the total times the trajectory is red or black in Fig. \[fig:trajAB\]). If $A$ and $B$ are metastable (i.e. if the trajectory commits to each of these sets and looses memory of its past before going back to the other set), $k_{A,B}$ and $k_{B,A}$ are the rates that enter the phenomenological mass-action law describing how the populations in $A$ and $B$ evolve in time. Notice that $\nu_R$ is related to $k_{A,B}$ and $k_{B,A}$ as $$\label{sec:nuRkABkBA}
  k_{A,B}=\frac{\nu_R }{\rho_A}, \qquad k_{B,A}=\frac{\nu_R}{ \rho_B},$$ where $ \rho_A$ is the fraction of time the trajectory is assigned to $A$ and $\rho_B$ the fraction of time it is assigned to $B$: $$\label{eq:rhoABrhoBA}
  \rho_{A}=\lim_{T\to \infty} \frac{T_A}{T}, \qquad 
  \rho_{B}=\lim_{T\to \infty} \frac{T_B}{T}.$$ By definition $\rho_A\le1$, $\rho_B\le1$ and $\rho_A+\rho_B=1$. The quantities $\nu_R$, $k_{A,B}$, $k_{B,A}$, $\rho_A$ and $\rho_B$ are the ones we will show how to compute exactly. To see how this can be done, it is useful to give first the TPT expressions for these quantities. These expressions involve the backward committor function, $q_-(\zz)$, which gives the probability that a trajectory observed at point $\zz$ is coming from $A$ last rather than from $B$ (i.e. that it is assigned to $A$ rather than $B$ using the jargon introduced above). By 