Coupling for Drifted Brownian Motion on an Interval with Redistribution from the Boundary *

We answer a question by Kolb and Wubker [7] on the threshold drift for Brownian Motion on an interval with redistribution from the boundary. We do that by constructing an efficient coupling.


Introduction
Consider an elliptic diffusion on a bounded domain D ⊂ R d , with infinitesimal generator L, that upon hitting the boundary ∂D is redistributed in D according to some prescribed probability measure (that could depend on the point of exit), restarting afresh, and repeated indefinitely.The redistribution is independent of the past.Under some standard smoothness assumptions on the boundary (and the redistribution measure as a function of the boundary point), the process is ergodic and converges exponentially fast to its invariant distribution in total variation.We call the resulting process diffusion with redistribution, or DR in short.This process has been studied by several authors: [4] [5] (analysis for BM with fixed deterministic redistribution), [3] [2](ergodicity, characterization, comparison), [8] (analysis of spectrum), [9] (holding times at boundary), and most recently [6] [7] (coupling approach).The DR is never reversible and the problems of characterizing and estimating the exponential rate of convergence are typically non-trivial.Unsurprisingly, the exponential rate is equal to the "spectral-gap" for −L with the nonlocal boundary conditions imposed by the redistribution, that is, the minimal real part among all non-zero eigenvalues, yet this result is far from trivial to prove.As for estimation and comparison with other quantities, the main obstacle is that the corresponding eigenvalue may not be real.Yet, if it is real, then the spectral gap is bounded below by the principal eigenvalue for −L with the Dirichlet boundary condition.This realness condition was shown to hold under certain conditions and in some concrete examples.This has led to the question whether the principal Dirichlet eigenvalue is a lower bound when the realness condition is relaxed, at least when the underlying diffusion process is reversible.The question was formulated by Pinsky and the author of this note in [2].
µ above which the spectral gap is constant, and the authors conjectured it is equal to √ 3 2πσ 2 .
Let X = (X t : t ≥ 0) be the DR process on the interval (0, ), with underlying diffusion generated by L σ,µ and redistribution from the boundary {0, } to 2 .We refer the reader to [3] for the construction of the process.We denote the corresponding probability and expectation with initial distribution ρ by P ρ and E ρ , and if ρ = δ x for some x ∈ (0, ), then we abbreviate and write P x and E x , respectively.Let P ρ (t) denote the distribution of X t under P ρ , and for distributions µ 1 , µ 2 , let d t (µ 1 , µ 2 ) = P µ1 (t) − P µ2 (t) T V , where • T V is the total variation norm, We write x for δ x when it appears as a parameter of d t (•, •), and define x,y d t (x, y). and We define the spectral gap Similarly, let and It is easy to see that the three parameters σ, µ, can be reduced to one parameter by scaling.Indeed, it follows directly from the definition of Σ σ,µ, that for 1 > 0, (1.1) The analogous identity holds for Σ D • as well.In fact, we will prove our results for σ = 1, = 2π, and derive the general statements from these scaling identities.
The eigenvalue is called the principal eigenvalue.
Here is a summary of the relevant results.
a. X has a unique stationary distribution ν, and and there exists an efficient coupling.
We briefly recall the notion of coupling and efficient coupling.In this paper we will only consider Markovian couplings.A Markovian coupling for X with initial distributions µ 1 and µ 2 is a Markov process ((X 1 t , X 2 t ) : t ≥ 0) on (0, ) × (0, ), such that the marginals (X 1 t : t ≥ 0) and (X 2 t : t ≥ 0) are copies of X, the distribution of X 1 0 is µ 1 , and the distribution of X 2 0 is µ 2 .Given a coupling, the coupling time (or meeting time) τ C is defined through As it is well known and easy to verify, the tail of the coupling time dominates d t (µ 1 , µ 2 ).
That is, As a result, couplings provide lower bounds on γ 1 (σ, µ, ).Coupling for X with initial distributions µ 1 , µ 2 is efficient if the coupling time decays at an exponential rate equal to γ 1 (σ, µ, ): The first part of the theorem (for a general domain, diffusion and redistribution) was proved in [3], and the second part was proved in [7].We refer the reader to [7] for the explicit formula for the ν.The proof of part b. of the theorem has led Kolb and Wubker to conjecture that We are ready to state our main result.
and there exists an efficient coupling.
As it turns out, the first statement follows from a straightforward eigenvalue calculation, see Proposition 1.4 below.But this does not provide any insight or explanation.The more substantial result is the existence of efficient coupling.This coupling both proves (1.4) and explains the origin of each of the terms.In addition, the proof gives upper and lower bounds on d t (x, y) for any x, y ∈ (0, ) in terms of tails of exit times of BM or drifted BM from an interval.
The coupling used to prove the theorem is fairly simple.When the two copies of the DR are exactly /2 units apart, they have the same increments.This guarantees that they meet when the copy in (0, /2) exits this interval.When the distance between the two copies is different than /2, then the (non-drifted) Brownian components of the increments are of the same magnitude but with opposite signs.The main idea is then to exploit the symmetry of the model to show that with this coupling, the first time when either the copies meet or are /2 apart (note that both events can occur after a large number of redistributions), coincides with the exit time for BM from an interval of length /4.
Suppose that under P x , the process Y = (Y t : t ≥ 0) is BM with diffusivity σ and drift µ, with Y 0 = x, and let τ = inf{t ≥ 0 : Since for all θ, E x e θτ = 1 + θ ∞ 0 e θt P x (τ > t)dt, it follows from (1.5) that From (1.2) we obtain Combining this with Theorem 1.2 gives: otherwise.
The analytic proof to (1.3) is an immediate consequence to the following standard calculation.
2 Proof of Theorem 1.2 We will reduce the problem from three parameters, σ, µ, to one parameter by scaling, using the identity (1.1) and its analog for Σ D σ,µ, .It follows that Drifted BM with redistribution Therefore if we prove the theorem for σ = 1 and = 2π, then for the general case we obtain: In light of the above, in the remainder of this section we will assume the diffusivity σ = 1, and = 2π.We will use b for the drift coefficient, and without loss of generality, assume b ≥ 0. We let P b x denote the probability measure under which the process Y = (Y t : t ≥ 0) is BM on R starting from x with diffusivity 1 and drift b.Recall that τ = inf{t ≥ 0 : |Y t | ≥ 2 }.We also denote by B = (B t : t ≥ 0) standard BM on R.

