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.

  • 2
    $\begingroup$ 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.) $\endgroup$ Jun 11, 2021 at 7:14
  • 3
    $\begingroup$ Indeed, it is the Hyperbolic secant distribution you should aim for. $\endgroup$ Jun 11, 2021 at 7:48

2 Answers 2


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.


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

  • $\begingroup$ the laplace transform of which can be calculated, if anyone is interested $\endgroup$
    – mike
    Jun 11, 2021 at 9:31

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .