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ID: past-exam-of-the-biological-physics-course-of-the-university-of-oxford/2015/5/c/solution

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Past exam of the biological physics course of the University of Oxford / 2015 / 5 / c / Solution by

Codex  0 2026-10-25

Interpret the irreversible limit as a large favorable drop for each forward step, so backward transitions during the observation can be neglected. Let the distinct forward rates be

k_{1}, k_{2} > 0

. The first waiting time has probability density

k_{1} e^{- k_{1} t_{1}}

; conditional on its completion, the second has independent probability density

k_{2} e^{- k_{2} (t - t_{1})}

. Integrating over the first completion time gives the convolution

p_{1, 2} (t) = \int_{0}^{t} k_{1} e^{- k_{1} t_{1}} k_{2} e^{- k_{2} (t - t_{1})} d t_{1} = \frac{k _{1} k _{2}}{k _{2} - k _{1}} (e^{- k_{1} t} - e^{- k_{2} t}), t \geq 0.

(1)

Hence

Pr (t < T_{1} + T_{2} < t + d t) = p_{1, 2} (t) d t + o (d t) .

(2)

The expression is symmetric under interchanging the rates, positive for either ordering, and integrates to one. Reverse rate constants do not enter this approximation; merely making all four constants different would not justify eliminating them without the downhill condition.

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