Final answer:
In an irreducible Markov chain, if P(j, j) > 0 for some state j, it signifies a self-loop at state j. Due to the self-loop, state j is aperiodic, and since the chain is irreducible, all states also become aperiodic. The correct implication of P(j, j) > 0 is an aperiodic state. Option A is the correct answer.
Step-by-step explanation:
To demonstrate if P(j, j) > 0 for some state j implies that all states in an irreducible Markov chain are aperiodic, we first need to understand what it means for states to be aperiodic. A state i is considered aperiodic if the greatest common divisor (GCD) of all the lengths of paths that start and end at state i is 1. When we have P(j, j) > 0, it indicates that there is a self-loop at state j, meaning the chain can remain in state j with a positive probability. This loop has a length of 1, and when combined with other possible path lengths that enter and leave state j, their GCD must be 1 since any number's GCD with 1 is 1. Therefore, state j is aperiodic.
Considering the chain is irreducible, every state can be reached from every other state, and therefore, the property of being aperiodic must also hold for every other state in the chain. Hence, if P(j, j) > 0 for some j, it implies that all states must be aperiodic, as they are all in communication with state j and can inherit its aperiodicity property.
Now, knowing that P(j, j) > 0 does not necessitate that the state is absorbing, transient, or necessarily in communication with others (apart from being in an irreducible chain), but it does denote that the state j is aperiodic. Therefore, the correct option is d. Aperiodic state.