# Hitting probability of a line

Consider a simple (nearest neighbor) random walk on a lattice $$\Bbb Z^2$$ which starts at the origin, is constrained to $$x\ge 0$$ halfplane, and stops when it hits the line $$x=n$$. Denote by $$p(n,k)$$ the probability the walks stops at the point $$(n,k)$$, where $$k \in \Bbb Z$$.

Question. Does $$p(n,tn)$$ converge to a Gaussian distribution? I imagine somebody already proved that, so I am especially interested in the reference to this result.

Note: If the constraint $$x\ge 0$$ is removed, the hitting probabilities have a Cauchy distribution, see e.g. page 155 in F. Spitzer, Principles of random walk (Second ed.), Springer, New York, 1976.

• The limit should have something to do with the Poisson kernel for the strip, which is definitely not Gaussian — it decays as $\exp(-|y|)$ rather than $\exp(-c |y|^2)$. See DOI:10.2307/2034126, for example. (I believe the invariance principle easily leads to the identification of the limit with the Poisson kernel, but I did not work out the details.) Jun 11, 2021 at 7:14
• Indeed, it is the Hyperbolic secant distribution you should aim for. Jun 11, 2021 at 7:48

By the reflection principle, the random walk in question can be substituted with one that does not have $$x\geq 0$$ constraint, but is terminated upon hitting the line $$x=n$$ or $$x=-n$$. Let $$X_n$$ be the vertical coordinate of the walker when it stops. Furthermore, let $$Y_n$$ be the vertical coordinate of the random walker that is stopped when hitting the line $$x=n$$. It has been pointed out in the original post that $$\frac{Y_n}{n}$$ converges to Cauchy distribution.
We now decompose the random variable $$Y_{n}$$ in the following way. Let the walker perform random walk until it hits $$x=n$$ or $$x=-n$$. If it hits the line $$x=n$$, then its height will follow the law of $$X_n$$. If it hits the line $$x=-n$$ instead, then let the walker perform random walk until it hits the line $$x=n$$ and its height will follow the law of $$X_n+Y_{2n}$$. That is to say, $$Y_n = \begin{cases}X_n & \text{ if x=n is visited first},\\ X_n+ Y_{2n} & \text{ if x=-n is visited first}, \end{cases}$$ and notice that both scenarios occur with probability $$\frac{1}{2}$$. By taking their characteristic functions, we get $$F_{Y_n}(t) \ = \ \frac{1}{2} F_{X_n}(t) \ + \ \frac{1}{2} F_{X_n}(t) F_{Y_{2n}}(t)$$ Since $$Y_n/n$$ converges in distribution to Cauchy distribution, the limit of the equation above is equal to $$e^{-|t|} \ = \ \frac{1}{2}\lim_{n \to \infty} F_{\frac{X_n}{n}}(t) \ + \ \frac{1}{2}\lim_{n \to \infty} F_{\frac{X_n}{n}}(t) \, e^{-2|t|}.$$ Solving this equation, we get $$\lim_{n \to \infty} F_{\frac{X_n}{n}}(t) \ = \ \frac{2}{e^{|t|}+e^{-|t|}},$$ which is the characteristic function for hyperbolic secant distribution.
Don't think so. Let $$\tau_n$$ be the stopping time. Replace by a brownian motion, which probably doesn't make any difference. Since the Y coordinate is independent of the stopping time, the distribution you are looking for can be represented as $$\sqrt \tau_n Z$$ where Z is a standard normal, and $$\tau_n$$ is the time when the X coord of an ordinary brownian motion hits $$\pm n$$. By scaling this is the same a $$n\sqrt \tau_1 Z$$, so $$\frac {Y_{\tau_n}} n \rightarrow \sqrt \tau_1 Z$$.