3.1. Distance to stationary distribution

The following theorem gives an upper bound for the distance to stationary distributions in terms of the eigenvalues given in the previous section.

As mentioned in the introduction, this upper bound is not obtained by eigenvalue techniques, symmetric matrices, etc. This is due to a coupling argument.

We provide two examples where we identify the eigenvalues of the transition matrix and apply the theorem above.

Example 1. (The complete graph, $K_n$ .) $K_n$ is the graph with n vertices where every vertex is connected to every other vertex. Let $X_k$ be a flat in L consisting of k vertices. Then $\mu(X_k, V(G)) = (\!-\!1)^{n-k}$ . By Theorem 2, the eigenvalue corresponding to flat $X_k$ is ${\binom{k}{2}}/{\binom{n}{2}}$ and the number of flats of size k is $\binom{n}{k}$ . Hence,

\begin{align*} d_\mathrm{TV}(t) \leq -\sum_{X \in L^*} \mu(X,V(G)) \lambda_X^t = & -\sum_{k=2}^{n-1}\sum_{X_k} \mu(X_k,V(G)) \left({\binom{k}{2}}\big/{\binom{n}{2}}\right)^t \\[5pt] & = -\sum_{k=2}^{n-1} (\!-\!1)^{n-k} \binom{n}{k} \left({\binom{k}{2}}\big/{\binom{n}{2}}\right)^t \\[5pt] & \leq n \bigg(1-\frac{2}{n} \bigg)^t + \sum_{i=2}^{n-2} (\!-\!1)^{i-1} \binom{n}{i} \bigg(1-\frac{i}{n} \bigg)^{2t} + o(n) \\[5pt] & = \mathcal{O}\bigg( n \bigg(1-\frac{2}{n} \bigg)^t\bigg), \end{align*}

provided that $t \geq \frac{1}{2}n \log n$ .

Example 2. (The complete bipartite graph, $K_{m,n}$ .) $K_{m,n}$ is a graph whose vertex set is partitioned into two sets of sizes m and n, where every vertex of one set is connected to every vertex of the other set. Take a flat $X_{k,l} \in L$ , where k and l denote the numbers of vertices from the set of size m and the set of size n respectively. The reader can verify using Theorem 2 that $\lambda_{X_{k,l}}={kl}/{mn}$ with multiplicity $\binom{m}{k} \binom{n}{l}$ . We can also show that $\mu(X_{k,l}, V(G)) = (\!-\!1)^{m+n-k-l}$ . So, by Theorem 3,

\begin{align*} d_\mathrm{TV}(t) \leq -\sum_{X \in L^*} \mu(X,V(G)) \lambda_X^t & = n \bigg(\frac{(n-1)m}{nm}\bigg)^t + m \bigg(\frac{n (m-1)}{nm}\bigg)^t \\[5pt] & \quad -\sum_{k=1}^{m-1}\sum_{l=1}^{n-1} (\!-\!1)^{m+n-k-l}\binom{m}{k} \binom{n}{l} \bigg(\frac{kl}{mn} \bigg)^t \\[5pt] & = n \bigg(1- \frac{1}{n} \bigg)^t + m \bigg(1- \frac{1}{m} \bigg)^t \\[5pt] & \quad + \sum_{j=1}^{m-1} (\!-\!1)^j \binom{m}{j} \bigg(1- \frac{j}{m} \bigg)^t \, \sum_{i=1}^{n-1} (\!-\!1)^{i} \binom{n}{i} \bigg(1- \frac{i}{n} \bigg)^t. \end{align*}

Suppose $n\geq m$ . Then, for $t \geq n \log n$ , we have $d_\mathrm{TV}(t) = \mathcal{O}(n(1 - ({1}/{n}))^t)$ .

A computationally tractable bound was obtained in [Reference Brown and Diaconis5] by a coarser version of the coupling argument in the proof of Theorem 3 in [Reference Bidigare, Hanlon and Rockmore3]:

(5) \begin{equation} d_\mathrm{TV}(t) \leq - \sum_{X \in L^*} \mu(X,V(G)) \lambda_X^t \leq \sum_{ \{ M:\, M \text{ co-maximal in } L \}} \lambda_{M}^t,\end{equation}

where M is a co-maximal flat in L if $M \prec X$ implies X is maximal in L. In edge flipping, co-maximal flats are simply obtained by removing a vertex from the flat of all vertices, since any flat is a subset of vertices.

