Subject CT4 April 2010 Question 4

[Pulit Chhajer]

For this Question why we cannot bridge the state space B and C?

 

[Andrew Martin]

Hello

This is due to the periodicity constraint. From the third piece of information, we know that part of the state space must look like:


B <- A -> C

However, as we're told it is periodic, we know there can not be an arrow from A to itself, so these probabilities must be 0.5 and there must be no more arrow out of State A.

Now, as we also know it is irreducible, we have to be able to get back to A from at least one of these states. It's arbitrary but let's choose state B:


B <- A -> C
↓......↑
→→→


Now consider state C, we need to have either an arrow to B or A in order for the MC to be irreducible. Let's consider the case when we have an arrow from C -> B and consider the possible return times to state B:

2: eg B -> A -> B
3: eg B -> A -> C -> B
4: eg B -> A -> B -> A -> B
5: eg B -> A -> C -> B -> A -> B
6,7,8,...

indeed every integer past this point. So, this would mean B is aperiodic, which is a problem. So we can't have C -> B here, ie PCB = 0. To ensure irreducibility, we must have PCA > 0. By symmetry this means we must also have PBC = 0.

Hope this helps!

Andy