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In Reichl, 2016; pg405 the author gives a derivation of the master equation - which I will outline below in my own words:

  • We start with:$$P(n,t+\Delta t)=\sum_m^M P(m,t)P(n,t+\Delta t\mid m,t)\tag{$\star$}$$
  • We then Taylor expand $P(n,t+\Delta t\mid m,t)$ as a function of $\Delta t$ to get: $$ P(n,t+\Delta t\mid m,t)=\delta_{m,n} \left[ 1-\Delta t \sum^M_{l=1} w_{m,l}(t)\right]+w_{m,n}(t)\Delta t+\ldots$$ where we have enforced conservation of probability
  • Sub this back into our expression for $P(n,t+\Delta t)$ , rearrange and take the limit as $\Delta t\rightarrow 0$ gives: $$\frac{\partial P(n,t)}{dt}=\sum_m^M\left[ P(m,t) W_{m,n}(t)-P(n,t) w_{n,m}(t)\right] $$

However many sources (including Reichl) hint at a restriction to Markovian process. Given that ($\star$) is simply true and everything else seems is standard I cannot see where the Markovian property comes into this - if it does? Please can someone explain this to me.

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    $\begingroup$ I don't think (*) is simply true. I think it being true is what makes a process Markov. $\endgroup$
    – The Photon
    Apr 24, 2018 at 20:00
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    $\begingroup$ Your first equation (*) contains the assumption that the system has "no memory" beyond the last step - i.e. that the time evolution of the probability is only dependent on the current value of the probability, and not on the history of the system. $\endgroup$ Apr 24, 2018 at 23:20
  • $\begingroup$ @ThePhoton I disagree. All ($\star$) is saying is that the probability you are in $n$ at $t+\Delta t$ is equal to the probability that you where at $1$ at $t$ and are now at $n$ plus the probability you where at $2$ at $t$ and are now at $n$. The fact that a process is Markovian or not does not change the definition of $P(n,t+\Delta t\mid m,t)$. $\endgroup$ Apr 25, 2018 at 3:43
  • $\begingroup$ @probably_someone See my comment above. $\endgroup$ Apr 25, 2018 at 3:43
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    $\begingroup$ If the process were not Markovian, there would be terms proportional to $P(m,t-\Delta t)$, $P(m,t-2\Delta t)$, $P(m,t-3\Delta t)$, etc. The fact that this initial expression does not depend on any additional history of the system is the definition of a Markovian process. $\endgroup$ Apr 25, 2018 at 3:46

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