The precision of the alternating bound compared to (5) is discussed in [Reference Chung and Hughes8] considering various examples of random hyperplane arrangements. In the examples that we study, the latter bound is as good as the former.

Now we can provide an upper bound for the rate of convergence of edge flipping in general connected graphs.

Theorem 4. Consider edge flipping on a connected graph G with n vertices and m edges, and let $\delta$ be the degree of the vertex with the minimum degree in G. For $c>0$ ,

\begin{equation*} c\frac{m}{\delta} \leq t_{\mathrm{mix}}\left(q\mathrm{e}^{-2c}\right) \leq \frac{m}{\delta}( \log n + 2c - \log q). \end{equation*}

Proof. Observe that the co-maximal flats obtained by removing any vertex with the minimum number of edges have the smallest eigenvalue among all co-maximal flats. Let $M^*$ denote this flat. Then the eigenvalue indexed by $M^*$ is

\begin{equation*} \lambda_{M^*} = \sum_{\mathrm{supp}\, F \subseteq M^{*}} w(F) = \frac{m- \delta}{m}=1-\frac{\delta}{m} \end{equation*}

by Theorem 2. By (5), we have

\begin{align*} d_\mathrm{TV}(t) \leq \sum_{ \{ M:\, M \text{ co-maximal in } L \}} \lambda_{M}^t \leq n \, \lambda_{M^{*}}^t \leq n \bigg( 1- \frac{\delta}{m}\bigg)^t \leq \mathrm{e}^{-tm/ \delta} \leq q\mathrm{e}^{-2c} \end{align*}

if $t \geq ({m}/{\delta})(\log n + 2c -\log q)$ .

For the lower bound, let us label a vertex with the minimal degree by x and take its color to be blue in the initial configuration without loss of generality. Let R be the set of all configurations where x is colored red. We have $\pi(R)=q$ because whenever an edge with an endpoint x is chosen, it is colored blue with probability p and red with probability q. Now let us look at $P^t(R)$ . The probability that the edge connecting x to the rest of the graph is not chosen before step t is

\begin{align*} \bigg(1-\frac{\delta}{m} \bigg)^t \geq \exp\bigg\{{-}t\bigg(\frac{m}{\delta} + \frac{m^2}{\delta^2}\bigg)\bigg\}. \end{align*}

If we take $t\leq cm/ \delta$ , then we have $P^t(R) \leq q(1-\mathrm{e}^{-2c})$ . Finally, using the second inequality in (4), we have $\|P^t-\pi\|_\mathrm{TV} \geq |P^t(R)-\pi(R)| \geq q\mathrm{e}^{-2c}$ .

We note that this bound is not optimal in either direction, as shown in the following section.

3.2. Rate of convergence for regular graphs

A graph G is called regular if every vertex of G has the same degree. It is called k-regular if the degree of the vertices is k. We show the following bounds on the mixing time of edge flipping for regular graphs, which are independent of the degree of the vertices.

The smallest-degree regular graph on n vertices is the cycle $C_n$ , and the largest-degree regular graph on n vertices is $K_n$ . We remark that the stationary distributions of edge flipping for these graphs are studied in [Reference Butler, Chung, Cummings and Graham6, Reference Chung and Graham7].

The derivation of these bounds shows that, as long as the number of vertices of the same degree is of order n, the principal terms in the bounds above remain the same.

Proof of the upper bound in Theorem 5. Suppose the degree of the vertices of G is equal to k. Let M denote a co-maximal flat. Then the corresponding eigenvalue is

\begin{equation*} \lambda_M = \sum_{\mathrm{supp}\, F \subseteq M} w(F) = \frac{({kn}/{2})-k}{{kn}/{2}} = 1-\frac{2}{n} \end{equation*}

by Theorem 2. The bound (5) gives

\begin{align*} \|P^t_C - \pi\|_\mathrm{TV} \leq n \bigg(1- \frac{2}{n} \bigg)^t \leq \mathrm{e}^{-c} \end{align*}

if we take $t \geq n\log n/2 + cn/2$ .

