Interpretation of the first passage time

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SUMMARY

The discussion focuses on the interpretation of the first passage time (FPT) in a stochastic system with three states, specifically the transition from state A to state B. The presence of a non-zero peak at time t=0 in the probability distribution indicates that there is a direct path from A to B, which follows a single exponential function. However, this peak does not imply that the system typically takes 0 time to transition; rather, it highlights the distinction between "most probable" and "expected" first passage times. The expected time accounts for all possible transitions, including those with longer durations, thus providing a more comprehensive understanding of the system's kinetics.

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  • Understanding of stochastic processes and kinetic models
  • Familiarity with probability distributions, particularly exponential functions
  • Knowledge of first passage time (FPT) concepts in statistical mechanics
  • Ability to interpret graphical data representations
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phyalan
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Hi all,
Suppose I have a kinetic model for a stochastic system of three states, as shown in the attachment. I solve for the probability distribution of the first passage time from A to B and I get the distribution shown on the right hand side.

I can understand that if there is a peak in the distribution, we can say there is some most probable first time(fpt) for the system to transit from A to B because in the path A->C->B, one can go back and fourth between A and C before reaching B. But how about the non-zero peak at t=0 in the distribution? I know it comes from the path A->B because this path has no intermediate stop, the distribution follows a single exponential function but I am confused about how to interpret it physically. Does it means that the system 'typically' takes 0 time to transit to B in this path?

And if I want to have a estimation of the most probable fpt, is taking the weighted mean of the two peaks with respect to their probabilities a reasonable approach?
 

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phyalan said:
Does it means that the system 'typically' takes 0 time to transit to B in this path?
In the graph you have shown, even though the highest probability is at time 0 it is still a small fraction if the whole. There are still all the rest to consider. You should distinguish between "most probable" and "expected". The expected time is not the same as the peak time. The expected time is the average of all the results.
 
phyalan said:
Does it means that the system 'typically' takes 0 time to transit to B in this path?
In the graph you have shown, even though the highest probability is at time 0 it is still a small fraction if the whole. There are still all the rest to consider. You should distinguish between "most probable" and "expected". The expected time is not the same as the time with highest probability. The expected time is the average of all the results, including those all the way out on the right hand tail of the distribution.
 
FactChecker said:
In the graph you have shown, even though the highest probability is at time 0 it is still a small fraction if the whole. There are still all the rest to consider. You should distinguish between "most probable" and "expected". The expected time is not the same as the time with highest probability. The expected time is the average of all the results, including those all the way out on the right hand tail of the distribution.

Yes, I know that. But what I mean is in this case, how can one interpret the most probable first passage time where you have a peak at time 0? The point is sometimes, in some systems, the distribution has to very long tail that makes the mean fpt carries less significant meaning in describing the kinetics. So I what to know how to make sense our of this case.
 
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