First-passage time distribution in Laplace space?

Question

I'm struggling to understand the reasoning between moving between two steps in a reaction scheme for a paper I am reading. For this (from the description), the probabilities over different paths are summed, and the first passage-time distribution of completing the irreversible step ($irr$) is written in Laplace space as:

$$\Psi_{irr}(s) = \sum^\infty_{n=0}(\frac{k_{post}}{s+k_{in}+k_{post}}\frac{K_{pre}}{s+K_{pre}+K_{irr}})^n\times\frac{k_{post}}{s+k_{in}+k_{post}}\frac{K_{irr}}{s+K_{pre}+K_{irr}}$$

$$=\frac{K_{irr}k_{post}}{(s+k_{in}+k_{post})(s+K_{pre}+K_{irr})-(K_{pre}k_{post})}$$

I'm trying to go through reviews for Laplace space but, for me the math is just not clicking. I don't get how the summation above, multiplied by the final irregular step, yields the second part of the expression? Any help or guidance on where to look would be greatly appreciated, as I've been struggling with this for a bit now. I get that it connects to markov models and transversing states according to probabilities defined by the rates $K_x$, maybe I'm missing something important related to Laplace space (which is a new area for me)?

Thank you.

Answer 1

I think that your issue has little to do with the Laplace transform. You just need to sum the geometric series. Recall that (for $|q|<1$): $$ \sum_{n=0}^\infty q^n = \frac1{1-q} $$ therefore: $$ \begin{align} \Psi_{irr}(s) &= \sum^\infty_{n=0}\left(\frac{k_{post}}{s+k_{in}+k_{post}}\frac{K_{pre}}{s+K_{pre}+K_{irr}}\right)^n\frac{k_{post}}{s+k_{in}+k_{post}}\frac{K_{irr}}{s+K_{pre}+K_{irr}} \\ &= \frac{k_{post}}{s+k_{in}+k_{post}}\frac{K_{irr}}{s+K_{pre}+K_{irr}}\sum^\infty_{n=0}\left(\frac{k_{post}}{s+k_{in}+k_{post}}\frac{K_{pre}}{s+K_{pre}+K_{irr}}\right)^n \\ &= \frac{k_{post}}{s+k_{in}+k_{post}}\frac{K_{irr}}{s+K_{pre}+K_{irr}}\frac1{1-\frac{k_{post}}{s+k_{in}+k_{post}}\frac{K_{pre}}{s+K_{pre}+K_{irr}}} \\ &= \frac{K_{irr}k_{post}}{(s+k_{in}+k_{post})(s+K_{pre}+K_{irr})-(K_{pre}k_{post})} \end{align} $$

Probabilisticaly, you interpret this as a geometric distribution, with the ratio $q$ being the failure and $1-q$ the success.

Hope this helps.