Abstract
We consider a discrete-time, continuous-state random walk with steps uniformly distributed in a disk of radius h. For a simply connected domain D in the plane, let \(\omega _h(0,\cdot ;D)\) be the discrete harmonic measure at \(0\in D\) associated with this random walk, and \(\omega (0,\cdot ;D)\) be the (continuous) harmonic measure at 0. For domains D with analytic boundary, we prove there is a bounded continuous function \(\sigma _D(z)\) on \(\partial D\) such that for functions g which are in \(C^{2+\alpha }(\partial D)\) for some \(\alpha >0\) we have
We give an explicit formula for \(\sigma _D\) in terms of the conformal map from D to the unit disk. The proof relies on some fine approximations of the potential kernel and Green’s function of the random walk by their continuous counterparts, which may be of independent interest.
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Acknowledgments
We would like to thank the anonymous referee for many valuable comments and suggestions. The research of T. Kennedy was supported in part by NSF Grant DMS-1500850.
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Appendix
Appendix
In this appendix, we prove the asymptotics for the potential kernel described in Lemma 4. We follow the methods introduced in Section 12 of [14] and [6]. Let \(\phi (\theta )\) be the characteristic function of the continuous-state random walk with \(h=1\), i.e.,
where \(\theta =(\theta _1,\theta _2)\) and \(X=(X^{(1)},X^{(2)})\) is uniformly distributed in the disk of radius 1.
Lemma 10
Proof
The first estimate in the lemma follows by Taylor expansion, while the second follows since
and the symmetry of \(\theta _1\) and \(\theta _2\). \(\square \)
The following lemma says our potential kernel is well defined.
Lemma 11
Proof
By applying the continuous inversion formula, the proof is similar to the proof of P1 in Section 12 of [14] if one can show \(\frac{1-e^{i\theta \cdot x}}{1-\phi (\theta )}\phi ^3(\theta )\in L^1(\mathbb {R}^2)\). The latter is true because of Lemma 10. \(\square \)
Let \(Q(\theta )=E(X\cdot \theta )^2=\frac{|\theta |^2}{4}\) and \(\psi (\theta )=1/(1-\phi (\theta ))-2/Q(\theta )\). Then Lemma 10 implies
Now we have all ingredients to prove Lemma 4.
Proof of Lemma 4
By Lemma 11 and the evenness of \(\phi \), we see that
By the estimates in Lemma 10 and (24), and the Riemann–Lebesgue lemma,
which is a constant contributing to \(C_0\) in the lemma.
This gives the first o(1) term
Let \(B:=B(0,\pi ):=\{z:|z|<\pi \}\) and \(B^c=\mathbb {R}^2{\setminus }B\). Then the first integral together with the attached multiplicative term in (25) can be written as the sum of the following two integrals
By the estimate in Lemma 10 and the Riemann–Lebesgue lemma we have
which leaves the second o(1) term
We rewrite \(I_1(x)\) as follows
By Lemma 10, the second integral in (30) is a constant contributing to \(C_0\) in the lemma, and by the Riemann–Lebesgue lemma the last integral in (30) gives the third o(1) term
and the first integral together with the attached multiplicative term in (30) is equal to (using the proof of P3 in Section 12 of [14])
where \(\gamma \) is Euler’s constant.
It is clear that \(\gamma +\ln \pi +\ln |x|+\ln (\sin \alpha )\) in (32) as a function of \(\alpha \) is integrable from 0 to \(\pi /2\), so the fourth o(1) term is
where the equality follows by reversing the procedure which led to (32).
Adding the four o(1) terms, i.e., (26) + (29) + (31) + (33), we get
Noting that \(\cos (x\cdot \theta )=\nabla \cdot {\mathbf {b}}(\theta )\) where \(\mathbf {b}(\theta )=\sin (x\cdot \theta )(x_1/|x|^2,x_2/|x|^2)\), the divergence theorem gives
We can apply the divergence theorem again to (35). As a result, we see that (34) has order \(O(|x|^{-2})\).
Therefore, the proof of Lemma 4 is complete if one can show \(\Delta _{h} a(x/h)=\frac{1}{\pi }I_{B(0,h)}(x)\). But the latter is easy to verify (note that \(a(x/h)=h^2a_h(x)\)). \(\square \)
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Jiang, J., Kennedy, T. The Difference Between a Discrete and Continuous Harmonic Measure. J Theor Probab 30, 1424–1444 (2017). https://doi.org/10.1007/s10959-016-0695-3
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DOI: https://doi.org/10.1007/s10959-016-0695-3