Stirring two grains of sand

Consider two unit balls in a $d$-dimensional flat torus with edge length $r$, for $d\geq 2$. The balls do not move by themselves but they are pushed by a Brownian motion. The balls never intersect---they reflect if they touch. It is proved that the joint distribution of the processes representing the centers of the balls converges to the distribution of two independent Brownian motions when $r\to \infty$, assuming that we use a proper clock and proper scaling. The diffusion coefficient of the limit process depends on the dimension. The positions of the balls are asymptotically independent also in the following sense. The rescaled stationary distributions of the centers of the balls converge to the product of the stationary (hence uniform) distributions for each ball separately, as $r\to\infty$.


Introduction
The word "stirring" in the title of this paper refers to a random change in a system of many bodies that is caused by a single agent that moves continuously and acts locally. This is in contrast to those stochastic flows where different parts of the moving medium are simultaneously "pushed" by different (although possibly correlated) random noises. In everyday life, stirring typically refers to activities such as stirring coffee in a cup or stirring paint in a bucket. In these situations, stirring the medium with a spoon or a stick causes the bulk of the liquid to move (in a circular fashion). Our model is closer to stirring sand in a sandbox with a stick. In this situation, sand grains are displaced locally and there is no overall motion of the bulk of the sand mass.
Stirring sand in a sandbox provided motivation for this project but our model is a simplification of the reality in (at least) two significant ways. First, we will consider only two "sand grains" represented by balls. This seems to be the crucial step in the analysis of the motion of many "sand grains" (see the remarks on [BCP13] below). Second, the stirring agent will be represented by an infinitely small particle performing Brownian motion. One may consider our results as a first step towards a more realistic model.
In our model, the stirring agent, represented by Brownian motion, is not affected by the motion of "sand grains." The two sand grains (balls) remain motionless except when they are pushed by the Brownian particle aside, when its trajectory hits their surfaces.
The problem that we will investigate is that of the evolution of the vector between the centers of the two balls. It is natural to guess that the motion of a each ball should be similar to that of Brownian motion on the local time scale. The crux of the problem is that the directions of the push that the balls receive from the Brownian particle are not independent. Therefore, even if the guess about the motion of a single ball is correct, that does not immediately imply that the limit distribution for the pair of the balls is a pair of independent Brownian motions. We will prove that this is in fact true and we will express this idea in two different ways, to be described below. The main technical challenge of the paper is to estimate the magnitude of the dependence between motions of the two balls.
We will separately prove the invariance principle for a single ball pushed by Brownian motion in dimension 2 in the whole space R 2 . This is meaningful because Brownian motion is recurrent in two dimensions so it will keep pushing the ball forever. We consider this simplified question separately to present a more or less straightforward proof. Many technical details obscure this part of the argument in the case of two balls or in higher dimensions.
In dimensions 3 and higher, the two balls and Brownian motion will be located in a torus because Brownian motion is transient in these dimensions (but the theorem will cover the two dimensional case as well). First, we will prove an invariance principle on the local time scale for the centers of the two balls. The limiting process is a pair of two independent Brownian motions. Next we will show that the rescaled stationary distributions for the two balls in a torus of diameter r converge to the product of the stationary (and hence uniform) distributions for the individual balls as r → ∞. At the end of Section 3 we will explain why the latter theorem does not immediately follow from the former.
The present paper is a part of a larger project. Our present model is "almost" equivalent to the model in which a ball with the center moving as a Brownian motion pushes two point-like particles. The equivalence is not complete because the two balls in our model cannot intersect (by assumption) and hence their centers are always at least two units apart. In the other model, the two point-like particles can come arbitrarily close. Their motion was partly analyzed in [BCP13], where it was proved that the distance between the two particles does not converge to 0 in a three dimensional torus. This is very close to proving recurrence for the two-particle process. The main results of the present article show, more or less, that the particles are independent on the large scale. Only one element of the program initiated in [BCP13] is still missing-the positive recurrence (as opposed to the mere recurrence) of the two particle motion. If this gap is filled then this will be, most likely, sufficient to prove Conjectures 1.5 and 1.6 in [BCP13].
The paper is organized as follows. The next section contains the formal description of the model. Section 3 presents the statements of the main results. It is followed by Section 4 with a review of excursion theory and some results on excursion laws. The motion of a single ball is analyzed in Section 5. Section 6 is devoted to estimates of local time. Section 7 gives estimates for hitting distributions. The main theorem on invariance principle is proved in Section 8. The theorem on convergence of the stationary distributions is proved in two sections, Sections 9 and 10, the first of which is devoted to the irreducibility of the process.

Preliminaries
2.1. Processes in R d . First we will consider the case when the Brownian motion and the balls are located in R d with d ≥ 2. We will consider two moving balls with radii 1 and centers X t and Y t , resp. Brownian motion will be denoted B t .
Let B(x, r) denote the open ball with center x and radius r and let S(x, r) = ∂B(x, r). The two moving balls will be denoted X t = S(X t , 1) and Y t = S(Y t , 1).
For x ∈ S(y, r), let n(S(y, r), x) be the unit outward normal vector to S(y, r) at x. We will now describe the effect of the push of B t on the trajectory of X t . Let us ignore the other ball Y t for the moment. We assume that |B 0 − X 0 | ≥ 1, a.s. By the results of [LS84], there exist a continuous process Z t taking values in B(X 0 , 1) c and a non-decreasing real valued continuous process ("local time") L X t such that L X 0 = 0, At this point, a better name for the process L X would have been L Z but L X was used in anticipation of (2.3) below. The process Z t is Brownian motion reflected on X 0 . We define X t by In this way, the ball X t is pushed by Brownian motion B t . Note that we have B t / ∈ B(X t , 1) for all t ≥ 0, ∞ 0 1 {Bt / ∈Xt} dL X t = 0, and Next we will consider the motion of the balls X t and Y t only, ignoring B t . Suppose that X t is a continuous process and |X 0 −Y 0 | ≥ 2. We do not want the balls X t and Y t to intersect. We apply the results of [LS84] once again. There exist a continuous process V t taking values in B(Y 0 , 2) c and a non-decreasing real valued continuous process ("local time") L ′ t such that L ′ 0 = 0, In this way, the ball Y t is pushed by the ball X t . Note that we have B(X t , 1)∩B(Y t , 1) = ∅ for all t ≥ 0 and ∞ 0 1 {|Xt−Yt|>2} dL ′ t = 0. Now we will describe the joint evolution of B t , X t and Y t . Suppose that |B 0 −X 0 | ≥ 1, . Assume without loss of generality that B t hits X 0 strictly before hitting Y 0 . Then we use (2.2) and (2.4) to define X t and Y t until the first time T 1 ≥ 0 such that B T 1 ∈ Y T 1 . Suppose that B T 1 / ∈ X T 1 . At this time we switch the roles of X t and Y t in the definitions (2.2) and (2.4). In other words, B t is now pushing the ball Y t and the ball Y t is pushing the ball X t . We define in this way processes X t and Y t for t ≥ T 1 until the first time T 2 ≥ T 1 such that B T 2 ∈ X T 2 . Suppose that B T 2 / ∈ Y T 2 . We continue in this fashion, i.e., we construct stopping times T 1 , T 2 , T 3 , . . . , such that T n = inf{t ≥ T n−1 : B t ∈ Y t } for odd n and T n = inf{t ≥ T n−1 : B t ∈ X t } for even n. The inductive definition is continued as long as B Tn / ∈ X Tn for odd n and B Tn / ∈ Y Tn for even n.