Lower bound on d t
The lower bound is very simple, but it suggests the couplings to be used for the upper bound.
Thus from Theorem 1.1, Lemma 2.1 and (1.5), we obtain the following: ). Proof.Define the coupled processes X 1 and X 2 on (0, 2π) by letting Let X be a DR starting at π. Let τ = inf{t : X 1 t ∈ {0, π}}.Then τ has the same distribution as τ π (defined above (1.5)),under P b 0 .Observe that at time τ , X 2 τ = π or 2π according to whether X 1 τ = 0 or X 2 τ = π.We continue the coupling for t ≥ τ by redefining X 1 and X 2 through: Thus X 1 and X 2 are copies of the DR process, meeting at time τ .Let f = 1 [0,π) .This gives To prove the second bound, observe that if f : R → R is piecewise continuous and πperiodic, then under P x , f (X t ) has the same distribution as f (x + B t + bt).Furthermore, for a fixed t, let g(u) = f (u + bt).Note that this transformation is one-to-one and onto from the set of piecewise continuous π-periodic functions to itself, and that under P x , the distribution of f (X t ) coincides with the distribution of g(x + B t ).Thus, if in addition 2 , y = π, and let g be the π-periodic function equal to Then τ has the same distribution as τ π/2 under P 0 0 .Observe that for t < τ , g(x + B t ) = 1, and g(y − B t ) = 0. We define the coupling ECP 19 (2014), paper 16.
Observe that either B 1 τ = B 2 τ = 3π 4 , or B 1 τ = π 4 and B 2 τ = π + π 4 .Thus, according to the construction of the coupling and the choice of g, it follows that g(B 1 t ) = g(B 2 t ) for all t ≥ τ .In particular,

An upper bound on d t
We will use the following well-known lemma, which is an immediate corollary to the eigenvalue expansion for the transition function of drifted BM (e.g.[1, p. 94, Theorem 5.9]).The lemma could be avoided, but helps us obtain tighter upper bounds on d t . , If (X, Y ) is a coupling for the DR starting from x and y respectively, then we denote the joint distribution by P x,y and the corresponding expectation by E x,y .Let τ C denote the coupling time, In particular, d t (x, y) ≤ P x,y (τ C > t).
From the triangle inequality, d t (x, y) ≤ d t (x, π) + d t (π, y) ≤ 2 sup x d t (x, π).Thus, (2.1) We will now obtain an upper bound on P θ,π (τ C > t) through coupling of two copies of the DR.As will be shown below, we only need to consider θ ∈ (π, 2π).The coupling is constructed and analyzed in the proof of Lemma 2.4 below.This coupling consists of two stages.
Stage 1.The two copies have Brownian increments which are of the same magnitude and opposite signs.We begin this stage with one copy at π and another copy at θ ∈ (π, 2π).We stop this stage at time T 1 , which is when either: (a) The two copies meet, and then τ C = T 1 ; or (b) The distance between the copies is π.In this case we move to the second stage.
The first stage continues as long as none of the above conditions is met.While in the first stage, the first redistribution event, if occurs, can only occur when the copy starting in (π, 2π) hits 2π − (this will be explained below).Since the initial distance is strictly less than π, and condition (b) was not met, at this time the second copy is in (π, 2π).As a result, at the redistribution, we again have one copy (the redistributed one) at π and another copy in (π, 2π).By induction, this holds for all redistributions during the first stage.
Thus, X 1 and X 2 are coupled by time τ , and τ is the exit time of BM with drift b, starting from θ from the interval (0, π).Note that a priori λ may not be real.This is crucial.The solution is of the form v(x) = e xA v(0).If A is diagonalizable, with eigenvalues λ 1 , λ 2 , then this implies u(x) = Ae λ1x + Be λ2x .Otherwise, u(x) = e λ1x (A + Bx).