0

According to the usual definition: "A Markov chain is absorbing if it has at least one absorbing state, and if from every state it is possible to go to an absorbing state." (taken for example from Darmouth College - slide #3)

Now I am looking for an example of a Markov chain having an absorbing state, but without the property that, starting from any state, one can reach this absorbing state... (meaning that the chain can't be defined as absorbing).

I'm sorry to say that I have no imagination so as to construct such an example.

Does someone have some? I have to say I would really appreciate it...

Andrew
  • 483

1 Answers1

2

Suppose $p(0,1)=p(1,0)=1$ and $p(2,3)=p(3,3)=1$. This chain with 4 states has an absorbing state at state 3, but is not an absorbing chain because there is no way to get from state 1 to state 3.

Addendum: Here is a variant which is a bit less disconnected: Suppose $p(0,1)=p(1,0)=1$ and $p(2,3)=p(2,1)=1/2$ and $p(3,3)=1$.

Yuval Peres
  • 21,955
  • Thanks @Yuval. Really nice. I thought that would have been more complicated. Nevertheless, your example looks very special, as the MC is made of 2 distinct parts; would it have been possible if some link had existed between the 2 parts? This is where I still don't see how we could do it... – Andrew Jul 02 '22 at 16:42
  • Well, if from any node you could reach the absorbing node, the chain would be absorbing. I added another example. Do you know how to accept an answer? – Yuval Peres Jul 02 '22 at 16:47
  • Yes. That's the kind of example I was looking for. – Andrew Jul 02 '22 at 20:35