Lemma 2.1. With probability 1, all stopping times T n , n ≥ 1, are well defined and lim n→∞ T n = ∞.
Proof. The claim in the lemma may be false for two different reasons. First, it is possible that for some random time T ∞ and n < ∞, Second, it may be that all T n 's are well defined and for some random time T ∞ we have P(T ∞ < ∞) > 0 and T ∞ = lim n→∞ T n , a.s.
We will analyze the second case first. Consider ω such that all T n 's are well defined and T ∞ = lim n→∞ T n < ∞. Consider an ε ∈ (0, 1) and let t 1 < T ∞ be so close to T ∞ that sup t 1 ≤s,t≤T∞ |B s − B t | < ε. Suppose that n 1 is so large that T 2n > t 1 for 2n ≥ n 1 . Consider any n such that T 2n > t 1 . We have By analogy, (2.6) and (2.7) hold on every interval of the form [T 2n+1 , T 2n+2 ) for 2n ≥ n 1 . Since ε can be taken arbitrarily small, This implies that lim t↑T∞ dist(X t , Y t ) = 2. It is easy to see that this claim and (2.8) are also true in the case represented by (2.5) because in that case we have B T∞ ∈ X T∞ ∩Y T∞ .
Note that the definition (2.3) of the local time L X t applies in the new context of one Brownian particle and two balls, for t < T ∞ . Let L Y t be the "local time" of B t on Y t , defined in a way analogous to L X t . Suppose that for some s 1 < T ∞ , Then it follows from (2.3) and an analogous formula for Y t that X t = X T∞ and Y t = Y T∞ for all t ∈ [s 1 , T ∞ ]. Since {B t } ∩ (B(X t , 1) ∪ B(Y t , 1)) = ∅ for t ∈ [s 1 , T ∞ ] and the circles X T∞ and Y T∞ are tangent at B T∞ , it follows that for some s 2 ∈ (s 1 , T ∞ ), the piece of Brownian path {B t , t ∈ [s 2 , T ∞ ]} stays inside a cone with vertex B T∞ and opening π/8. This event has probability 0 according to a theorem in [Bur85].
Next consider the case when (2.9) is false. Let s 3 < T ∞ be so large that dist(B t , X t ) ≤ 1.01 and dist(B t , Y t ) ≤ 1.01 for t ∈ [s 3 , T ∞ ]. Simple geometry shows that if dist(B t , X t ) ≤ 1.01, dist(B t , Y t ) ≤ 1.01, and B is pushing X or Y at time t then the distance between X and Y is increasing. We have assumed that (2.9) is false so the amount of push on the interval [s 3 , T ∞ ] is strictly positive and we conclude that lim inf t↑T∞ dist(X t , Y t ) > 2 on the event {T ∞ < ∞}, contradicting our earlier claim.
Lemma 2.1 completes the construction of our model on R d .

2.2.
Processes on a torus. We will denote the flat d-dimensional torus R d /(rZ d ) with edge length r by T d r . Technical difficulties with the definition of the joint distribution of B, X and Y on R d are local in nature so we will not go into details of the analogous construction on T d r . We limit ourselves to the remark that if r > 4 then the evolution of the process {(B t , X t , Y t ), t ≥ 0} on T d r is the natural analogue of the evolution of this process on R d .
We will say that x = y (mod r) for x, y ∈ R d and r ∈ R if x = y + rz for some z ∈ Z d . We can identify T d r with [0, r) d in the obvious way. If Brownian motion B t is defined on T d r then we define its "unfolded" versions B t by requiring that it is a continuous process on R d , B 0 = B 0 and B t = B t (mod r) for all t ≥ 0. Processes X t and Y t are defined in an analogous way.

Main results
Our first theorem is concerned with the motion of a single ball pushed by Brownian motion in two dimensions. Since the two-dimensional Brownian motion is recurrent, we do not need to place the processes in a torus to obtain a meaningful result. On the technical and conceptual side, the proof of the first theorem is much simpler than those of the other main results so it is natural to state and prove this result first.
Recall the definition of the vector of three process (B t , X t , Y t ) on a torus from Section 2.2. We will sometimes emphasize the dependence on r in the notation by writing (B r t , X r t , Y r t ) and similarly (B r t , X r t , Y r t ). Let L X be defined as in (2.3), let L Y be defined in an analogous way, and let L t = L X t + L Y t for t ≥ 0. Let σ t be the inverse local time, i.e., σ t = inf{s ≥ 0 : L s ≥ t}. Let C d = (d − 1)d and note that C d > 0 for d ≥ 2. We have C d = √ 2 for d = 2 so the normalization in the following theorem seems to contradict the normalization in Theorem 3.1. There is no contradiction because the two theorems use different local time clocks; the local time clock in the next theorem is twice as fast as that in Theorem 3.1, on average.
Theorem 3.2. Suppose that d ≥ 2, and for each r > 4 we have converge weakly to standard 2d-dimensional Brownian motion when n → ∞ and r → ∞.
In two dimensions, the last theorem could be stated for {(B t , X t , Y t ), t ≥ 0} defined on (R 2 ) 3 . The proof of Theorem 3.2 would apply almost verbatim in that setting so this version of the theorem is omitted.
Theorems 3.1-3.2 are concerned with processes run with the local time clock as it is more meaningful than the standard clock in this context. The next theorem is stated for processes run with the usual clock but the proof shows that the result is equally true for processes run with the local time clock.
Let ν d r be the uniform probability measure on T d r . Theorem 3.3. Suppose that d ≥ 2, r > 4, and {(B r t , X r t , Y r t ), t ≥ 0} is defined on T d r . (i) The process (B r t , X r t , Y r t ) has a unique stationary distribution L d r . (ii) The distributions of (X r 0 , Y r 0 )/r under L d r converge to ν d 1 × ν d 1 as r → ∞. It follows immediately from translation invariance of the process (B r t , X r t , Y r t ) that the distribution of X r 0 under L d r is ν d r , and the same remark applies to Y r 0 . Hence, the essence of Theorem 3.3 (ii) is that the two components of (X r 0 , Y r 0 ) are asymptotically independent.
