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.



Hitting probability of a line
As Timothy Budd has commented above, the limiting distribution is hyperbolic secant distribution. Here is a proof.


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.