For the proof of the lower bound in Theorem 5 we use the Wilson’s method: [Reference Wilson18] showed that an eigenvector is a good candidate for a test statistic as the variance of the associated eigenfunction can be estimated inductively from the transition probabilities of the Markov chain. Assuming the second-order estimate for the eigenfunction, a lower bound can be obtained as follows.

Lemma 1. ([Reference Levin, Peres and Wilmer11, Theorem 13.5].) Let $X_t$ be an irreducible, aperiodic, time-homogenous Markov chain with state space $\Omega$ . Let $\Phi$ be an eigenfunction associated with eigenvalue $\lambda > \frac{1}{2}$ . If, for all $x \in \Omega$ , $\mathbb{E}((\Phi(X_1) - \Phi(x))^2 \mid X_0 = x) \leq R$ for some $R > 0$ , then

\begin{equation*} t_{\mathrm{mix}}(\varepsilon) \geq \frac{1}{2 \log (1 / \lambda)} \bigg(\log\bigg(\frac{(1 - \lambda)\Phi(x)^2}{2R}\bigg) + \log\bigg(\frac{1 - \varepsilon}{\varepsilon}\bigg)\bigg) \end{equation*}

for any $x \in \Omega$ .

The maximum eigenvalue is 1 with right eigenfunction $\phi_{0}(C)=1$ for all $C \in \mathcal{C}$ . The eigenvectors of eigenvalues corresponding to co-maximal flats in hyperplane arrangements are identified by Pike as follows.

Theorem 6. ([Reference Pike14, Theorem 3.1.1].) For each $i \in \{1, \dots, n \}$ , the Markov chain defined on the chambers by the transition probabilities in (2) has the right eigenfunction

\begin{equation*} \phi_i(C)= \begin{cases} -\sum_{\substack{F \in \mathcal{F} \\ \sigma_i(F)=+}} w(F), & \sigma_i(C)=-, \\[12pt] \quad \sum_{\substack{F \in \mathcal{F} \\ \sigma_i(F)=-}} w(F), & \sigma_i(C)=+, \end{cases} \end{equation*}

corresponding to eigenvalue $\lambda_i=\sum_{\sigma_i(F)=0} w(F)$ . Moreover, $\phi_0, \phi_1, \ldots, \phi_n$ are linearly independent, where $\phi_0$ is the eigenvector for the largest eigenvalue 1.

In our problem, the eigenvectors have a simpler expression as

(6) \begin{equation} \begin{aligned} & \phi_i(C) = \begin{cases} - p, & \sigma_i(C)=-, \\[5pt] \quad q, & \sigma_i(C)=+, \end{cases} \end{aligned}\end{equation}

for $i=1, \dots, n$ . Each i represents a vertex, and we can read off the color of i in a chamber by the eigenfunction $\phi_i$ . These eigenfunctions are not quite distinguishing on their own. In fact, Wilson’s lemma applied to any $\phi_i$ gives a lower bound of order n only. Nevertheless, if an eigenvalue corresponding to a co-maximal flat has algebraic multiplicity larger than one, linear combinations of eigenvectors can be used in Wilson’s lemma. In the case of regular graphs, where each vertex has the same degree, we have the second largest eigenvalue

\begin{equation*} \lambda=\sum_{\sigma_1(F)=0} w(F)= \frac{kn/2-k}{kn/2}=1-\frac{2}{n}\end{equation*}

associated with the co-maximal flat obtained by removing the vertex indexed by 1. Of course, it has multiplicity n considering all other co-maximal flats and equality of degrees. Therefore, from Theorem 6, the eigenspace for the second largest eigenvalue is spanned by eigenvectors (6) for $i=1, \dots, n$ . If we consider the linear combination $\Phi = \sum_{i=1}^n \phi_i$ , it is easy to see that $\Phi(C) = \Phi(C')$ if and only if they have the same number of blue vertices. If the color configuration C has k blue vertices, then $\Phi(C)= k -pn$ . So, $\Phi$ is a statistic that counts the number of blue vertices up to an additive term which guarantees $\mathbb{E}(\Phi(\pi))=0$ , where the expectation is taken with respect to the stationary distribution $\pi$ , i.e. $\mathbb{E}(\Phi(X_t \mid X_0=C)) = \lambda^{t} \Phi(C) \rightarrow 0$ . We can use this statistic to obtain the lower bound by Wilson’s method. Since, at each step, at most two vertices can change color, we have

(7) \begin{equation} \mathbb{E}[(\Phi(X_1)- \Phi(C))^2 \mid X_0=C)] \leq 4, \end{equation}

so that the condition in Lemma 1 is satisfied with $R=4$ .

