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I currently try to model nucleation as an absorbing Markov chain. I have an idea how to do that but, however, I cannot convince myself that it is correct.

The state space consists of the number of atoms $n$ in a cluster which grows/shrinks upon attachment/detachment of monomeres/atoms until a critical size $n^*$ is reached (absorbing state in the chain). The transition probabilities for growing $P_{a,n} = a_n * t_n$ or shrinking $P_{b,n} = b_n *t_n$ (with $P_{a,n} + P_{b,n} = 1$) are a function of $n$. The time, needed for a transition is then $t_n = 1/(a_n + b_n)$.

It is known that the expected number of steps to reach an absorbing state can be calculated by

$$\vec{n}_{exp} = \mathbf{N}_{n^*-1, n^*-1} \cdot \vec{1}$$

where $N = (\mathbf{I} - \mathbf{Q})^{-1}$ is the fundamental matrix and $\vec{n}_{exp}$ contains the expected number of steps starting at the $i$-th position in the chain. But, I'm interested in the expected time until $n^*$ is reached. Therefore, I would exchange $\vec{1}$ with the vector containing the times needed for a transition (grow or shrink) $\vec{t} = [t_1, t_2, \dots t_{n^*-1}]$.

$$\vec{t}_{exp} = \mathbf{N}_{n^*-1, n^*-1} \cdot \vec{t}$$

However, as I'm not an expert in the field of Markov chains I cannot convince myself that this is correct. Could anybody help me to clarify? Many thanks in advance.

Physikuss.

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  • $\begingroup$ I think you'll need to be a bit more specific about the meaning of these times $t_i$ and their relationship to the Markov chain. Is the "time duration" of a step in the chain fixed (with the $t_i$ measuring something like the average time / number of steps spent before transitioning out of state $i$), or does a Markov step take variable time $t_i$ depending on the state $i$ it's in? (In the latter case your equation is correct.) $\endgroup$
    – joriki
    Commented Jul 27, 2015 at 10:04
  • $\begingroup$ So smaller clusters cannot combine into larger ones or break up into smaller ones in your model? $\endgroup$
    – Hans Engler
    Commented Jul 27, 2015 at 10:45
  • $\begingroup$ I added some explanation about the time and yes, only single atoms/monomeres are involved. So only $n \rightarrow n+1$ and $n \rightarrow n-1$ is allowed. $\endgroup$
    – Physikuss
    Commented Jul 27, 2015 at 11:46
  • $\begingroup$ Unfortunately I understand the roll of the $t_i$ less and not more after you edited. From what you've written, it seems that they're redundant and should simply be dropped, so I suspect there's an aspect you haven't told us about yet. In my experience, progress in clarifying posts is achieved more efficiently by concretely answering concrete questions that have been posed in the comments than by going off and making changes without specifically answering the questions. Your edit doesn't seem to address the relationship between the numbering of the Markov steps and the time that has elapsed. $\endgroup$
    – joriki
    Commented Jul 27, 2015 at 12:03
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    $\begingroup$ @ joriki: You are right. Each Markov step takes a different time. The expression for the $t_i$ does not matter at this point. So the question is, how can I get the expectation for the time elapsed time which would be the sum of all timesteps $t_i$ for each step that has been taken to reach $n^*$. $\endgroup$
    – Physikuss
    Commented Jul 27, 2015 at 13:39

1 Answer 1

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Your expression is correct. The $i$-th entry of $\vec t_{\text{exp}}$ is the sum over all states $j$ of the expected number of times state $j$ is visited, starting in state $i$ and ending with absorption, multiplied by the time one Markov chain step takes in state $j$.

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