False:
Start at n=0.
If n even let P(n->(n+1))=0.5 P(n->(n+2))=0.5
In n odd let P(n->n)=1
Then odd numbers are trivially recurrant, even numbers trivially transient.
Am i right in thinking that it is impossible to have a Markov chain with an infinite number of transient states and an infinite number of positive recurent states??????
where a positive-recurrent state, is a recurrent state where the expected no. of transition to return is finite....
If anyone could let me know whether I am correct and this is impossible I would be very grateful...
P(a->b) = probability that Markov chain goes from state a to state b.
The expected number of steps before an odd number returns to itself is one, which is less than infinity, and so the odd numbers are positive recurrent.
According to my understanding of the definitions, this is a counter example.
Matrix:
0 0.5 0 0.5 0
0 0.5 0.5 0 0
0 0.5 0.5 0 0
0 0 0 0.5 0.5
0 0 0 0.5 0.5
should do it... its basically a similar set up to the one I posted earlier.
Hope you can read that.
Not quite sure why you need 5 states though.
Is the 2x2 identity matrix a markov chain?
If it is, does that not have 2 stationary distributions?