We will explain, in an informal way, why Theorem 3.3 does not immediately follow from Theorem 3.2. Suppose that processes X and Y satisfy the conclusion of Theorem 3.2. It is conceivable that the process |X r σt − Y r σt | on T d r has a positive drift of order 1/r because this drift would disappear in the limit theorem for {n −1/2 (X r nσt − Y r nσt ), t ∈ [0, t 1 ]} for any fixed t 1 < ∞. The process X r σt − Y r σt needs about r 2 units of time to reach the stationary distribution. On this time scale, the drift of size 1/r would move X σt and Y σt about r 2 /r = r units apart, relative to the analogous situation without a drift. Since the effect of the drift on this time scale is comparable with the diameter of the torus, it is conceivable that under the stationary distribution X r and Y r would be typically farther apart than two random vectors with the joint distribution ν d r × ν d r .

Excursion theory
This section contains a brief review of excursion theory needed in this paper. See, e.g., [Mai75] for the foundations of the theory in the abstract setting and [Bur87] for the special case of excursions of Brownian motion. Although [Bur87] does not discuss reflected Brownian motion, all results we need from that book readily apply in the present context.
Let P x denote the distribution of Brownian motion starting from x and let E x be the corresponding expectation. For a domain (open connected set) D ⊂ R d , let P x D denote the distribution of Brownian motion starting from x ∈ D and killed upon exiting D.
An "exit system" for excursions of reflected Brownian motion Z from ∂D is a pair (L * t , H x ) consisting of a positive continuous additive functional L * t of Z and a family of "excursion laws" {H x } x∈∂D . Let ∆ denote the "cemetery" point outside D and let C be the space of all functions f : [0, ∞) → D ∪ {∆} which are continuous and take values in D on some interval [0, ζ), and are equal to ∆ on [ζ, ∞). For x ∈ ∂D, the excursion law H x is a σ-finite (positive) measure on C, such that the canonical process is strong Markov on (t 0 , ∞), for every t 0 > 0, with the transition probabilities P D . Moreover, H x gives zero mass to paths which do not start from x. We will be concerned only with the "standard" excursion laws; see Definition 3.2 of [Bur87]. For every x ∈ ∂D there exists a unique standard excursion law H x in D, up to a multiplicative constant.
Excursions of Z from ∂D will be denoted e or e s , i.e., if s < u, Z s , Z u ∈ ∂D, and The following is a special case of the exit system formula of [Mai75]. For every x ∈ D, every bounded predictable process V t and every positive universally measurable function f : C → [0, ∞) that vanishes on excursions e t identically equal to ∆, we have Here and elsewhere H x (f ) = C f dH x . Intuitively speaking, (4.1) says that the right continuous version E t+ of the process of excursions is a Poisson point process on the local time scale with variable intensity H .
The normalization of the exit system is somewhat arbitrary. For example, if (L * t , H x ) is an exit system and c ∈ (0, ∞) is a constant then (cL * t , (1/c)H x ) is also an exit system. One can even make c dependent on x ∈ ∂D. Theorem 7.2 of [Bur87] shows how to choose a "canonical" exit system; that theorem is stated for the usual planar Brownian motion but it is easy to check that both the statement and the proof apply to reflected Brownian motion. According to that result, we can take L * t to be the continuous additive functional whose Revuz measure is a constant multiple of the surface area measure dx on ∂D and H x 's to be standard excursion laws normalized so that for any event A in the σ-field generated by the process on an interval [t 0 , ∞), for any t 0 > 0.
Recall the local time L X from the Skorokhod representation of reflected Brownian motion given in (2.1). In the present context L X will be called L Z . The Revuz measure of L Z is the measure dx/(2|D|) on ∂D, i.e., if the initial distribution of Z is the uniform probability measure µ on D, then for any Borel set A ⊂ ∂D. It has been shown in [BCJ06] that L * t = L Z t , i.e., (L Z t , H x ) is an exit system if excursion laws H x are defined as in (4.2). 4.1. Excursions crossing spherical shells. We will calculate the "probability" under excursion law that an excursion starting at the inner boundary of a spherical shell hits the outer boundary.
Let A be the event that the process hits S(0, b) before hitting S(0, 1). The function x → P x (A) is harmonic in D with boundary values 1 on S(0, b) and 0 on S(0, 1).
In the two-dimensional case we have P The last formula holds with b = ∞. We have in that case 4.2. Expected lifetimes of excursions. We will derive estimates for expected excursion lifetimes in the exterior of two balls in a torus. Let Lemma 4.1. Suppose that d ≥ 2, r > 4, x 1 , x 2 ∈ T d r , and D = T d r \(B(x 1 , 1)∪B(x 2 , 1)). Let (L t , H x ) denote the exit system for reflecting Brownian motion in D, normalized as in (4.2).
(i) There exist r 1 and c 1 such that if Proof.
Step 1. In this proof, B will denote reflected Brownian motion in D. Let D := T d r \ B(x 1 , 1) and let H x denote an excursion law in D normalized as in (4.2). We will consider various "large" spheres, balls, etc. We will always tacitly assume that their diameters are smaller than r, the edge of the torus.
Every positive harmonic function in D is integrable by the results of [Arm72]. Since the density of the expected occupation measure for an excursion law for reflected Brownian motion in D is (a constant multiple of) the Poisson (Martin) kernel, it follows that for some For an excursion e t , let S a = inf{s ≥ 0 : e t (s) ∈ S(x 1 , a)}. The argument given in the last paragraph implies that for every a > 1 there exists α(a) < ∞ such that for all x ∈ S(x 1 , 1), Standard arguments show that (4.3) implies that, a.s., The middle expression in (4.9) contains 2 in front of s d /(2|D|) because the boundary of D consists of two spheres.
Step 2. Let D k = B(x 1 , 2 k ) \ B(x 1 , 1) and let f k (x, · ) be the density with respect to the surface area measure µ k on ∂D k of the hitting distribution on ∂D k of Brownian motion starting from x ∈ D k . We apply the Harnack inequality in the domain B(x 1 , 2 k−1 · 3/2) \ B(x 1 , 2 k−1 · 3/4) to see that there exists c 3 > 0, independent of k, such that for all z 1 , z 2 ∈ S(x 1 , 2 k−1 ) and y ∈ ∂D k , By the strong Markov property applied at the hitting time of S(x 1 , 2 k−1 ), for z ∈ D k ∩ B(x 1 , 2 k−1 ) and y ∈ S(x 1 , 2 k ), We now apply [BTW89, Lem. 6.1] (see [BK98, Lem. 1] for a better presentation of the same estimate) to see that (4.11)-(4.12) imply that there exist constants C j , 1 ≤ j ≤ k − 1, such that for every 1 ≤ j ≤ k − 1 and all z 1 , z 2 ∈ S(x 1 , 2 k−j ) and y ∈ S(x 1 , 2 k ), Moreover, C j ∈ (0, 1), C j 's depend only on c 3 , and 1 − C j ≤ e −c 4 j for some c 4 > 0 and all j. Hence, for z 1 , z 2 ∈ S(x 1 , 2) and y ∈ S(x 1 , 2 k ), Step 3. We will first prove part (ii) of the lemma. Fix an arbitrarily small ε > 0. Assume that |x 1 − x 2 | > 2 m+1 and choose m so large that, in view of (4.13) and the strong Markov property applied at the time S 2 , for all z 1 , z 2 ∈ S(x 1 , 1) and y ∈ S(x 1 , 2 m ).
