Two-Dimensional Brownian Random Interlacements

Abstract

We introduce the model of two-dimensional continuous random interlacements, which is constructed using the Brownian trajectories conditioned on not hitting a fixed set (usually, a disk). This model yields the local picture of Wiener sausage on the torus around a late point. As such, it can be seen as a continuous analogue of discrete two-dimensional random interlacements (Comets et al. Commun. Math. Phys. 343, 129–164, 2016). At the same time, one can view it as (restricted) Brownian loops through infinity. We establish a number of results analogous to these of Comets and Popov (Ann. Probab. 45, 4752–4785, 2017), Comets et al. (Commun. Math. Phys. 343, 129–164, 2016), as well as the results specific to the continuous case.

Appendix

Proof of Proposition 3.1.

Let a = ∥x∥. Note that,

$$ \frac{1}{2\pi}{\int}_{0}^{\pi} \ln (a^{2}+1-2a\cos\theta) d\theta -\ln a = \frac{1}{2\pi}{\int}_{0}^{\pi} \ln (1+a^{-2}-2a^{-1}\cos\theta) d\theta, $$

and that the claim is obviously valid for a = 0, so it remains to prove that

$$ I(a):= {\int}_{0}^{\pi} \ln (a^{2}+1-2a\cos\theta) d\theta = 0 \qquad \text{ for all } a\in(0,1]. $$

By the change of variable 𝜃 → π − 𝜃 we find that $I(a)= {\int }_{0}^{\pi } \ln (a^{2}+1+2a\cos \theta ) d\theta$, and so

$$ \begin{array}{@{}rcl@{}} I(a) &=& \frac{1}{2} {\int}_{0}^{\pi} \ln \left( (a^{2}+1)^{2}-4a^{2}\cos^{2}\theta\right) d\theta\\ &=& {\int}_{0}^{\pi/2} \ln \left( (a^{2}+1)^{2}-4a^{2}\cos^{2}\theta\right) d\theta. \end{array} $$

(4.23)

Then, using the same trick as above (change the variable $\theta \to \frac {\pi }{2}-\theta$ so that the cosine becomes sine), we find

$$ \begin{array}{@{}rcl@{}} I(a) &=& \frac{1}{2} {\int}_{0}^{\pi/2} \ln \left[\left( (a^{2}+1)^{2}-4a^{2}\cos^{2}\theta\right) \left( (a^{2}+1)^{2}-4a^{2}\sin^{2}\theta\right) \right] d\theta\\ &=& \frac{1}{2} {\int}_{0}^{\pi/2} \ln \left[(a^{2}+1)^{4} - 4a^{2}(a^{2}+1)^{2} + 16a^{4}\cos^{2}\theta\sin^{2}\theta \right] d\theta\\ &=& \frac{1}{2} {\int}_{0}^{\pi/2} \ln \left[(a^{2}+1)^{4} - 4a^{2}(a^{2}+1)^{2} + 4a^{4}\sin^{2} 2\theta \right] d\theta\\ &=& \frac{1}{2} {\int}_{0}^{\pi/2} \ln \left[(a^{2}+1)^{4} - 4a^{2}(a^{2}+1)^{2} + 4a^{4} - 4a^{4}\cos^{2} 2\theta \right] d\theta\\ &=& \frac{1}{2} {\int}_{0}^{\pi/2} \ln \left[\left( (a^{2}+1)^{2} - 2a^{2}\right)^{2} - 4a^{4}\cos^{2} 2\theta \right] d\theta\\ &=& \frac{1}{2} \times \frac{1}{2}{\int}_{0}^{\pi} \ln \left[(a^{4}+1)^{2} - 4a^{4}\cos^{2} \theta \right] d\theta \end{array} $$

using Eq. 4.23, and we finally arrive to the following identity:

$$ I(a) = \frac{I(a^{2})}{2}. $$

(4.24)

This implies directly that I(1) = 0; for a < 1 just iterate Eq. 4.24 and use the obvious fact that I(⋅) is continuous at 0. □

We have to mention that other proofs are available as well; see [7, Ch. 20].