Middle edge perturbation for birth and death chains with symmetry

Remark 4.17. In case 2, i.e., when p = q, the result stated above can be improved by using the Nash inequality techniques developed in [25]. This leads to bounds on the merging times of order N 2 instead N 2 log N . Using singular values, we can get a hint that this is the case by considering the average ℓ 2 distance and the first inequality in Theorem 3.2. Indeed, in the case p = q, comparison with the simple random walk on the stick yields control not only of the top singular value, but of most singular values. Namely, if K i ∞ 1 is a sequence in Qp, p, r, ε, p, r ∈ [ε, 1 − 2ε] and µ = ν ≡ 1N + 1, µ n = ν K 0,n , then the j-th singular value σ j K i , µ i−1 is bounded by σ j K i , µ i−1 ≤ 1 − Cε −1 j N 2 . Using this in the first inequality stated in Theorem 3.2 together with stability yields 1 N + 1 2 X x, y K 0,n x, y µ n y − 1 2 ≤ η if n ≥ D εN 2 1 + log + 1η. 5 Stability of some inhomogeneous birth and death chains Recall that a necessary condition for a family Q of irreducible aperiodic Markov kernels to be c-stable is that, for any pair of kernels in Q, the associated stationary measures π, π ′ satisfy c −2 ≤ ππ ′ ≤ c 2 . In less precise terms, all the stationary measures must have a similar behavior which we refer to as the stationary measure behavior of the family. The goal of this section is to provide examples of c-stable families of Markov chains that allow for a great variety of stationary measure behaviors. Because we lack techniques to study c-stability, providing such examples is both important and not immediate. The examples presented in this section are sets of “perturbations” of birth and death chains having a center of symmetry. Except for this symmetry, the stationary measure of the birth and death chain that serves as the basis for the family is arbitrary. Hence, this produces examples with a wide variety of behaviors.