Let T A = inf{t ≥ 0 : B t ∈ A} for any set A. We will now treat m as a fixed number (we will consider r a variable and we will let r → ∞) so that (4.8) yields H x (ζ ∧ S 2 m ) = H x (ζ ∧ S 2 m ) = c 6 for some c 6 < ∞ and all x ∈ S(x 1 , 1). We obtain for z 1 , z 2 ∈ S(x 1 , 1), 1). The last estimate may be now written as for z ∈ S(x 1 , 1). By symmetry, the estimate holds also for all z ∈ S(x 2 , 1). We can use the lower bound in (4.14) to derive the analogous lower estimate, so for z ∈ ∂D, Note that |D|/r d → 1 as r → ∞. By (4.10) and (4.15), for large r, For similar reasons, The estimates (4.16)-(4.17) show that, Since ε > 0 is arbitrarily small and r can be made large, this and (4.15) imply part (ii) of the lemma. Next we prove part (i) of the lemma. Note that the argument given in this step remains valid if we replace all occurrences of excursion laws H with H and drop the assumption on the distance between x 1 and x 2 . The only difference is that we would need (4.10) with an extra constant 2 on the left hand side because we would use the exit system in D rather than D. Hence, we have H x (ζ) ≤ c 7 r d for large r, that is, an inequality analogous to the upper estimate in part (ii). Part (i) follows because H x (ζ) ≤ H x (ζ).

Motion of a single ball
Proof of Theorem 3.1. Our starting point is the following equation, analogous to (2.1), where X 0 is replaced with S(0, 1). We will consider a continuous process Z t (reflected Brownian motion) taking values in B(0, 1) c and local time It follows from the recurrence of two-dimensional Brownian motion that lim t→∞ L Z t = ∞.
Comparing the above setup with the statement of the theorem and (2.1)-(2.3), we conclude that it will suffice to prove that processes {n −1/2 L Z nσ Z t , t ≥ 0} converge weakly to Brownian motion as n → ∞.
We will first compute the distribution of Z σ Z t assuming that Suppose that B 0 = 0 and let (a, M a ) be the (random) location of B at the hitting time of the line ∂K a . It is well known that the processes {W 2 σ W t , t ≥ 0} and {M t , t ≥ 0} have the same distribution. Hence, for a fixed t, the distribution of W σ W t is the same as the harmonic measure on ∂K t with the base point at (0, 0). The complex analytic function z → exp(−z) maps W onto a time change of Z and the local time is conformally invariant, so Z σ Z t is distributed as the image of the distribution of W σ W t under the map z → exp(−z). By the earlier remarks and conformal invariance of harmonic measure under the map z → exp(z − t), the distribution of Z σ Z t is the same as the harmonic measure in B(0, 1) relative to the base point exp(−t). The density of this harmonic measure with respect to the uniform probability measure on S(0, 1) at a point z ∈ S(0, 1) is Let θ = arg(z) ∈ [0, 2π) for z ∈ S(0, 1) and let µ t (dz) be the distribution of Z σ Z t . We will need a formula for S(0,1) cos(θ)µ t (dz). Note that for any fixed ϕ, the function z → cos(θ − ϕ) is equal to the harmonic function z = (z 1 , z 2 ) → z 1 cos(ϕ) + z 2 sin(ϕ) on S(0, 1). Since µ t (dz) is the harmonic measure with the base point exp(−t), it follows that Recall the normal vector n and write n = (n 1 , n 2 ). The strong Markov property applied at σ Z t , invariance of the transition probabilities of Z under rotations about 0, and (5.4) imply for s > t, with the convention θ = arg(z), (5.5) Let U x denote the uniform probability distribution on S(x, 1). Assume that B 0 has the distribution U 0 . Then Z σ Z t has the same distribution. Hence, by (5.5), It follows that For j = 1, 2, and n = 1, 2, . . . , let Suppose that B 0 has the uniform distribution on S(0, 1). Then, for every t, Z σ Z t also has the uniform distribution on S(0, 1). By the strong Markov property, the sequence {L 1,Z (n) , n ≥ 0} is strictly stationary. It follows from (5.3) and the strong Markov property that the sequence {L 1,Z (n) , n ≥ 0} is ϕ-mixing in the sense of [Bil68, Sect. 20] with ϕ n ≤ c 1 e −n for some c 1 < ∞. Hence, by [Bil68,Thm. 20 converge in distribution to Brownian motion with some diffusion coefficient, as n → ∞. · v, t ≥ 0}, for every vector v of unit length. The last observation can be rephrased by saying that every linear combination of the processes L 1,Z σ Z t and L 2,Z σ Z t satisfies the same type of invariance principle, with possibly different normalizing constant. Applying the strong Markov property at σ Z ns , we can prove that for any unit vector v, converge jointly and the first component of the limit is Brownian motion independent of the other two components. Hence, converges to a process whose each component is Brownian motion and whose increments are Gaussian and independent. We conclude that the limit process is Gaussian and it is a constant multiple of Brownian motion, by rotation invariance. It remains to identify the diffusion coefficient. The formula (5.6) implies that lim u→∞ Var L 1,Z σ Z u /u = 1. In general, the existence of a weak limit for a sequence of random variables does not imply that the variance of the limit is the limit of variances but this is the case in the setting of [Bil68, Sect. 20] so we conclude that the diffusion coefficient is 1.

Estimates for the local time
Recall the process Z defined in (2.1). Let X 0 = S(0, 1) in that equation so that the process Z takes values in R d \ B(0, 1). We will identify L Z with L X that appeared in (2.1). Let Processes {U t , t ≥ 0} and {Z t , t ≥ 0} have the same distribution. This claim is a slight modification of [Pas02, Thm. 2.3], where a "scaling coupling" was constructed. The above construction is related to "perturbed Bessel processes," see, e.g., [DWY98]. Let L U t be the local time of U on S(0, 1). Then, informally speaking, dL U = −dM/M and, therefore, L U t = − log M t . The last formula can be verified rigorously, for example, by using excursion theory.
