A COUNTEREXAMPLE FOR THE OPTIMALITY OF KENDALL-CRANSTON COUPLING

We construct a Riemannian manifold where the Kendall-Cranston coupling of two Brownian particles does not maximize the coupling probability.


Introduction
Given two stochastic processes X t and Y t on a state space M , a coupling Z t = (Z (1) t , Z (2) t ) is a process on M × M so that Z (1) or Z (2) has the same distribution as X or Y respectively. Of particular interest in many applications is the distribution of the coupling time T (Z) := inf{t > 0 ; Z for all s > t}. The goal is to make the coupling probability P[T (Z) ≤ t] as large as possible by taking a suitable coupling. When X and Y are Brownian motions on a Riemannian manifold, Kendall [3] and Cranston [1] constructed a coupling by using the Riemannian geometry of the underlying space. Roughly speaking, under their coupling, infinitesimal motion ∆Y t ∈ T Yt M at time t is given as a sort of reflection of ∆X t via the minimal geodesic joining X t and Y t . Their coupling has the advantage of controlling the coupling probability by using geometric quantities such as the Ricci curvature. As a result, Kendall-Cranston coupling produces various estimates for heat kernels, harmonic maps, eigenvalues etc. under natural geometric assumptions. On the other hand, there is the question of optimality. We say that a coupling Z of X and Y is optimal holds for any other couplingZ. Though Kendall-Cranston coupling has a good feature as mentioned, in general there is no reason why it should be optimal.
The Kendall-Cranston coupling is optimal if the underlying space has a good symmetry. For example, in the case M = R d , the Kendall-Cranston coupling (Z (1) , Z (2) ) is nothing but the mirror coupling. It means that Z t ) up to the time they meet, where Ψ is a reflection with respect to a hyperplane in R d so that Ψ(X 0 ) = Y 0 . It is well known that the mirror coupling is optimal. Indeed, it is the only coupling which is optimal and Markovian [2]. More generally, the same result holds if there is a sort of reflection structure like a map Ψ on R d (see [4]).
In this paper, we show that the Kendall-Cranston coupling is not optimal in general.
Theorem 1.1 For each t > 0, there is a complete Riemannian manifold M where the Kendall-Cranston coupling of two Brownian motions X · and Y · with specified starting points is not optimal.
The proof of Theorem 1.1 is reduced to the case t = 1 by taking a scaling of Riemannian metric. We construct a manifold M in the next section and prove Theorem 1.1 in section 3.
Notation: Given a Riemannian manifold N we denote by B N r (x) or simply B r (x) the open ball in N of radius r centered at x. Given a Brownian motion (X t ) t≥0 on N we denote by τ A = inf {t > 0 : X t ∈ A} the hitting time of a set A ⊂ N . We remark that, throughout this article, τ A always stands for the hitting time for the process (X t ) t≥0 even when we consider a coupled motion (X t , Y t ) t≥0 .

Construction of the manifold
We take three parameter R > 0, ζ > 0 and δ > 0 such that ζ < R/4 and δ < ζ/3. Let C = R × S 1 be a cylinder with a flat metric such that the length of a circle S 1 equals ζ. For simplicity of notation, we write z = (r, θ) for z ∈ C where r ∈ R and θ ∈ (−ζ/2, ζ/2] such that the Riemannian metric is written as dr 2 + dθ 2 . If appropriate, any θ ∈ R will be regarded mod ζ and considered as element of (−ζ/2, ζ/2]. We put and write ∂ 1,0 := ∂B C δ ((0, ζ/2)) as well as ∂ 1,2 := {−R} × S 1 (see Fig.1). Let C ′ be a copy of C. Then we put analogously We endow M with a C ∞ -metric g such that (M, g) becomes a complete Riemannian manifold and: (i) g| M1 coincides with the metric on M 1 inherited from C, where d is the distance function on M .

