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When considering a chain of states from 0 to N in continuous time with constant up and down going transition rates, the jumping or waiting times are exponentially distributed. Now consider the following time dependent up rate and constant down rate,

$f_+(t) = \alpha (1-e^{- \beta t})$
$f_-(t) = \alpha.$

What is the best course of action to compute the jumping time distribution for such a case? Also, the memoryless property should not hold, due to the time dependence. It would not be correct to call it a Markov or Poisson process, wouldn't it?

Nicouh
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1 Answers1

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Note that if you know the state of the system at some time $t>0$, say $n(t)=n\in[0,N]$, the rates $f_-$ and $f_+$ are all the information you need to build realizations of the process at future times ($t'>t$). Therefore, the process is Markovian since you don't need any information about the state of the system at times $s$ before the time $t$ (i.e. $s<t$) to compute future states. In general, Markovianity is different than time-independent.

In mathematical terms, your process with time-dependent rates fulfills $$P\left(n(t')=m|n(t)=n;n(s)=k\right)=P\left(n(t')=m|n(t)=n\right) \, \forall s<t,$$ and therefore, is Markovian.

To compute the jumping time distribution I would suggest to understand the proof for time-indemendent rates, you can use similar arguments for your process.

To simulate the process, you can use Lewis thinning algorithm, see e.g. Lewis' paper.

Javi
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