Let e k be the k-th vector in the usual orthonormal basis for R d and For all t ∈ (0, ∞), the distributions of σ U t and σ M t are defective because there may be no s such that L U s ≥ t or M s = e −t . The distribution of U σ U t is the same as the distribution of B σ M t /|B σ M t |. We will give the value 0 to these and similar quantities in our calculations when σ U t or σ M t are undefined. The distribution of B σ M t is the same as the hitting distribution of S(e −t , 0). Suppose that B 0 = e 1 . Then, by the Kelvin transformation (see [PS78, Thm. 3.1, p. 102]), the distribution of B σ M t is the same as the hitting distribution of S(e −t , 0) by Brownian motion starting from the point e −2t e 1 times the probability that Brownian motion starting from e 1 will hit S(0, e −t ). The last probability is equal to e −t(d−2) . Since |B σ M t | = e −t , we see that the (defective) distribution µ t (dz) of B σ M t /|B σ M t | is the same as e t(2−d) times the hitting distribution of S(0, 1) by Brownian motion starting from the point e −t e 1 .
For any fixed unit vector v ∈ S(0, 1), the function z → v · z is harmonic. It follows that Let U x denote the uniform probability distribution on S(x, 1). Note that U 0 is the harmonic measure in B(0, 1) with the base point at 0. The function Let n = (n 1 , n 2 , . . . , n d ) and for j = 1, . . . , d, and n ≥ 1, We have by the strong Markov property applied at σ U t , rotation invariance of reflected Brownian motion and (6.2), for s > t and v ∈ S(0, 1), Now assume that B 0 is uniformly distributed over S(0, 1). Then U σ U t has the defective distribution e t(2−d) U 0 . We obtain from (6.3) and (6.4), for s > t,

It follows that
Recall the notation from the beginning of this section and (6.1). Consider any b > 1 and let The expectation on the last line is calculated under the assumption that B 0 is distributed uniformly on S(0, 1).
Recall that B 0 has the uniform distribution on S(0, 1). Then, for every n, Z Sn also has the uniform distribution on S(0, 1). By the strong Markov property, the sequence { L 1,Z (n) , n ≥ 0} is strictly stationary. One can prove that the sequence { L 1,Z (n) , n ≥ 0} is ϕ-mixing in the sense of [Bil68, Sect. 20] with ϕ n ≤ c 1 e −n for some c 1 < ∞ using the formula (5.3) and the method employed in the proof of (4.13). Let T (t) be the largest T n ≤ t. By [Bil68,Thm. 20.1], for some c 1 , , t ≥ 0} converge to Brownian motion in distribution, as n → ∞. The distribution of L Z Tn − L Z Sn is exponential with mean log b, by (4.4). Hence | L 1,Z (n) | is majorized by an exponential random variable with mean log b. This implies that {c 1 n −1/2 L 1,Z σ Z nt , t ≥ 0} converge to Brownian motion in distribution, as n → ∞. This we already know from the proof of Theorem 3.1. The point of the present argument is that this time we divided the time axis into subintervals of independent lengths with exponential distributions with mean log b. Since the limit process is the same in both cases, the variances must match and, therefore, lim b→∞ λ 2 (d, b)/ log b = 1 because the contributions from the cross terms will disappear in the limit, for the same reason why we have ϕ-mixing.
(ii) Suppose that d ≥ 3. It follows from (4.6) that L Z ∞ has the exponential distribution with mean 1/(d − 2). This and the formula (6.1) show that the family {L 1,Z T b } b>1 is uniformly integrable. We have lim b→∞ L 1,Z T b = L 1,Z ∞ , a.s., so part (ii) of the lemma follows from Lemma 6.1.
(iii) For any starting point B 0 = x / ∈ B(0, 1), the distribution of Z σ Z 1 is absolutely continuous with respect to the uniform probability measure on S(0, 1), according to the proofs of Theorem 3.1 and Lemma 6.1. Let c 1 = c 1 (d) be the maximum of the corresponding Radon-Nikodym derivative and note that c where θ is the usual Markov shift. By the strong Markov property applied at σ Z 1 and parts (i) and (ii) of this lemma, for all x ∈ S(0, 1), Let c 2 = 2 + 2c 1 λ 2 (d, b). Then for large b, 2 ) ≤ 1/4. An application of the strong Markov property at times σ Z 2kc 2 shows that, for k ≥ 1, P x (L 1,Z T b ≥ 2kc 2 ) ≤ 1/4 k , and this implies that for some c 3 > 0 and all a > 0, This implies that for every ε, β > 0, .
For a fixed d ≥ 3, the quantities λ 1 (d, b) and λ 2 (d, b) have limits in (0, ∞) as b → ∞. This and the fact that the series is summable for any c 4 ∈ (0, ∞) imply that the limit in (6.8) is equal to 0. This proves part (iii) for d ≥ 3.
For d = 2, lim b→∞ λ 2 (d, b)/ log b = 1. If 2 k+1 ≥ ε √ b β and n ≥ k + 1, then the ratio of the two consecutive terms in the series on the right hand side of (6.8), corresponding to indices k = n and k = n + 1, is equal to The last expression is greater than 1/2 for sufficiently large b, so for large b, the series on the right hand side of (6.8) is bounded by twice its first term, and this implies that

Hitting distribution estimates
For an open set D and a point x ∈ D, let µ D x (dy) be the harmonic measure on ∂D with the base point x.
We have A(x 1 ) ∪ A(y 1 ) = T d r so the volume of each of these sets is equal to or greater than r d /2 (the volumes are equal by symmetry). We define some subsets of R d as follows, The problem is invariant under translations so we may and will assume that x 1 and y 1 are positioned in such a way in Let B denote Brownian motion in R d starting from z 1 ∈ S(x 1 , b) ∪ S(y 1 , b) and for K ⊂ R d let T B (K) be the first hitting time of K. Let U 1 = 0 and T ′ 1 = T d r , and note that B U 1 ∈ T ′ 1 . Let T 1 be a d-dimensional cube in R d with the same center as T ′ 1 but with edge length equal to 3r. For k ≥ 2, let U k = inf{t ≥ U k−1 : B t ∈ ∂T k−1 } and let x is not unique then we choose one of the x's in an arbitrary way). Let T k be the d-dimensional cube in R d with the same center as T ′ k but with the edge length equal to 3r. By our assumption on the positions of x 1 and y 1 , the distance from B U k to S(x 1 , 1) ∪ S(y 1 , 1) is greater than or equal to b − 1 for all k, a.s. Suppose that z / The following estimate is standard, The set S(x 1 , 1) ∩ T k−1 consists of 3 d copies of S(x 1 , 1), each one of them at a distance greater than or equal to b − 1 from B U k−1 . This, the strong Markov property and the last estimate imply that Since the d-dimensional Lebesgue measure of A(x 1 ) ∩ T k−1 is greater than (3r) d /2, we have for some c 4 , For the same reason The above estimates imply that For any x ∈ ∂A(x 1 ), we have P x (T B (S(x 1 , 1)) ≤ T B (S(y 1 , 1))) = 1/2, so P z 1 (T B (S(x 1 , 1)) ≤ T B (S(y 1 , 1))) ≤ 1/2 + c 5 α(d, b). (S(y 1 , 1)) ≤ T B (S(x 1 , 1))) ≤ 1/2 + c 5 α(d, b).
