What Is the Expected First-Passage Time in a Markov Chain?

PhyzX
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Homework Statement


Markov process has probabilities p_{j,j+1} = 1-p_{j,0} = (\frac{j+1}{j+2})^k for j=0,1,2,...
If T_j = min[n>1 : X_n=j]
What is E[T(j)|X_0=j] for j=0,1,2,...?

Homework Equations

The Attempt at a Solution


I figured that T(j)|X_0=j is j+1 but don't know how to work out the expectation of this
 
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PhyzX said:

Homework Statement


Markov process has probabilities p_{j,j+1} = 1-p_{j,0} = (\frac{j+1}{j+2})^k for j=0,1,2,...
If T_j = min[n>1 : X_n=j]
What is E[T(j)|X_0=j] for j=0,1,2,...?

Homework Equations

The Attempt at a Solution


I figured that T(j)|X_0=j is j+1 but don't know how to work out the expectation of this?

Your statement makes no sense: ##j+1## is an integer number, while ##T(j)|X_0 = j## is a random variable. What do you really mean, here?

Also: get rid of that pesky "?", because the way you have written it, you are asking us whether or not you know how to do something.
 
Ray Vickson said:
Your statement makes no sense: ##j+1## is an integer number, while ##T(j)|X_0 = j## is a random variable. What do you really mean, here?

Also: get rid of that pesky "?", because the way you have written it, you are asking us whether or not you know how to do something.
I think I don't quite understand the question. My interpretation of ##T(j)|X_0 = j## is that it is equal to the minimum number of steps needed to return to state ##j##. Looking at the transition probabilities I figured that the chain should immediately return to state 0 and then climb back up to state ##j## hence taking ##j+1## steps. My thinking might be entirely wrong though.
 
PhyzX said:
I think I don't quite understand the question. My interpretation of ##T(j)|X_0 = j## is that it is equal to the minimum number of steps needed to return to state ##j##. Looking at the transition probabilities I figured that the chain should immediately return to state 0 and then climb back up to state ##j## hence taking ##j+1## steps. My thinking might be entirely wrong though.

It is entirely wrong.

If ##X_0 = j##, where can the chain go next? It can go to state 0 or state j+1. If it had gone to 0 it can flip numerous times back and forth between state 0 and states 1,2, ..., j-1, then go to j in the next step after that. If it had gone to state j+1, it can next go to j+2 or to 0; once at 0 the behavior just described takes over. If at j+2 it next goes either to j+3 or down to 0, etc. etc.

Getting the exact probability distribution of the first return-time to j is not easy, but there are (in your textbook, or notes, or on-line) some equations which, when solved, spit out the expected return time as part of their solution. I have not looked at the details in this case, so I don't know how easy it would be to solve the equations.

However, there is a more serious issue: can we be sure that the chain does, indeed, return to state j with probability 1? If so, can we be sure the expected return time is finite? The point is that ##p_{j,j+1} \to 1 ## as ##j \to \infty##, so the farther out we are, the less chance there is of going back. There are test criteria available that allow you to deal with some such cases, but how easy they are to apply in this case is something I have not investigated.
 
Thank you, I understand what the question is asking now. Would you happen to know what these equations might be called so that I can search them up.
 
PhyzX said:
Thank you, I understand what the question is asking now. Would you happen to know what these equations might be called so that I can search them up.

Are you telling me that your course notes and/or textbook does not have such material? I am surprised, since it is so very fundamental to the subject. However, you can Google "expected first-passage time" or "expected first-passage time + markov chain".
 
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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