This question is regarding Constructing the equilibrium ensemble of folding pathways from short off-equilibrium simulations which uses Markov State Model to find the equilibrium protein folding pathways from off-equilibrium simulations.

I find the notation a bit hard to follow. How does Frank Noe and others derived equation 1?

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In the supplementary information they don't really explain and aren’t careful with the math notations:

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I do see how a rate equals $k_{AB} = P(A^C|A)/P(B^C \cap A^C)$, but I don’t see how $\sum_{i=1}^m \pi_i q_i^{-}$ gives $P(B^C \cap A^C)$ which is the probability to be outside of $A$ and not in $B$.


Let us start by defining all the terms in Eq. 1 carefully, as the OP does not do so. I have always found it easier to interpret such models in the context of particles flowing from state to state, and that is the picture that I will try to present here.

  • $T_{ij}$ measures the probability of going from state $i$ to state $j$ within a time $\tau$.
  • $\pi_i$ is the equilibrium probability of being in state $i$.
  • $q_i^+$ is the probability that, starting at state $i$, a particle will reach the set of states $B$ before the set of states $A$.
  • $q_i^-$ is the probability that, ending at state $i$, a particle has more recently encountered $A$ than $B$. (At equilibrium, all trajectories are reversible, so we can alternately think of $q_i^- = 1 - q_i^+$ as the probability that, starting at state $i$, a particle will reach $A$ before $B$, but that is irrelevant for this problem.)
  • $k_{AB}$ is the normalized rate of particle flow from $A$ to $B$. In other words, given particles that start out in $A$, $k_{AB}$ is the fraction of particles per unit time that end up in $B$.

Now we simply need to interpret the numerator and denominator of Eq. 1. To think about these quantities in the particle picture, let us multiply top and bottom by the number of particles $N$ evolving in the system, defining $N_i = N\pi_i$, so that $$k_{AB} = \frac{\sum_{i\in A}\sum_{j\notin A}N_iT_{ij}q_j^+}{\tau\sum_{i\in\text{all states}}N_iq_i^-}.$$ In the numerator $$\sum_{i\in A}\sum_{j\notin A}N_iT_{ij}q_j^+,$$ the quantity $N_iT_{ij}$ is the number of particles that leave $A$ (going from $i$ in $A$ to $j$ not in $A$) in a time $\tau$. A fraction $q_j^+$ of these particles will end up in $B$. Now, in steady state, this must also be the number of particles that are arriving at $B$ from $A$, or otherwise we would have an accumulation of particles in the system.

In the denominator $$\tau\sum_{i\in\text{all states}}N_iq_i^-,$$ the quantity $\sum_{i\in\text{all states}}N_iq_i^-$ is the number of particles in the system that started out in $A$, which serves to normalize the number of particles in the numerator, and $\tau$ is the time interval in which this all occurs.

Putting all this together, then, $$k_{AB} = \underbrace{\frac{1}{\sum_{i\in\text{all states}}N_iq_i^-}}_\text{normalized}\underbrace{\frac{1}{\tau}}_\text{rate of}\underbrace{\sum_{i\in A}\sum_{j\notin A}N_iT_{ij}}_\text{particle flow from A}\underbrace{q_j^+}_\text{to B.},$$ as desired.

  • $\begingroup$ Can you explain this phrase: "$q_i^-$ is the probability that, ending at state $i$, a particle has more recently encountered $A$ than $B$." $\endgroup$
    – 0x90
    Feb 6 '19 at 1:24
  • $\begingroup$ @0x90: In other words, $q_i^-$ is the probability that a trajectory currently at state $i$ started out in a state associated with $A$. $\endgroup$ Feb 6 '19 at 2:26
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    $\begingroup$ So here is where this formalism isn't sound $q^{−i}=1−q{+i}$ defined where the particle will end up and $q^{+i}$ where it will start. These events are not complementary. $\endgroup$
    – 0x90
    Feb 6 '19 at 2:34
  • $\begingroup$ @0x90: I have addressed this point in my answer. In equilibrium, where the trajectories are reversible, the events are indeed complementary. $\endgroup$ Feb 6 '19 at 6:28

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