Recall that for an open set D and a point x ∈ D, µ D x (dy) is the harmonic measure on ∂D with the base point x and U x denotes the uniform probability distribution on S(x, 1).
Lemma 7.2. Suppose that x 1 , y 1 ∈ T d r and dist(x 1 , y 1 ) > 2b. Let D = T d r \ (B(x 1 , 1) ∪  B(y 1 , 1)). There exists c 1 such that for b > 4 and r > 8b there exists a ∈ (0, 1] such that for z ∈ S(x 1 , b) ∪ S (y 1 , b) there exists a probability distribution D on S(x 1 , 1) satisfying Heuristically, (7.3)-(7.4) say that the exit distribution from D, normalized so that its restriction to S(x 1 , 1) is a probability measure, is very close to the uniform distribution on S(x 1 , 1). Proof.
Step 1. By translation invariance we may and will suppose that x 1 = 0.
Suppose that the lemma has been proved for all z ∈ S(0, b). If z ∈ S(y 1 , b) then one can apply the strong Markov property at the hitting time of S(0, b) and use standard arguments to extend the claim to z ∈ S(y 1 , b). Hence, we will assume that z ∈ S(0, b). Let x 0 (dx)/U 0 (dx) for x ∈ S(0, 1). By rotational symmetry, f (x) is a function of x 1 only. We will show that f (x) is a non-decreasing function of x 1 . Suppose that x, y ∈ S(0, 1) and x 1 > y 1 and let M be the (d − 1)dimensional hyperplane such that x and y are symmetric with respect to M. Note that M passes through the origin and, therefore, D 1 is symmetric with respect to M. The points x 0 and x are on the same side of M. If we start a Brownian motion at x 0 and it hits M then it has the same chance of exiting D 1 at x and y, by symmetry. But Brownian motion starting at x 0 can exit D 1 at x without hitting M, so f (x) ≥ f (y).

Invariance principle
Proof of Theorem 3.2.
Step 1. Suppose that b > 10, r > 10b 2 and let a be as in Lemma 7.2. Recall that the state space for each of the processes B, X and Y is T d r . Assume

Recall (2.3) and let
. Note that for k < k 1 , X U k = X 0 + 1≤n≤k ∆ n L X , and a similar formula holds for Y .
The strong Markov property and Lemma 7.2 imply that for n ≥ 1 such that j(X, n)− 1 < k 1 , the conditional distribution of B T j(X,n) given F U j(X,n)−1 is equal to aU X U j(X,n)−1 + (1−a)D X n , where D X n is a probability distribution on X U j(X,n)−1 , determined by the values of X U j(X,n)−1 , Y U j(X,n)−1 and B U j(X,n)−1 . Recall formulas (5.1)-(5.2) and for x ∈ S(0, 1) let , assuming that B 0 = x. Let D b = S(0,1) D b x U 0 (dx) and let {M 1 n } n≥1 be a sequence of i.i.d. random vectors with distributions D b . Let {δ n } n≥1 be i.i.d. random variables with P(δ n = 1) = 1 − P(δ n = 0) = a. We assume that {δ n } n≥1 and {M 1 n } n≥1 are independent. We will define another process {M 2 n } n≥1 . Before doing so, we note that we can and will assume that {M 1 n } n≥1 , {M 2 n } n≥1 and {δ n } n≥1 are defined on the same probability space as (B, X, Y ). For every n ≥ 1, let M 2 n be a random vector with the conditional distribution S(0,1) D b x D X n (dx) given F U j(X,n)−1 . Let M n = δ n M 1 n + (1 − δ n )M 2 n for n ≥ 1. It is elementary to check that the sequences {M n } n≥1,j(X,n)−1<k 1 and {∆ n L X } n≥1,j(X,n)−1<k 1 have the same distributions.
Recall λ 1 (d, b) from (6.5). Let {N b t , t ≥ 0} be a Poisson process with the rate (expected number of jumps per unit of time) equal to λ 3 (d, b) := b 4 λ 1 (d, b). We assume that N b is independent of {δ n M 1 n } n≥1 . Let {N 1 t , t ≥ 0} be a continuous time pure jump process with values in R d , starting from 0, with jump times matching those of N b . For n ≥ 1, the n-th jump of N 1 is equal to δ n M 1 n /b 2 . Let Π be the covariance matrix equal to the unit diagonal matrix times 2/((d − 1)d). We will use the invariance principle in the form given in [JS03, Thm. IX 4.21] to show that the processes {N 1 t , t ≥ 0} converge weakly to Brownian motion with the covariance matrix Π as b → ∞. To apply [JS03, Thm. IX 4.21], one needs to check three conditions. Their condition (iii) is concerned with the initial distributions and it is clearly satisfied in our case-the initial distributions converge to the delta function at 0. Condition (i) is an assumption on the asymptotic form of the expectation and variance of the jumps. The jumps of N 1 are symmetric so the expected value of the jumps is zero. Note that the first coordinate (M 1 n ) 1 of M 1 n has the same distribution as L 1,Z T b in Lemma 6.2. Hence, by Lemma 6.2 (i)-(ii) and (7.4), the variance of the first component of the limit is equal to The covariance structure of the limit is represented by a constant multiple of the unit diagonal matrix because of the rotational symmetry of M 1 n . Finally, the Lindeberg-Feller-type condition (ii) in [JS03, Thm. IX 4.21] has been verified in Lemma 6.2 (iii). We conclude that processes {N 1 t , t ≥ 0} converge weakly to Brownian motion with the covariance matrix Π.
We define { M 1 n } n≥1 , { M 2 n } n≥1 , { δ n } n≥1 and { M n } n≥1 relative to Y in the same way as {M 1 n } n≥1 , {M 2 n } n≥1 , {δ n } n≥1 and {M n } n≥1 were defined relative to X. We can and will assume that { M 1 n } n≥1 and { δ n } n≥1 are independent of {M 1 n } n≥1 and {δ n } n≥1 . We assume that all processes and { M n } n≥1 are defined on the same probability space as (B, X, Y ).
Recall the process N b . We assume that N b is independent of { δ n M 1 n } n≥1 . Let { N 1 t , t ≥ 0} be constructed from { δ n M 1 n } n≥1 and N b in the same way as {N 1 t , t ≥ 0} was constructed from {δ n M 1 n } n≥1 and N b . Note that we use the same Poisson process N b in both cases.
