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.
 A: 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.
A: 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 $\frac {Y_{\tau_n}} n \rightarrow \sqrt \tau_1 Z$.