Proof of the lower bound in Theorem 5. For the lower bound, since the second eigenvalue has multiplicity n given that the graph is regular, the sum of eigenvectors is an eigenvector of the second largest eigenvalues above. Take the initial state to be the blue monochromatic graph, for which $\Phi(C)= n-pn=qn$ . Then, using Lemma (1),

\begin{align*} t_{\mathrm{mix}}(\varepsilon) \geq \frac{n}{4} \bigg( \log \bigg(\frac{q^2n^2}{4n}\bigg) + \log \bigg( \frac{1- \varepsilon}{\varepsilon}\bigg)\bigg) \geq \frac{1}{4}n \log n - Cn \end{align*}

for some constant C depending only on p and $\varepsilon$ . We take $\varepsilon=\mathrm{e}^{-c}$ to conclude the proof.

We also note, with thanks to the anonymous referee who pointed this out, that the construction of eigenfunctions of edge flipping is analogous to the construction of eigenfunctions for the simple random walk on the hypercube. This also justifies the lower bound on the leading order of $\frac{1}{4} n \log n$ .

3.3. Exact rate of convergence for complete graphs

In Section 3.2 we established a lower bound of $\frac{1}{4}n \log n$ for the mixing time of edge flipping in regular graphs with n vertices. However, the upper bound obtained from eigenvalues differs by a factor of two. So we can only give an interval for the exact rate of convergence. Yet some earlier studies on similar processes suggest that the lower bound captures the correct rate of convergence; see [Reference Nestoridi12], for instance. We can verify that in the case of complete graphs. First, we formally define what we mean by the ‘exact rate’.

1. (i) $\lim\limits_{n\rightarrow \infty} {w_n}/{t_n} = 0$ ,

2. (ii) $\lim\limits_{\gamma \rightarrow - \infty} \liminf\limits_{n\rightarrow \infty} d_\mathrm{TV}(t_n + \gamma w_n) = 1$ ,

3. (iii) $\lim\limits_{\gamma \rightarrow \infty} \limsup\limits_{n\rightarrow \infty} d_\mathrm{TV}(t_n + \gamma w_n) = 0$ .

We will use a coupling argument to show that edge flipping in complete graphs shows a cutoff at time $\frac{1}{4}n \log n$ with a window of size n. The coupling time of two copies of a process is defined as $\tau\;:\!=\; \min\{t\colon X_t =Y_t\}$ . Writing $d(\!\cdot\!)$ for $d_\mathrm{T V}(\!\cdot\!)$ , the coupling time is related to the mixing time by

(8) \begin{equation} d(t) \leq \max_{x,y \in \Omega} \mathbb{P} (\tau > t \mid X_0=x,Y_0=y).\end{equation}

See, for example, [Reference Levin, Peres and Wilmer11, Theorem 5.4].

Proof of Theorem 1. Consider the linear combination of eigenvectors denoted by $\Phi$ in Section 3.2. We already showed that it counts the number of blue vertices. Now, if we view edge flipping on $K_n$ as a random walk on the hypercube $(\mathbb{Z} / 2 \mathbb{Z})^n$ where at each step two coordinates are replaced either by 1s with probability p or by 0s with probability q, then $\Phi$ is just the Hamming weight, the sum of the 0–1 coordinates. See [Reference Levin, Peres and Wilmer11, Section 2.3] for the definition of a hypercube random walk. An $n \log n$ upper bound is obtained for a lazy random walk on the hypercube by a coupling and a strong stationary time argument, which can be found in [Reference Levin, Peres and Wilmer11, Sections 5.3 and 6.4], respectively. This bound is later improved by monotone coupling in Section 18.2 of the same book following the observation that the problem can be reduced to convergence of Hamming weights, which can be considered as a lazy Ehrenfest urn problem. The same reduction argument is valid in the case of edge flipping in complete graphs, as follows. We claim that if the initial configuration is monochromatic, then

