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If a markov chain which has many states but only one state has a self-loop edge, then does it mean that the markov chain is aperiodic? Or every state in the markov chain has to have self-loop? For instance, in the following markov-chain, only state 2 has a self-loop edge, would this edge make the markov-chain aperiodic? :

Example of markov-chain

John Miller
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  • A Markov chain with only one state is rather trivial (and obviously aperiodic and irreducible). In general, a Markov chain with a finite number of states satisfying $P(X_1 = i | X_0 = i) > 0$ for every $i$ (which I think is what you mean by every state having a self-loop) is clearly aperiodic (but may be reducible). – Pedro M. Apr 13 '15 at 16:51
  • Sorry for the confusion, what I meant that a markov chain has several state but only state has a self-loop edge. So can we call this markov chain aperiodic? – John Miller Apr 14 '15 at 09:47
  • It will depend on the specific Markov chain (it may be aperiodic or not). Do you have a specific example in mind? – Pedro M. Apr 14 '15 at 11:26
  • @PedroM. I have put the markov-chain in the question. Is it aperiodic? Or how can I make sure that it is aperiodic or not? I trully appreciate your help! – John Miller Apr 14 '15 at 12:31
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    Every irreducible Markov chain with at least one loop is aperiodic. – Did Apr 14 '15 at 12:43
  • @Did Thanks for clear and brief answer – John Miller Apr 14 '15 at 13:17

3 Answers3

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Yes, the Markov chain you gave is aperiodic. To see this, you can try proving that whenever the underlying graph is strongly connected (or, in other words, the MC is irreducible) and contains at least one self-loop, then the Markov chain is aperiodic.

Pedro M.
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The definition of aperiodic is that every state is aperiodic, meaning that $\gcd\{t: \ P(X_t=s | X_0=s)>0\}=1$ for all states $s$. You can easily verify this to be the case for your Markov chain with one state and a self-loop.

Alex R.
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  • Sorry, I have edited my question. What I meant that a markov chain can have several states but only state has a self-loop edge. So can we call this markov chain aperiodic? – John Miller Apr 14 '15 at 09:48
  • Every irreducible Markov chain with at least one loop is aperiodic. – Did Apr 14 '15 at 12:44
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The Markov chain is irreducible, aperiodic. To see this, note $ 1 \leadsto 3 $ as $ p_{1,3} = 0.4 > 0 $, $ 3 \leadsto 2 $ as $ p_{3,2} = 0.1 > 0 $, $ 2 \leadsto 4 $ as $ p_{2,4} = 0.87 > 0 $, $ 4 \leadsto 5 $ as $ p_{4,5} = 0.75 > 0 $, $ 5 \leadsto 6 $ as $ p_{5,6} = 0.5 > 0 $, $ 6 \leadsto 7 $ as $ p_{6,7} = 0.3 > 0 $ and finally $ 7 \leadsto 1 $ as $ p_{7,1} = 1 > 0 $.

Since periodicity is a class property, enough to show any state has period $1$. Note $ p_{2,2} = 0.1 > 0 $, so the period of $ 2 $ must be $1$.

Rana
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