Comparison of coupling probabilities
Let M be the manifold constructed above (with suitably chosen parameters R, ζ and δ) and fix two points x = (0, ζ/6) ∈ M 1 and y = (0, ζ/3) ∈ M 2 . In this paper, the construction of Kendall-Cranston coupling is due to von Renesse [5]. We will try to explain his idea briefly. His approach is based on the approximation by coupled geodesic random walks {Ξ k } k∈N starting in (x, y) whose sample paths are piecewise geodesic. Given their positions after (n − 1)-th step, one determines its next direction ξ n according to the uniform distribution on a small sphere in the tangent space and the other does it as the reflection of ξ n along a minimal geodesic joining their present positons. We obtain a Kendall-Cranston coupling (X t , Y t ) by taking the (subsequential) limit in distribution of them. We will construct another Brownian motion (Ŷ t ) t≥0 on M starting in y, again defined on the same probability space as we construct (X t , Y t ) such that P (X and Y meet before time 1) < P X andŶ meet before time 1 .
In other words, if Q denotes the distribution of (X, Y ) andQ denotes the distribution of (X,Ŷ ) then Our construction of the processŶ will be as follows. We define a map Φ : M 1 → M 2 by Φ((r, θ)) = (−r, ζ/2 − θ) and then put Note that τ ∂1,2 = T holds when τ ∂1,2 ≤ τ ∂1,0 underQ. Set H = S 1 × {0} ⊂ M 0 ⊂ M . For z 1 , z 2 ∈ M and A ⊂ M , minimal length of paths joining z 1 and z 2 which intersect A is denoted by d(z 1 , z 2 ; A). We define a constant L 0 by Proof. First we show L 0 < R. Let z 1 = (R, 0) ∈ M 1 and z 2 = (R, ζ/2) ∈ M 2 . Obviously there is a path of length 2R joining z 1 and z 2 across ∂ 1,2 . Thus we have d(z 1 , z 2 ; ∂ 1,2 ) ≤ 2R. By symmetry of M , where the second equality follows from the third and fourth properties of g and the last inequality follows from the choice of δ. These estimates imply L 0 < R. Next, let z ′ 1 = (R − ζ, θ) ∈ M 1 and z ′ 2 = (R − ζ, ζ/2 − θ) ∈ M 2 . In the same way as observed above, we have Note that the length of a path joining z ′ 1 and z ′ 2 which intersects both of ∂ 1,2 and H is obviously greater than d(z ′ 1 , z ′ 2 ; H). Thus, in estimating d(z ′ 1 , z ′ 2 ; ∂ 1,2 ), it is sufficient to consider all paths joining z ′ 1 and z ′ 2 across ∂ 1,2 which do not intersect H. Such a path must intersect both {δ} × S 1 ⊂ M 1 and {−δ} × S 1 ⊂ M 1 (see Fig.3). Thus we have Hence, the conclusion follows. Fig.4). Let Ψ : M ′ → M ′ be the reflection with respect to H. For instance, for z = (r, θ) ∈ M ′ 1 , Ψ(z) = (r, ζ/2−θ) ∈ M ′ 2 . Note that Ψ is an isometry, Ψ • Ψ = id and {z ∈ M ′ ; Ψ(z) = z} = H. Let X ′ be the given Brownian motion starting in x and now stopped at ∂M ′ , i.e. X ′ t = X t∧τ ∂M ′ . Define a stopped Brownian motion starting in y by Y ′ t = Ψ(X ′ t ) for t < τ H and by Y t = X t for t ≥ τ H (that is, the two Brownian particles coalesce after τ H ). Then we can prove the following lemma.
Proof. Note that the minimal geodesic in M joining z and Ψ(z) must intersect H for every The remark after the definition ofQ implieŝ By lifting X t to R 2 , the universal cover of C, Here P R 2 and P R denote the usual Wiener measure for Brownian motion (starting at the origin) on R 2 or R, resp. For simplicity, we write τ R instead of τ {R} .
Next we give an upper estimate of Q [T ≤ 1]. Let E := τ ∂1,0 < 1 ∧ τ ∂M ′ . Then In order to collide two Brownian motions starting at X τ ∂M ′ and Ψ(X τ ∂M ′ ), either of them must escape from the flat cylinder of length 2(L 0 − δ) where its starting point has distance L 0 − δ from the boundary. This observation together with the strong Markov property yields By Lemma 3.2 and the definition of ζ and δ, we have L 0 − δ ≥ R − ζ − δ > 2R/3. Thus By lifting X t to R 2 , we have Here the last inequality follows from Lemma 3.2. Consequently, we obtain Now take R > 3 √ 2 log 2. After that we choose ζ so small that P R [τ R−ζ < 1] ≈ P R [τ R < 1]. Finally we choose δ so small that P R 2 τ A(δ) < 1 ≈ 0. Then Proposition 3.1 follows from (3.1) and (3.2).