Transition Matrix ( Markov Chain Monte Carlo)

mjt042
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1. -Find a regular transition matrix that is not time reversible, i.e., doesn't satisfy the
balance equations?

2.Pi,j=0≠Pj,ifor some i and j
My understanding from Markov Chain Monte Carlo is that for the transition matrix to be regular the matrix has to have all positives entries and each row will add up to one. I was thinking the trick to this problem for it not satisfy the balance equation would be to take the transpose of the transition matrix. I was hoping someone could give me a hint if I am on the right track of thinking and where to go from there.
Thanks
 
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mjt042 said:
1. -Find a regular transition matrix that is not time reversible, i.e., doesn't satisfy the
balance equations?




2.Pi,j=0≠Pj,ifor some i and j



My understanding from Markov Chain Monte Carlo is that for the transition matrix to be regular the matrix has to have all positives entries and each row will add up to one. I was thinking the trick to this problem for it not satisfy the balance equation would be to take the transpose of the transition matrix. I was hoping someone could give me a hint if I am on the right track of thinking and where to go from there.
Thanks

Your understanding is incorrect: P can be regular and have lots of zero entries. The regularity requirement is that some power ##P^n## has all positive entries. Also, your equation ##P_{ij} = 0 \neq P_{ji}## for some ##i,j## is likely not enough, nor is it needed.

You will probably get nowhere by taking the transpose of a transition matrix, since that will rarely give back a transition matrix again; if it does we call the transition matrix "doubly stochastic", and such transition matrices are rare.

Why not just try some more-or-less random (regular) transition matrices? They are unlikely to be reversible.
 
Thanks for your help.
 
I am still a little confused on how a regular transition matrix could not be reversible. Also what is meant by the statement Pij=0≠Pji for some i,j. Thanks
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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