Assume that I have a particle doing random walk in $2D$, independent in each direction. It moves one step in both directions at each time step, i.e., if it starts at $(x,y)=(2.5,2.5)$, at $N=0$, it ends up at one of the $(1.5,1.5)$, $(3.5,1.5)$, $(1.5,3.5)$, $(3.5,3.5)$ at $N=1$, all with equal probability.
The random walk starts at $(x,y)=(0.5,0.5)$. If it hits $x=0$, it is absorbed with a probability of $P_A$. If it is not absorbed it is reflected to $x=0.5$. If it hits $y=0$, it is absorbed with a probability of 1.
I am trying to find the probability that it is absorbed at $N=3$, i.e., $P(N=3)$.
My approach is as follows:
For $N=1$, $$ P(N=1) = \frac{1}{2} + \frac{1}{4} \times P_A, $$ where the first part comes from hitting $y=0$ regardless of the movement in the $x$ direction. The second part comes from the probability of going up in $y$ and going left in $x$, i.e., the probability of not hitting in either direction.
For $N=2$, $$ P(N=2) = 0 + \color{red}{\frac{1}{4} \times (1-P_A)} \times \frac{1}{2} \times P_A, $$ where the first part is due to the particle not being able to hit at $y=0$ at $N=2$. The red part of the second part denotes the probability of the particle hitting at $x=0$ but not getting absorbed at $N=1$. Black part is the probability of hitting $x=0$ again at $N=2$.
For $N=3$, I worked on each axis separately. $P_Y(N=3)$ stands for the probability that it hits $y=0$ and $P_X(N=3)$ stands for the probability that it doesn't hit $y=0$ and it hits $x=0$. Thus, we have $$ P(N=3) = P_Y(N=3) + P_X(N=3). $$
I thought that \begin{align} \tag{1} P_Y(N=3) &= \frac{1}{8}(1-P_X(N=1)-P_X(N=2)),\\ &=\frac{1}{8}(1-\frac{1}{4}\times P_A-\frac{1}{8} \times (1-P_A)\times P_A) \end{align} which is incorrect according to the Monte Carlo simulations. My logic is that, $P_Y(N=3)=1/8$ for $P_A=0$, i.e., regardless of the movement on $x$, it is absorbed at $N=3$ only if it follows the $0.5 \rightarrow 1.5, \rightarrow 0.5\rightarrow 0$ path in the $y$ axis. Thus, I thought that normalising for the absorption in $x$ axis would give me quick results.
The $x$ axis part is just tedious, boring and unimportant for the question. I'll include this for the sake of completeness, but this part holds with the Monte Carlo simulations.
For $P_X(N=3)$, I counted all possibilities separately. The paths that end up at $x=0.5$ at the end of $N=2$ are
- $0.5 \xrightarrow{\text{no-abs}} 0.5\xrightarrow{\text{no-abs}}0.5$
- $0.5 \longrightarrow 1.5\longrightarrow 0.5$
Thus, the probability that the particle is at $x=0.5$, i.e., adjacent to the boundary at the end of $N=2$ is
$$ P(x=0.5, N=2) = \left(\frac{1}{4}(1-P_A)\times \frac{1}{2}(1-P_A)\right) + \frac{1}{4}\times \frac{1}{2}. $$
The $y$ coordinate is either at $y=2.5$ or $y=0.5$, with equal probability. Hence $P_X(N=3)$ becomes
$$ P_X(N=3) = \color{red}{\frac{1}{2}\times \frac{1}{4} P(x=0.5, N=2) \times P_A} + \frac{1}{2}\times \frac{1}{2} P(x=0.5, N=2) \times P_A. $$
The red part covers the situation where $(x,y)=(0.5,0.5)$ at the end of $N=2$ and the rest covers the situation where $(x,y)=(0.5,2.5)$ at the end of $N=2$. The former has a multiplier $\frac{1}{4}$ to exclude the probability that the particle hits $y=0$.
Although it looks simple, my results for the $P_Y(N=3)$ diverges from Monte Carlo simulations while $P_X(N=3)$ holds. According to the simulation results, my approach overestimates $P_Y(N=3)$. This is either due to double counting some scenario or due to (1) being totally wrong.
Why do I get wrong results by simple renormalisation? I might be missing something extremely simple but I tried too long to proceed myself.
Thanks in advance!