproved:

a) If a $x\in S$ with $P(x,x)>0$ exists, then $X$ is aperiodic.

b) If $X$ is aperiodic, then there exists a $x\in S$ with $P(x,x) >0$.

c) If $S$ is finite and $X$ is periodic with period $2$, then $-1$ is an eigenvalue of $P$.

d) If $S$ is finite and $-1$ is an eigenvalue of $P$, then $X$ is periodic with period $2$.

I think the first three statements are true and the last one is false, but I have no idea how to prove or disprove them.

Tino
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  • What if $P(x,x)=1$? – Alex Jun 19 '20 at 18:01
  • what sort of background knowledge do you have here? You should be able to generate a counterexample for (b), btw. – user8675309 Jun 19 '20 at 18:01
  • @user8675309 I only know the definition of periodic and aperiodic Markov chains. I have no idea for counterexample. Can you help please? – Tino Jun 19 '20 at 18:25
  • What text are you using? This is basically impossible to answer in a meaningful way without some understanding of what is considered background knowledge. hints (i) for (b) try construct a transition matrix that is positive everywhere except zero on the diagonal. (ii) for (d) I'd suggest looking at the companion matrix associated with $x^n-1$. – user8675309 Jun 19 '20 at 20:27
  • @user8675309 I am using the lecture notes of my professor. Regarding (b): Is the counterexample of a $3\times 3$ transition matrix, where all entries are $\frac{1}{2}$ except the zeros on the diagonal, suitable? Unfortunately I don't understand your hint (ii). How can I recognise from a transition matrix if the Markov chain is periodic? – Tino Jun 19 '20 at 23:22
  • your example for (b) works. For hint (ii) https://en.wikipedia.org/wiki/Companion_matrix and consider what it looks like for different values of $n$ (in particular n=2, n=3 and n=4) again with the polynomial $x^n-1$. This shouldn't be hard but it will take a little work on your end... permutation matrices give the simplest examples of (periodic) markov chains. – user8675309 Jun 19 '20 at 23:26
  • @user8675309 I have looked up what a companion matrix is, but I am not sure how to write down the matrix for different values of $n$ with the polynomial $x^n-1$. Is it correct that every companion matrix with the polynomial mentioned before is a quadratic matrix, where all entries are zero except the entry in the upper right corner? I haven't understood the idea of the counterexample for (d) yet. – Tino Jun 20 '20 at 09:05
  • @user8675309 Is it possible for (d) to find a quadratic primitve matrix with eigenvalue of $-1$, which is aperiodic and has not the period of $2$? Does the following matrix work as a counterexample for (d)? $\begin{bmatrix} 0&1&0\ 1&0&0\ 0&0&1 \end{bmatrix}$ – Tino Jun 20 '20 at 15:06
  • Can someone show me the proof for (a) and especially (c)? Or is there any literature, where I can find the proofs? – Tino Jun 20 '20 at 15:09
  • your suggestion for (d) is reducible so it cannot be right. The case I've suggested you look into is irreducible. – user8675309 Jun 20 '20 at 17:26
  • @user8675309 I really cannot understand your suggestion for (d). Can you show me, what you meant? – Tino Jun 20 '20 at 18:03
  • @user8675309 Another suggestion I would have is the following matrix: $\begin{bmatrix} 0&1&0\ \frac{1}{2}&0& \frac{1}{2}\ 1&0&0\end{bmatrix}$. Would this be correct for (d)? If not, then I have really no ideas anymore. – Tino Jun 20 '20 at 18:21
  • using your $P=\begin{bmatrix} 0&1&0\ \frac{1}{2}&0& \frac{1}{2}\ 1&0&0\end{bmatrix}$ then $\text{trace}(P)=0 = \lambda_1+\lambda_2+\lambda_3 = 1 + \lambda_2+\lambda_3$ so if $\lambda_2 = -1$ then $\lambda_3 =0$ but $\det(P) \neq 0$ -- contradiction. All I'm suggesting is you look at https://math.stackexchange.com/questions/1380717/is-it-possible-to-find-a-companion-matrix-of-a-polynomial-which-is-also-hermitia and consider $a_0=-1$, $a_i = 0$ for $1\leq i \leq n-1$ and try it for n=3 and n=4. It appears that you may be missing some linear algebra pre-requisites here (re: eigenvalues). – user8675309 Jun 20 '20 at 20:02
  • @user8675309 I have tried it for $n=3$ and $n=4$, but I am not sure if they are correct. For $n=3$ I got the following matrix: $\begin{pmatrix} 0&0&1\ 1&0&0\ 0&1&0 \end{pmatrix}$. For $n=4$: $\begin{pmatrix} 0&0&0&1\ 1&0&0&0\ 0&1&0&0\ 0&0&1&0 \end{pmatrix}$. – Tino Jun 20 '20 at 20:42
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    @user8675309 At my point of view the matrix for $n=4$ has an eigenvalue of $-1$, it is periodic, but the period is not $2$, but $4$. – Tino Jun 20 '20 at 20:55
  • exactly so the n=4 case gives you a counter example for (d). Now (a) is true but I don't know an easy direct explanation -- it's really a greatest common divisor concept for countable state chains. (c) can be considered as irreducible period 2 finite chain means $\lim_{k\to \infty}P^{2k}$ exists so any eigenvalues of $(P^2)$ with modulus 1 must be equal to 1 -- i.e. $P$ can only have modulus one eigenvalues of ${-1,1}$ and if $P$ didn't have -1 as an eigenvalue then it would have period 1. – user8675309 Jun 21 '20 at 02:54

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