The Hitting Distribution of a Line Segment for Two-Dimensional Random Walks

Abstract

Asymptotic estimates of the hitting distribution of a long segment on the real axis for two-dimensional random walks on $\mathbf{Z}^2$ of zero mean and finite variances are obtained: Some are general and exhibit its apparent similarity to the corresponding Brownian density, while others are so detailed as to involve certain characteristics of the random walk.

Appendices

Appendices

(1)  Let D be the complement of the line segment with edges at $\pm 1$:

$$\begin{aligned} D=\mathbf{C}\setminus \{s: -1\le s\le 1\} \end{aligned}$$

and denote the Poisson kernel (density of harmonic measure) for D by $h_D(z, s\pm i0)$. Putting $h^{[-1,1]}_z(s)= h_D(z, s + i0)+h_D(z, s - i0)$, we have

$$\begin{aligned} h^{I(n)}_z(s) = h^{[-1,1]}_{z/n_*} (s/n_*)/n_* . \end{aligned}$$

We compute $h_D(z, s\pm i0)$ by using the conformal invariance of harmonic measures. The function $z =\frac{1}{2} (w+w^{-1})$ univalently maps the exterior of the unit circle onto D. Denote by f(z) its inverse map, which may be represented by

$$\begin{aligned} f(z)=z+\sqrt{z^2-1},\quad z\in D \end{aligned}$$

with the standard choice of a branch of the square root (so that $f(\pm s)=\pm s\pm \sqrt{s^2-1}$ for $s>1$ and $f(s\pm i0)=s \pm i\sqrt{1-s^2}$ for $-1<s<1$). As $w=f(z)$ moves on a circle centered at the origin counter-clockwise starting at a point $R>1$, z describes the ellipse $[2x/(R+R^{-1})]^2+[2y/(R-R^{-1})]^2=1$ (which surrounds the segment $-1\le s\le 1$ and shrinks to it as $R\downarrow 1$) rotating also counter-clockwise and starting at the point $f(R) =\frac{1}{2}(R+R^{-1}) \in (1, R)$. (Cf. [1]:p.94 or [10]:p.269). The Poisson kernel for the exterior of the unit circle is given by

$$\begin{aligned} K(Re^{i\theta }, e^{i\theta '})=\frac{R^2-1}{2\pi (R^2-2 R \cos (\theta -\theta ')+1)},\quad R>1. \end{aligned}$$

Put

$$\begin{aligned} \theta (z)=\arg f(z). \end{aligned}$$

Then, for $-1\le s\le 1$, $\theta (s \pm i0)=\pm \arccos s\in (-\pi ,\pi )$, so that $|d\theta (s\pm i0)|=\mathrm{d}s/\sqrt{1-s^2}$; thus, the conformal invariance shows that

$$\begin{aligned} h_D(z, s\pm i0)= & {} \frac{1}{2\pi }\cdot \frac{|f(z)|^2-1}{|f(z)|^2-2|f(z)|\cos \,[\, \theta (z)-\theta (s\pm i0) ]+1}\cdot \frac{1}{\sqrt{1-s^2}} \nonumber \\= & {} \frac{1}{2\pi }\cdot \frac{|f(z)|^2 -1}{|f(z)|^2-2[s \,\mathfrak {R}f(z) \pm \sqrt{1-s^2}\, \mathfrak {I}f(z)]+1}\cdot \frac{1}{\sqrt{1-s^2}}, \nonumber \\ \end{aligned}$$

(38)

which, for $z=x \in \mathbf{R}\setminus [-1,1]$, reduces to

$$\begin{aligned} h_x^{[-1,1]}(s) =2h_D(x, s+i0) = \frac{\sqrt{x^2-1}}{\pi \, |x-s|}\cdot \frac{1}{\sqrt{1-s^2}}. \end{aligned}$$

Let $Q=(\sigma _{ij})$ be a $2\times 2$ matrix that is symmetric and positive definite and $\tilde{h}_D(z, s \pm i0)$ the corresponding hitting density for the process $Q^{1/2} B_t$, a two-dimensional Brownian motion of mean zero and the covariance matrix tQ. Then for $z\in D$,

$$\begin{aligned} \tilde{h}_D(z, s\pm i0) = h_{D}(\tilde{z}, s\pm i0), \quad \tilde{z} =(x-\omega y) +i \lambda y, \end{aligned}$$

(39)

where $\omega =\sigma _{12}/\sigma _{22}$ and $\lambda =\sigma ^2/\sigma _{22} =\sqrt{\sigma _{11}/\sigma _{22}-\omega ^2}$. If z is real, the identity above follows immediately from the rotation invariance of the standard Brownian motion. In view of the strong Markov property of S, its full validity is deduced from the identity thus restricted in conjunction of the corresponding identity for the Poisson kernel of the upper half-plane (see 5 below).

