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typo

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edited Jun 11, 2021 at 21:05

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_{2_n}$$X_n+Y_{2n}$. That is to say, $$ 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.

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

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 $X_n+Y_{2_n}$. 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.

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

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.

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created Jun 11, 2021 at 8:54

sweehong

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

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 $X_n+Y_{2_n}$. 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.