Interpretation of the first passage time

AI Thread Summary
The discussion centers on interpreting the probability distribution of first passage time (fpt) from state A to state B in a stochastic system with three states. A peak at time zero indicates a direct transition from A to B without intermediate stops, but it raises questions about its physical interpretation, particularly whether it suggests that the system typically takes zero time to transition. Participants emphasize the distinction between "most probable" and "expected" fpt, noting that the expected time accounts for the entire distribution, including longer tails. The conversation also touches on the appropriateness of using a weighted mean of the peaks to estimate the most probable fpt. Ultimately, understanding the significance of the zero peak in the context of the overall distribution is crucial for accurate interpretation.
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.
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...

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