(2)   In Sect. 5 (at the end of it), we have used the following lemma.

Lemma 11

For $x, s \in (-n_*, n_*)$ with $s\ne x$,

$$\begin{aligned} \frac{1}{(s-x)^2}+ \int _{|\xi |\ge n_*} \frac{1}{(\xi -x)^2} h^{I(n)}_\xi (s)\mathrm{d}\xi = \frac{n_*^2 -xs}{ (s-x)^2\sqrt{(n_*^2-x^2)(n_*^2-s^2)}}. \end{aligned}$$

(40)

Proof

By the scaling property, we may suppose the interval I(n) to be $[-1,1]$. Let $h_D$ be as in (1) and $h_z(s)$ be the Poisson kernel on the upper half-plane: $h_z(s) = y/\pi (y^2+(s-x)^2)$. Then,

$$\begin{aligned} h_z^{[-1,1]}(s) =h_D(z,s + i0)+h_D(z,s - i0) = h_z(s) + \int _{|\xi |>1} h_z(\xi )h_\xi ^{[-1,1]}(s) \mathrm{d}\xi , \end{aligned}$$

which shows $\lim _{y\downarrow 0}\pi y^{-1} h_{x+iy}^{[-1,1]}(s)$ equals L.H.S. of (40). The lemma therefore follows if we verify that if $|x|<1$ and $|s|<1$,

$$\begin{aligned} \lim _{y\downarrow 0} \frac{\pi [h_D(x+iy, s+i0) +\pi h_D(x+iy, s-i0)] }{y} = \text{ R.H.S. } \text{ of } \text{(41) }. \end{aligned}$$

(41)

If $w=- (1-x^2 +y^2) +i 2xy$ and $\phi = \pi - \arg w \in (-\pi /2,\pi /2)$, then

$$\begin{aligned} |f(z)|^2 =(x+|w| ^{1/2}\sin \phi )^2 +(y+|w| ^{1/2} \cos \phi )^2 \end{aligned}$$

and we see that $y^{-1}(|f(z)|^2 -1) \rightarrow 2/\sqrt{1- x^2}$. In view of (38), this shows that

$$\begin{aligned} \lim _{y\downarrow 0} \frac{\pi h_D(x+iy, s\pm i0) }{y} = \frac{1}{2(1-\cos (\theta _x\mp \theta _s))\sqrt{(1-x^2)(1-s^2)}} \end{aligned}$$

where $\cos \theta _t=t$ with $\theta _t \in (0,\pi )$ for $-1<t<1$. Now (41) follows from the identity

$$\begin{aligned} \frac{1}{1-\cos (\theta _x - \theta _s)}+ \frac{1}{1-\cos (\theta _x + \theta _s)} = \frac{2-2 \cos \theta _x \cos \theta _s}{(\cos \theta _x - \cos \theta _s)^2} =\frac{2(1-xs)}{(x-s)^2}. \end{aligned}$$

$\square$

(3)   Let $(X_n)$ be the imbedded walk on the real axis mentioned in Sect. 1. In other words, $(X_n)$ is the one-dimensional random walk with the transition probability $p^X(x,y)= H_0(y-x)$, where $H_0(x)$ be the hitting distribution of the real line for our random walk starting at the origin as being introduced in (31). Let $\mu (x)$, $x\ge 0$ be a renewal function for the ascending ladder height variables of the walk $(X_n)$ and $\nu$ its dual; they are normalized so as to satisfy $\lim _{x\rightarrow \infty }\mu (x)/\nu (x)=1$ and given by

$$\begin{aligned} \mu (x)=\frac{\sqrt{\pi }e^{\theta _+}}{\sigma }(v_0+\cdots +v_x),\quad \nu (x)=\frac{\sqrt{\pi }e^{-\theta _+}}{\sigma }(u_0+\cdots +u_x) \end{aligned}$$

for $x=0, 1, 2, \ldots$ ([11]:p. 212), where if $c=\exp \Big (-\sum _{k=1}^\infty \frac{1}{k}P_0[X_k=0]\Big )$,

$$\begin{aligned} \theta _+=\sum _{k=1}^\infty \frac{1}{k}\Big (\frac{1}{2}-P_0[X_k>0]\Big )+\frac{1}{2} \log c\,=\, \frac{1}{2} \sum _{k=1}^\infty \frac{1}{k}\Big (P_0[X_k<0]-P_0[X_k>0]\Big ), \end{aligned}$$

(42)