(9) \begin{equation} \| P^{t} - \pi \|_\mathrm{TV} = \| \Phi(X_t) - \pi_{\Phi} \|_\mathrm{TV} , \end{equation}

where $X_t$ is the chamber at step t, and $\pi_{\Phi}$ is the stationary distribution for the number of blue vertices. To argue for this, let us denote the set of chambers with k blue vertices by $\mathcal{C}_k=\{C \in \mathcal{C} \colon \Phi(C)=k \}$ . Starting from a monochromatic configuration, $X_t$ is equally likely to be any of the chambers with the same number of blue vertices by symmetry. Thus, we have

\begin{equation*} \sum_{C \in \mathcal{C}_k} |P(X_t=C) - \pi(C) | = \bigg| \sum_{C \in \mathcal{C}_k} P(X_t=C) - \pi(C) \bigg| = | \mathbb{P}(\Phi(X_t)=k) - \pi_{\Phi}(k) |. \end{equation*}

The equality in (9) shows that we can study the convergence of $\Phi(X_t)$ to its stationary distribution to obtain an upper bound for the mixing time of the edge flipping on $K_n$ . We define a coupling with three stages to study the mixing time. Let $X_t$ and $Y_t$ be two edge flipping processes where the initial states are monochromatic of opposite colors. We denote the number of blue vertices in each process by $W_t=\Phi(X_t),Z_t=\Phi(Y_t)$ , and take $W_0=0$ and $Z_0=n$ . From now on, we will only deal with $W_t$ and $Z_t$ . Let us first define $\Delta_t\;:\!=\;Z_t-W_t$ for $t=0,1,\ldots$

First stage: Coupon collector’s problem For the first stage of the coupling $(W_t,Z_t)_{t=0}^{\infty}$ we define the stopping time $\tau_1$ as $\tau_1=\min_t \big\{ \Delta_t \leq \big\lceil\sqrt{n}\big\rceil \big\}$ ; then we define the rule of coupling up to this time. If $t < \tau_1$ , the two chains move identically in the sense that we choose the same two vertices in the underlying graphs and color them the same.

The first part suggests that the process up to time $\tau_1$ is twice as fast as the coupon collector’s problem where you stop collecting more coupons if the remaining number of coupons is $\big\lceil\sqrt{n}\big\rceil$ . Let us define a process $\tilde{\Delta}_t$ for all t with transition probabilities

\begin{align*} \tilde{\Delta}_{t+1} = \tilde{\Delta}_t + \begin{cases} \hphantom{-}0 & \mathrm{with\ probability}\ \binom{n-\tilde{\Delta}_t}{2} / \binom{n}{2}, \\[5pt] -1 & \mathrm{with\ probability}\ \tilde{\Delta}_t(n-\tilde{\Delta}_t) \mathbin{/} \binom{n}{2}, \\[5pt] -2 & \mathrm{with\ probability}\ \binom{\tilde{\Delta}_t}{2} \mathbin{/} \binom{n}{2} \end{cases} \end{align*}

and $\tilde{\Delta}_0=n$ . Note that $\tilde{\Delta}_t=\Delta_t$ for $t \leq \tau_1$ . Calculating the conditional expectation, we have $\mathbb{E}[\tilde{\Delta}_{t+1}\mid\tilde{\Delta}_t] = (1-({2}/{n}))\tilde{\Delta}_t$ . Therefore,

\begin{equation*} \mathbb{E}[\tilde{\Delta}_t ] = \bigg(1-\frac{2}{n}\bigg)^t \tilde{\Delta}_0 \leq n \mathrm{e}^{-{2t}/{n}}. \end{equation*}

By Markov’s inequality, $\mathbb{P}(\tau_1 > s) = \mathbb{P}[\tilde{\Delta}_s > \sqrt{n}] \leq \sqrt{n}\mathrm{e}^{-{2s}/{n}}$ . If we take $s=\frac{1}{4}n \log n+\gamma_1 n$ , we obtain

(10) \begin{equation} \mathbb{P}\left(\tau_1 > \frac{1}{4}n \log n+\gamma_1 n \right) < \mathrm{e}^{-\gamma_1}. \end{equation}