Let {(W X t , W Y t ), t ≥ 0} be a pair of independent d-dimensional Brownian motions, each with variance 2/((d − 1)d). It follows from independence of { δ n M 1 n } n≥1 and t , t ≥ 0} be a continuous time pure jump process with values in R d , starting from 0, with jump times matching those of N b . For n ≥ 1, the n-th jump of N 2 is equal to (1 − δ n )M 2 n /b 2 . We obtain from (6.7) that for b > 2 and d ≥ 2, E |M 2 n | 2 ≤ d 2 (2 + 2c 1 λ 2 (d, b)) ≤ c 2 + c 3 log b. According to (7.4), P(1 − δ k = 0) ≤ c 4 b 1−d . It follows that, for large b, for n ≥ 1, j(X, n) − 1 < k 1 . The process N X,b has about λ 3 (d, b) jumps per unit of time. We bound the difference between the processes R t and b −2 L X (σ X b 4 t ) as follows. sup By (4.4), (4.5) and (6.5), the distribution of L X U j(X,n) − L X T j(X,n) is exponential with mean 1/λ 1 (d, b). Hence, For the same reason we have Note that the last two formulas still hold if we replace λ 3 (d, b) with any constant multiple of λ 3 (d, b). This and the weak convergence of We will now discuss a few technical points that were partly swept under the rug in the proof so far. First, we have just made a claim of convergence of some processes on the half-line although the construction of stopping times used in the proof stops at U k 1 . We assumed that dist(X 0 , Y 0 ) > b 2 . At time U k 1 , the processes X and Y are about 2b units apart. After rescaling by b −2 , this corresponds to the time when b −2 X and b −2 Y starting at a distance greater than 1 come closer than 2/b units apart. Since d-dimensional Brownian motion does not hit a fixed point for d ≥ 2, this time goes to infinity in probability as b → ∞. This justifies the assertion that convergence holds on the whole time half-line [0, ∞).
We can drop the assumption that dist(B 0 , X 0 ) ≥ b and dist(B 0 , Y 0 ) ≥ b as follows. The assumption is satisfied at the time U 1 so the invariance principle holds for the post-U 1 process. The amount of local time and the maximal displacement of the processes on the interval [0, U 1 ] can be easily estimated using the same methods that were used to estimate ∆ n L X . The estimates show that the initial part of the process, on the time interval [0, U 1 ], will disappear in the limit of rescaled processes.
We can now change the clocks from σ X t and σ Y t to the common clock σ t due to (8.2). We conclude that It is straightforward to check that this implies the theorem under the assumption that for each n, dist(X 0 , Y 0 ) > n −1/2 . We note that the same proof would apply if for some fixed c 8 > 0 and all n we assumed that dist(X 0 , Y 0 ) > c 8 n −1/2 .
We also note that our estimates are uniform in the sense that they do not depend on the initial positions of X, Y and B. We will make this claim more precise. Recall that the Prokhorov metric is a way to metrize weak convergence. For every T, ε, c 9 > 0 there exists n 1 such that for all n ≥ n 1 , all x 1 , y 1 and z 1 such that dist(x 1 , y 1 ) ≥ c 9 n −1/2 , dist(x 1 , z 1 ) ≥ 1 and dist(y 1 , z 1 ) ≥ 1, if X 0 = x 1 , Y 0 = y 1 and B 0 = z 1 then the Prokhorov distance between T ]} and standard (2d)-dimensional Brownian motion on [0, T ] is less than ε.
Step 2. We will show that for any p 0 < 1 and t 0 > 0 there exist n 0 and γ > 0 such that for any n ≥ n 0 and any starting point (B 0 , X 0 , Y 0 ) satisfying the usual conditions |B 0 − X 0 | ≥ 1, |B 0 − Y 0 | ≥ 1 and |X 0 − Y 0 | ≥ 2, the process |n −1/2 (X σnt − Y σnt )| will become greater than γ in at most t 0 units of time with probability greater then p 0 . Let The process A t is not Markovian but the process (A t , X σt −B σt ) is. We will write P x,y to denote the distribution of {(A t , X σt −B σt ), t ≥ 0} starting from (A 0 , X σ 0 −B σ 0 ) = (x, y). The last remark in Step 1 and standard Brownian estimates show that there exist p 1 , p 2 > 0 and k 2 such that for k ≥ k 2 and |y| ≥ 1, Note that we can take p 1 to be any number less than 1/2 for any d, so we will assume that p 1 = 3/8. By applying the Markov property at times j2 2k , j = 1, 2, . . . , and (8.4), we see that for some c 10 and k ≥ k 2 , We will show that there exists c 11 > 0 such that for all k ≥ 0, x ∈ B(0, 2 k ) and |y| ≥ 1, The proof will be based on induction. For all k, let where θ denotes the usual Markovian shift operator. Suppose that (8.6) holds for k 2 , k 2 + 1, . . . , k − 1 (the value of c 11 will be specified later). In particular, we assume that (8.6) holds for k − 1 and x ∈ S(0, 2 k−1 ). Then, by (8.5) and (8.6), E x,y T 3 k ≤ c 10 2 2k + c 11 2 2(k−1) .
The distribution of K is majorized by the geometric distribution with mean 1/p 1 , by  (8.3). This, the strong Markov property applied at T 4 j 's and (8.7) imply that E x,y T 1 k ≤ (c 10 2 2k + c 11 2 2(k−1) )/p 1 .
To complete the inductive step, we need to find c 11 such that the last expression is less than or equal to c 11 2 2k . In other words, we want to have (c 10 2 2k + c 11 2 2(k−1) )/p 1 ≤ c 11 2 2k . (8.9) The following inequality is equivalent, Since p 1 = 3/8, we can choose c 11 so large that (8.10) and, therefore, (8.9) hold. We combine this with (8.8) to conclude that E x,y T 1 k ≤ c 11 2 2k which concludes the inductive step.
To initialize the inductive proof, it suffices to show that (8.6) holds for k = k 2 (then (8.6) holds for all k ≤ k 2 with c 11 replaced by c 11 2 2k 2 ). We only sketch the proof. If |A 0 | ≤ 2 k 2 , it is easy to construct a deterministic smooth trajectory such that if we use it as the driving path in place of B t then |A t | will exceed 2 k 2 +1 in no more than 2 2k 2 units of time. By the support theorem (see [Bas95, Thm. I.6.6]) and the continuity of the Skorokhod map (see [LS84, Thm. 1.1]), with probability p 3 > 0 not depending on the starting point, if the driving process B t is Brownian motion then |A t | will exceed 2 k 2 in no more than 2 2k 2 +1 units of time. Applying the Markov property at times j2 2k 2 +1 , j ≥ 1, we conclude that the expected value of the time when |A t | exceeds 2 k 2 is bounded by 2 2k 2 +1 /p 3 . This implies (8.6) for k = k 2 (but we may have to enlarge c 11 ).