$v_0=u_0=1/\sqrt{c}$ and $\sqrt{c} v_k$ (resp. $\sqrt{c} u_k$) equals the probability that the ascending (resp. descending) ladder height process visits k (resp. $-k$) ([11]:p.202, 203). Then, $\mu$ and $\nu$ are positive solutions of the Wiener–Hopf integral equations associated with the kernels $p^X(x,y)$ and $p^X(y, x)$ ($x, y\ge 0$), respectively, to the effect that for $x= 0, 1, 2,\ldots$,

$$\begin{aligned} \mu (x)=\sum _{k=0}^\infty \mu (k) H_0(x-k) \quad \text{ and } \quad \nu (x)= \sum _{k=0}^\infty \nu (k)H_0(-x+k) \end{aligned}$$

(43)

(see [11]: p. 332 for uniqueness of positive solutions).

(4)   Suppose that $E_0[\,|Y|^2\log |Y|\,]<\infty$ (as in Theorem 5). Put $\sigma _j^2= E_0[(S_1^{(j)})^2]$ ($j=1,2$) and $\sigma _{12}= E_0[S_1^{(1)}S_1^{(2)}]$. Then, putting for $s, x,y\in \mathbf{Z}$

$$\begin{aligned} H_z(s) = P_z[\exists j, S_n=s \quad \text{ and }\quad S_k\notin \mathbf{R}\quad \text{ for }\quad 1\le k<j] \quad (z=x+iy) \end{aligned}$$

it is shown in [15] (Theorem 1.2) that as $|x-s|+|y|\rightarrow \infty$

$$\begin{aligned} H_z(s) = \frac{\, 1\vee \sigma ^2_{2}a(y)}{\pi \Vert s- z\Vert ^2}(1+o(1)). \end{aligned}$$

(44)

(Note that $\sigma _2^2 a(y) \ge |y|$ [11]:P28.8, P31.1.) Here

$$\begin{aligned} \Vert z\Vert ^2 = \sigma ^{-2} (\sigma _{2}^{2} x^2- 2\sigma _{12}xy+\sigma _{1}^{2}y^2) \end{aligned}$$

and $a(y), y\ne 0$ is the potential function of the one-dimensional walk $S^{(2)}_n$. Denote the Poisson kernel on the upper half-plane by $h_z(s)$ as in (2). Putting $\lambda = \sigma ^2/\sigma _2^2$ and $\omega = \sigma _{12}/\sigma _2^2$, we have $\Vert z\Vert ^2 = [(x+\omega y)^2 + (\lambda y)^2]/\lambda$, and for $y\ne 0$, (44) is written as

$$\begin{aligned} H_z(s) =\frac{1\vee \sigma _2^2a(y)}{ |y|} h_{x-\omega y +i\lambda y}(s)(1+o(1)). \end{aligned}$$

(45)

Using this, we can readily deduce from Theorems 2 and 2’ that as $|z|\wedge |z-s| \rightarrow \infty$

$$\begin{aligned} H^{I(n)}_{z}(s) =\frac{ \sigma _2^2a(y)}{|y|} \mu (-n+s)\nu (-n+s) \sqrt{n_*^2 -s}\, h^{I(n)}_{x-\omega y+i\lambda y}(s)(1+o(1)) \end{aligned}$$

(46)

for $y\ne 0$.

(5)   In view of Donsker’s invariance principle, the relation (45) (resp. (46)) incidentally shows that $h_{x-\omega y +i\lambda y}(s)$ (resp. $h^{[-1,1]}_{x-\omega y +i\lambda y}(s)$) is the density of the hitting distribution of the real line (resp. the interval I(n)) for the process $Q^{1/2}B_t$.

We give a direct algebraic verification. We may replace $B_t$ by $UB_t$ with any orthogonal matrix U and choose U so that the matrix $Q^{1/2}U$ sends the real line to itself. A simple algebraic manipulation leads to

$$\begin{aligned} (Q^{1/2}U)^{-1} = \frac{\sigma _2}{\sigma ^2} \left( \begin{array}{c@{\quad }c} 1 &{} -\omega \\ 0 &{} \lambda \end{array} \right) . \end{aligned}$$

Let $z=x+iy$, $\tilde{z} = x-\omega y +i\lambda y$ and $c=\sigma _2/ \sigma ^2$. Noting that $c\tilde{z}$ corresponds to z, we then find, with the notation analogous to $\tilde{h}_D(z,s)$ in (39), that

$$\begin{aligned} \tilde{h}_{z}(s) = c h_{c\tilde{z}}(cs), \end{aligned}$$

of which the right-hand side equals $h_{\tilde{z}}( s)$, yielding the analogue of (39) as desired.