Second stage: Coupling with lazy simple random walks Note that if we continued with the first coupling until the two chains meet, this expression would give an upper bound of order $\frac{1}{2}n \log n$ , which is the same as we obtained from eigenvalues. To improve this rate, whenever $t\geq \tau_1$ , we let the chains move according to the second rule where we choose edges independently for two graphs but color them with the same color. We will show that it only needs $\mathcal{O}(n)$ steps to hit zero with positive probability. Given that $t \geq \tau_1$ ,

(11) \begin{equation} \Delta_{t+1}=\Delta_t + \begin{cases} -2 & \mathrm{with\ probability}\ \binom{n-W_t}{2}\binom{Z_t}{2} / \binom{n}{2}^2, \\[5pt] -1 & \mathrm{with\ probability}\ \big[\binom{n-W_t}{2} Z_t(n-Z_t) + W_t(n-W_t)\binom{Z_t}{2}\big] / \binom{n}{2}^2, \\[5pt] \hphantom{-}1 & \mathrm{with\ probability}\ \big[W_t(n-W_t)\binom{n-Z_t}{2} + \binom{W_t}{2}Z_t(n-Z_t)\big] / \binom{n}{2}^2, \\[5pt] \hphantom{-}2 & \mathrm{with\ probability}\ \binom{W_t}{2}\binom{n-Z_t}{2} / \binom{n}{2}^2, \\[5pt] \hphantom{-}0 & \mathrm{otherwise}. \\[5pt] \end{cases} \end{equation}

We need to be more precise and show that the terms in the difference $\Delta_t$ are concentrated around pn, recalling that p was the probability of coloring a chosen edge blue.

Proof. It is not difficult to see that $\mathbb{E}[W_t] \rightarrow pn$ as $t \rightarrow \infty$ . To show the concentration result, we will use a second-order estimate. First, consider

(12) \begin{equation} W_{t+1} = W_t + \begin{cases} -2 & \mathrm{with\ probability}\ q \binom{W_t}{2} / \binom{n}{2}, \\[5pt] -1 & \mathrm{with\ probability}\ q W_t(n-W_t) / \binom{n}{2}, \\[5pt] \hphantom{-}1 & \mathrm{with\ probability}\ p W_t(n-W_t) / \binom{n}{2}, \\[5pt] \hphantom{-}2 & \mathrm{with\ probability}\ p \binom{n-W_t}{2} / \binom{n}{2}, \\[5pt] \hphantom{-}0 & \mathrm{otherwise}, \\[5pt] \end{cases} \end{equation}

where we take $\binom{z}{n}=0$ unless z is a positive integer. Let

(13) \begin{equation} \mathcal{F}_t=\sigma(W_{0}, W_1,\ldots, W_t) \end{equation}

be the $\sigma$ -field generated by $W_{0},W_{1},\ldots, W_t$ . Computation gives $\mathbb{E}[W_{t+1}\mid\mathcal{F}_t] = 2p + (1 - ({2}/{n}))W_t$ . We can evaluate this sum recursively to obtain

(14) \begin{equation} \mathbb{E}[W_t] = pn - \bigg(1-\frac{2}{n}\bigg)^{t+1}pn > pn - p\sqrt{n} , \end{equation}

provided that $t > \frac{1}4 n \log n$ and $W_0=0$ as assumed earlier.

Secondly, we bound the variance of $W_t$ by applying a crucial step in the proof of Wilson’s Lemma 1 (see [Reference Levin, Peres and Wilmer11, Section 13.2]), which gives

(15) \begin{equation} \mathrm{Var}(W_t) \leq \frac{R}{1-\lambda} = \frac{4}{1-(1-({2}/{n}))}=2n, \end{equation}

by (7). Then, by (14) and using (15) in Chebyshev’s inequality,

\begin{equation*} \mathbb{P}\big(|W_t-pn| \geq (p+r\sqrt{2})\sqrt{n} \big) \leq \frac{1}{r^2} \end{equation*}

for $r>0$ . This proves the result for $V_t=W_t$ . The proof for $V_t=Z_t$ follows exactly the same line of argument. By symmetric replacement of parameters, we have

\begin{equation*} \mathbb{E}[Z_t]=pn -\bigg(1-\frac{2}{n}\bigg)^{t+1}pn + \bigg(1-\frac{2}{n}\bigg)^{t} Z_0 < pn +q\sqrt{n}, \end{equation*}

where we take $Z_0=n$ . Since the bound in (7) is true for all initial states, we have the same variance bound. The rest follows from Chebyshev’s inequality.

