# Probability of a random walk with Laplace-distributed steps leaving a corridor

This is a problem that I am looking to solve for a project in a radiative transfer class. The project statement is to "Determine the distribution of path lengths of photons leaving a cloud."

The general version of this will probably have to be done with a Monte Carlo code (i.e. scattering in all directions, preferential scattering directions, etc) but I think I have a special case that can be done mostly analytically.

The setup is that a photon is incident upon a cloud, and in the cloud it is scattered either up or down after taking a step $$s_n \sim \text{LAPLACE}(0,1)$$. The top of the cloud is at z=0 and the bottom of the cloud is at z=Z. The photon's position at step N is given by $$S_N = \sum s_n$$ with $$S_0=0$$. In addition, we have $$T_N = \sum |s_n|$$, the total travel distance. If $$S_N<0$$ the photon is reflected, and if $$S_N>Z$$ the photon is transmitted. Otherwise, it remains in the cloud and another step is taken. It seems to me that $$S_N$$ is a martingale and the first $$N$$ where the photon leaves the cloud is a valid stopping time. I am not sure how to use that information should it be relevant, my stochastic processes class has been a bit hard to follow recently.

The questions I want to answer are (these feed into each other)

1. What is the probability that the photon is reflected/transmitted at step $$N$$?

2. What is the distribution of the particle's travel distance $$T_N$$ given that it has left the cloud either through reflection or transmission? What is the distribution of the travel distance given just that a photon is reflected/transmitted, independent of N? (Note: I am not super concerned with the distribution of $$S_N$$ but if you see an easy way forward let me know. It isn't needed for the project but might make a good note).

For the first question: At step 1, the photon is reflected with probability 1/2 and transmitted with probability $$\frac{1}{2}e^-Z$$. $$T_1$$ is exponentially distributed in both cases. At step 2, we can apply the Law of Total Probability and Bayes Theorem to see that

$$P(\text{reflected at step 2}|s_1=x) = P(S_2<0|s_1=x) \\ =P(s_1+s_2<0|s_1=x)=P(s_2<-x)=\frac{1}{2}e^{-x}\\$$

and therefore that

$$P(\text{reflected at step 2}|0

A similar calculation shows that

$$P(\text{transmitted at step 2}|0 and Monte-Carlo simulations back up these calculations over a wide range of values of $$Z$$. Additionally, I have found that $$T_N|\text{reflected} \sim \Gamma(N,1)$$ as it is a sum of exponentials (also backed up by simulation). What is less clear to me is the distribution of $$T_N|\text{transmitted}$$. Simulations seem to suggest that it is something like an exponential but I have struggled to show anything.

Now for the probability of reflection/transmission after 3 steps: This is where things get annoying. Going the same route as above, we have that

$$P(S_3<0|s_1=x,s_2=y) = P(s_3<-(x+y)|0

and then that

$$P(S_3<0|S_1,S_2 \in [0,z)) = \int_{\mathbb{R}^2}P(S_3<0|s_1=x,s_2=y) dP(s_1=x,s_2=y|0

I determined the differential by noting that $$dP(s_1=x|0 and, since we need $$s_2 \in [-s_1,Z-s_1)$$ I wrote that the differential is given by

$$\frac{e^{-x}dx}{1-e^{-Z}} \frac{e^{-|y|}dy}{\int_{-x}^{Z-x}e^{-|y|}dy} = \frac{e^{-x}dx}{1-e^{-Z}} \frac{e^{-|y|}dy}{2-e^{-x}-e^{Z-x}}$$

Therefore, the integral ought to be equivalent to the integral (noting that $$0)

$$\frac{1}{2(1-e^{-Z})}\int_0^Z\int_{-x}^{Z-x}\frac{e^{-x}e^{-(x+y)}e^{-|y|}}{2-e^{-x}-e^{x-Z}} dy dx$$

Since $$x, evaluating the integral in $$y$$ gives

$$\int_{-x}^0 1 dy +\int_0^{z-x}e^{-2y}dy = x+\frac{1}{2}(1-e^{2(x-Z)})$$

and so the whole integral becomes $$\frac{1}{4(1-e^{-Z})}\int_0^Z \frac{(2x+1)e^{-2x}-e^{-2z}}{2-e^{-x}-e^{x-Z}} dx$$ which can be evaluated numerically (I used the MATLAB quadgk function). This agrees okay with what I have simulated, but not as well as before which makes me suspect that I am slightly off somewhere. Doing the same for the probability of transmission, we have that

$$P(S_3>Z|s_1=x,s_2=y) = P(s_3>Z-x-y) = \int_{Z-x-y}^{\infty}\frac{1}{2}e^{-|t|}dt$$

and as above

$$P(S_3>Z|S_1,S_2 \in [0,z)) = \int_{\mathbb{R}^2}P(S_3>Z|s_1=x,s_2=y) dP(s_1=x,s_2=y|0

This integral becomes

$$\int_0^Z\int_{-x}^{Z-x}\frac{e^{y-z}e^{-|y|}}{2(1-e^{-z})(2-e^{-x}-e^{x-Z})}dydx\\ = \frac{e^{-z}}{4(1-e^{-z})} \int_0^Z \frac{2(Z-x)-e^{-2z}}{2-e^{-x}-e^{x-z}}dx$$

This especially doesn't agree with my simulations. I attached a figure showing the calculated transmission and reflection probabilities for the third step (dots) along with the theoretical values derived here (lines). The transmission calculation is especially bad for relatively thin layers, and the reflection calculation gets worse for thick layers, although it is comparatively better.

My suspicion is that for any $$N$$ the probability of transmission and reflection at $$N$$ can be computed by an induction like argument due to the conditional probabilities, but I want to get over the $$N=3$$ hump. Any ideas? And any ideas on how to determine the distribution of $$T_N$$?

Thank you very much!

• Some "progress": The statement that $T_N|\text{refl} \sim \Gamma(N,1)$ doesn't seem to be quite correct. For an infinitely thick cloud it may be but the limiting distribution seems to depend on $Z$. Apr 3 at 21:34