5.1 Middle edge perturbation for birth and death chains with symmetry

For N ≥ 1, a general birth a death chain on [−N , N ] is described by Qx, y =    p x if y = x + 1 q x if y = x − 1 r x if y = x and y 6= −N , N otherwise with p N = q −N = 0 and 1 = p x + q x + r x for all x ∈ [−N , N ]. We consider the case when for all x ∈ [−N , N ] p x = q −x and r x = r −x . 1 This immediately implies that q = p . The kernel Q has reversible stationary distribution, νx = c x Y y=−N +1 p y−1 q y 1480 where c is the appropriate normalizing constant. Moreover, one checks that ν−x = νx, x ∈ [−N , N ]. For ε ∈ R, let ∆ ε be the non-Markovian kernel defined by ∆ ε x, y =    ε if x, y = 0, 1 −ε if x, y = 0, −1 otherwise 2 For ε ∈ −q , q , the perturbation Q ε = Q + ∆ ε of Q is a Markov kernel and has stationary distribu- tion ν ε x =      c ε Q z y=−N +1 p y−1 q y q −ε q −ε if z c ε Q y=−N +1 p y−1 q y q q −ε if z = 0 c ε Q z y=−N +1 p y−1 q y q +ε q −ε if z 3 where c ε is a normalizing constant. Using the facts the P ν ε x = 1 and 1 it follows that q q − ε c ε = c. This implies that ν ε x =    νz − ενz q if z ν0 if z = 0 νz + ενz q if z 0. Definition 5.1. Fix ε ∈ [0, q and set Q N Q, ε = {Q δ : δ ∈ [−ε, ε]}. Definition 5.2. Fix ε ∈ [0, q . Let S N ν, ε be the set of probability measures µ on [−N , N ] satisfying the following two properties: 1 for x ∈ [−N , N ], there exist constants a µ,x such that a µ,x = a µ,−x and µx =    νx + a µ,x if x νx − a µ,x if x ν0 if x = 0; 2 for all x ∈ [0, N ], ν ε x ≥ µx ≥ ν −ε x and ν ε −x ≤ µ−x ≤ ν −ε −x. where ν ε and ν −ε are defined in 3. Theorem 5.3. Fix ε ∈ [0, q . Let µ ∈ S N ν, ε. Then for any δ ∈ [−ε, ε] we have µQ δ ⊆ S N ν, ε. 1481 Proof. Let µ ∈ S N ν, ε. 1 Set a = 0, for any x ∈ [1, N − 1] we have that µQx = p x−1 µx − 1 + r x µx + q x+1 µx + 1 = νx + p x−1 a x−1 + r x a x + q x+1 a x+1 and µQ−x = p −x−1 µ−x − 1 + r −x µ−x + q −x+1 µ−x + 1 = ν−x − p −x−1 a −x−1 − r −x a −x − q −x+1 a −x+1 = ν−x − p x−1 a x−1 + r x a x + q x+1 a x+1 . At the end-points −N and N we get that µQN = µN 1 − q N + µN − 1p N −1 = νN + a N 1 − q N + a N −1 p N −1 and µQ−N = µ−N 1 − p −N + µ−N + 1q −N +1 = ν−N − a −N 1 − p −N − a −N +1 q −N +1 = νN − a N 1 − q N + a N −1 p N −1 . Using similar arguments, it is easy to verify that µQ0 = ν0. To check that µQ δ satisfies 1 of Definition 5.2 we note that µ∆ δ x =    δµ0 if x = 1 −δµ0 if x = −1 otherwise. The desired result now follows from the fact that for any x ∈ [−N , N ] µQ δ x = µQx + µ∆ δ x. 2 For any δ ∈ [−ε, ε] we have that Q δ = Q ε − ∆ ε + ∆ δ = Q ε − ∆ ε−δ . It follows that ν ε Q δ = ν ε − ν ε ∆ ε−δ . We get that ν ε Q δ x =    ν ε 1 − ε − δν ε if x = 1 ν ε −1 + ε − δν ε 0 if x = −1 ν ε x otherwise. 1482 and ν −ε Q δ x =    ν −ε 1 + ε + δν −ε if x = 1 ν −ε −1 − ε + δν −ε 0 if x = −1 ν −ε x otherwise. Since δ ∈ [−ε, ε], we have that for x ∈ [0, N ] ν ε Q δ x ≤ ν ε x and ν −ε Q δ x ≥ ν −ε x 4 and also that, ν ε Q δ −x ≥ ν ε −x and ν −ε Q δ −x ≤ ν −ε −x. 5 By property 2 of µ being in S N ν, ε we get that for any x ∈ [1, N ] ν ε Q δ x ≥ µQ δ x ≥ ν −ε Q δ x and ν ε Q δ −x ≤ µQ δ −x ≤ ν −ε Q δ −x. Equations 5 and 4 imply that for x ∈ [1, N ] ν ε x ≥ µQ δ x ≥ ν −ε x and ν ε −x ≤ µQ δ −x ≤ ν −ε −x. The inequalities above, along with the first part of the proof give the desired result. Theorem 5.4. Fix ε ∈ [0, q . The set Q N Q, ε is q +ε q −ε -stable with respect to any measure µ ∈ S N ε, ν. Proof. Let µ ∈ S N ν, ε, and set µ n = µ K 0,n . By Theorem 5.3 it follows that for any n ≥ 0 and any x ∈ [0, N ] ν ε x ≥ µ n x ≥ ν −ε x and ν ε −x ≤ µ n −x ≤ ν −ε −x. So we have that for any x ∈ [−N , N ] min z ν ε z νz ≤ µ n x νx ≤ max z ν ε z νz . Recall that c ε q q − ε = c. It follows that max z ν ε z νz = c ε c q + ε q − ε = q + ε q , min z ν ε z νz = c ε c = q − ε q . We get q − ε q ≤ µ n x νx ≤ q + ε q which implies, q − ε q + ε ≤ µ n x µ x ≤ q + ε q − ε . 1483

5.2 Example: the binomial chain