I have come across this question and am fairly unsure how to demonstrate this proof:
Prove or provide an explicit counterexample. If a Markov chain on a state space $S$ has a state $i$ with period $d_i$, then $p_{ii}^{(d_i)} > 0$.
Could anyone help? Thanks in advance!
The period $d_i$ of a state $i$ is the gcd of the number of steps it is possible to return to the state in, ie to leave the state $i$ and return, the process must have completed some multiple of $d_i$ steps. If this greatest common divisor is also the minimum number of steps, then certainly $p_{ii}^{\left( d_i \right)}>0$, since this is possible so the chain returns with positive probability.
However it may not always be the case that the gcd is the minimmm number of steps. For a counter example then, suppose we have a Markov Chain on 5 states drawn as a 2-cycle (2 nodes including the initial state) and a 3-cycle (3 nodes including the initial state) with the one common state being the initial state. We will choose the transition probabilities to be deterministic at all states except the initial state, where there is some non-zero chance of going to the 2-cycle and a non-zero chance to go to the 3-cycle.
If it goes to the 2-cycle, then with probability 1 it returns to the initial state for a total of 2 steps. However if it goes to the 3-cycle, then it must first finish the cycle to return to the initial state after a total of 3 steps. These are the only two possible lengths as once the first step is taken, the Markov Chain is deterministic. The gcd of 2 and 3 is 1, but the minimum length is $2>1$ so $$p_{00}^{\left( d_0\right)}= p_{00}^{\left( gcd\left(2,3\right)\right)}= p_{00}^{\left(1\right)}=0,$$ since it is not possible to return to the initial state after just one step.
I hope this helps! Stay safe