Recall that we fixed a p 0 < 1 at the beginning of the proof. It follows from (8.6) that for all k ≥ 0, x ∈ B(0, 2 k ) and |y| ≥ 1, P x,y (T 1 k ≥ c 11 2 2k /(1 − p 0 )) ≤ 1 − p 0 . By scaling, the claim made at the very beginning of Step 2 follows if we take γ = ((1 − p 0 )t 0 /c 11 ) 1/2 . It remains to combine the claims proved in Steps 1 and 2. According to Step 2, irrelevant of the starting position of X, Y and B, the process |n −1/2 (X σnt − Y σnt )| will reach a small fixed distance in a small fixed time. After that time, we use the invariance principle in the form proved at the end of Step 1.

Irreducibility
The argument presented in this section is a straightforward adaptation of the proof of [BBCH10, Thm. 6.1] so we will omit many details.
Let P z,x,y denote the distribution of (B t , X t , Y t ) starting from (z, x, y).
Proof. Suppose that we replace Brownian motion B with a continuous function {A t , t ≥ 0} in (2.1)-(2.4). These equations have solutions according to [LS84]. Let (A t , X A t , Y A t ) be the resulting triplet of processes. We proved in Section 2.1 that the processes are defined until the accumulation time of visits of A to the unit spheres centered at X A and Y A . We will consider only functions A such that there is no such finite accumulation time.
Fix some u 1 , x 1 , y 1 ∈ T d r and assume that for every pair of these points, the distance between them is greater than 5. Consider any It is elementary to see that one can find a continuous function {A t , t ≥ 0} and a time t 1 < ∞ not depending on u 1 , x 1 , y 1 , A 0 , X 0 , Y 0 (but possibly depending on d and r) such that there exists t 2 ≤ t 1 with the property that A t 2 ∈ B(u 1 , α), X A t 2 ∈ B(x 1 , α) and Y A t 2 ∈ B(y 1 , α). We briefly justify this claim. If the spheres X 0 and Y 0 touch or are very close to each other then the function A has to start by "pushing them apart." Then A has to push the spheres in the right direction, one at a time. By the continuity of the Skorokhod map (see [LS84, Thm. 1.1]), there exists ε 1 > 0 such that if a continuous function The support theorem (see [Bas95, Thm. I.6.6]) implies that for any and Y t 2 ∈ B(y 1 , 2α) with positive probability. It is easy to see that the last claim implies that P(B t 1 ∈ B(u 1 , 3α), X t 1 ∈ B(z 1 , 3α), Y t 1 ∈ B(y 1 , 3α)) > 0. (9.1) Let ∠(v, w) denote the angle between vectors v and w and recall that e k is the k-th vector in the usual orthonormal basis for R d . Let C j (δ 0 ) = {v ∈ R d : ∠(e j , v) ≤ δ 0 }. Fix δ 0 > 0 so small that for any v j ∈ C j (2δ 0 ), j = 1, . . . , d, the vectors {v j } are linearly independent. Let C j,X t = X t + (C j (δ 0 ) ∩ X t ); this set is a small spherical cap on X t , with center in the direction e j from X t .
Let F X be the event that all of the following conditions hold: (i) Brownian motion B visits the (random and time dependent) sets C j,X t , j = 1, 2, . . . , d, in this order, between times t 1 and 2t 1 ; (ii) B does not visit any other part of X t ∪ Y t during [t 1 , 2t 1 ]; (iii) the local time L X increases less than 1/(2d) when B is hitting C j,X We define C j,Y t and F Y in an analogous way except that B is required to visit C j,Y t 's during [3t 1 , 4t 1 ].
Let F B be the event that all of the following conditions are satisfied: (i) B hits B(u 1 , 1) between 4t 1 and 5t 1 ; (ii) B does not hit X ∪ Y between the last visit to C d,X t during [t 1 , 2t 2 ] and the first visit to C 1,Y t during [3t 2 , 4t 2 ]; (iii) B does not visit X t ∪ Y t between the last visit to C d,Y t during [3t 2 , 4t 2 ] and hitting of B(u 1 , 1). The probability of F X ∩ F Y ∩ F B is strictly positive due to the support theorem and excursion theory. Let and note that K j,X , K j,Y ∈ C j (δ 0 ) for all j = 1, . . . , d.
Proof of Theorem 3.3 (i). If there were more than one invariant measure, at least two of them (say, µ and ν) would be mutually singular by Birkhoff's ergodic theorem [Sin94]. However, we have shown in Lemma 9.1 that there exists a strictly positive measure ψ which is absolutely continuous with respect to any transition probability, so that in particular, ψ ≪ µ and ψ ≪ ν. Since µ ⊥ ν by assumption, there exists a set Γ such that µ(Γ) = 0 and ν(Γ c ) = 0. Therefore, one must have ψ(Γ) = ψ(Γ c ) = 0 which contradicts the fact that the measure ψ is non-zero.

Stationary measure
Proof of Theorem 3.3 (ii). For a measure µ and function f , let µ(f ) denote the integral of f with respect to µ. Fix a continuous non-negative function f : (T d 1 ) 2 → R and note that f is bounded, by compactness of (T d 1 ) 2 . Let W denote Brownian motion on (T d 1 ) 2 with the covariance matrix equal to the unit diagonal matrix times 2/((d − 1)d) and let E w be the corresponding expectation, assuming that W 0 = w. Standard coupling methods show that W converges to the stationary distribution uniformly in w, that is, for every ε > 0 there exists t 0 such that for all t ≥ t 0 and all w ∈ (T d 1 ) 2 , the Prokhorov distance between the distribution of W t and the uniform distribution on (T d 1 ) 2 is less than ε. Fix an arbitrarily small ε 1 > 0. By convergence of W to the stationary distribution and the ergodic theorem, there exists t 1 so large that for any w ∈ (T d 1 ) 2 , Fix t 1 satisfying the above estimate. Let E z,x,y denote the expectation corresponding to the distribution of (B t , X t , Y t ) defined on (T d r ) 3 starting from (z, x, y). Recall from the last paragraph of Step 1 of the proof of Theorem 3.2 that the convergence of {n −1/2 (X r σnt − X r 0 , Y r σnt − Y r 0 ), t ∈ [0, t 1 ]} to {W t , t ∈ [0, t 1 ]} is uniform in the starting points of X, Y and B. It follows that there exists r 1 such that for all r ≥ r 1 , z = B 0 ∈ T d 1 , and (x, y) = w ∈ (T d 1 ) 2 such that |rx − rz| ≥ 1, |rz − ry| ≥ 1 and |rx − ry| ≥ 2, we have E rz,rx,ry 1 t 1 t 1 0 f ((X σ r 2 s , Y σ r 2 s )/r)ds − E w 1 t 1 t 1 0 f (W s )ds < ε 1 /2.