Next, we present a maximal inequality for $W_t$ in the second stage of the coupling.

Lemma 3. If $t\geq \frac{1}{4}n \log n$ and k is a positive real number independent of n, then

\begin{align*} \mathbb{P}\Big(\max\limits_{\tau_1 \leq t \leq \tau_1+kn }|V_t -pn| > n^{1/2+\beta}\Big) \leq 4n^{-\alpha}\end{align*}

for any $\beta>0$ and $\alpha \in (0,\beta)$ , where $V_t$ can be replaced by $W_t$ or $Z_t$ .

Proof. Since $W_t$ and $Z_t$ are symmetric cases as in Lemma 2, we only prove the result for $W_t$ . By Doob–Meyer decomposition, we construct a martingale from $\{W_{\tau_1+s}\}_{s \geq 0}$ with respect to the $\sigma$ -fields: $\mathcal{G}_s\;:\!=\;\mathcal{F}_{\tau_1+s}$ for all $s=0,1,2,\ldots$ , where $\mathcal{F}_t$ is as defined in (13) for all $t=0,1,2,\ldots$ Since $\mathbb{P}(\tau_1 < \infty)=1$ by (10), $M_1=W_{\tau_1+1}-W_{\tau_1}$ is well-defined. For $s\geq 2$ , we let

(16) \begin{equation} M_s=W_{\tau_1+s} +\frac{2}{n}\sum_{i=1}^{s-1}W_{\tau_1+i} -W_{\tau_1} -2sp. \end{equation}

From (12) and (14), we can prove the martingale property: $\mathbb{E}[M_{s+1}|\mathcal{G}_{s}]=M_s$ . Observe that the transition probabilities of the martingale depend on $W_{\tau_1}$ . We will fix $W_{\tau_1}$ later in the proof.

In order to bound the variance of the martingale, we consider the variance of the differences. The computations give

\begin{equation*} \mathbb{E}[(M_{s+1}-M_s)^2]=C_1(p)\frac{\mathbb{E}[W_{\tau_1+s}^2]}{n^2} + C_2(p)\frac{\mathbb{E}[W_{\tau_1+s}]}{n}+C_3(p) \leq C(p), \end{equation*}

where C(p) and the other constants in the expression are independent of n and s. Therefore, by the orthogonality of martingale differences $\mathbb{E}[M_s^2] = \mathrm{Var}(M_s) \leq C(p)s$ . Then, by the Doob–Kolmogorov maximal inequality [Reference Grimmett and Stirzaker10, Theorem 2 in Section 7.8],

(17) \begin{equation} \mathbb{P}\Big(\max\limits_{1 \leq t \leq kn}|M_t|\geq a\Big) \leq \frac{\mathbb{E}[M_{kn}]}{a^2} = \frac{C(p)kn}{a^2} \end{equation}

for $a>0$ .

Next, we show that, for a of order $n^{1/2+\beta}$ and large enough n, $|W_{\tau_1+s}-pn|\geq a$ implies $|M_{s}|\geq n^{-\beta/2}a$ (not optimal) with high probability. Let A be the event that

(18) \begin{equation} |W_{\tau_1}-pn|\leq n^{1/2+\beta/2}, \end{equation}

which has probability greater than $1-4n^{-\beta}$ by Lemma 2. The following argument is conditioned on A. Let us fix an outcome $W_{\tau_1}=W^*$ in A. Note that both the stopping time $\tau_1$ and $W_{\tau_1}$ are determined. We take $n > 2^{4/\beta}$ . Suppose

(19) \begin{equation} |W_{\tau_1+s}-pn| > n^{1/2+\beta}. \end{equation}

From (18) and (19), we get $|W_{\tau_1+s}-W_{\tau_1}| > \frac{1}{2}n^{1/2+\beta}$ . There are two cases for (19): $W_{\tau_1+s} > pn+n^{1/2+\beta}$ or $W_{\tau_1+s} < pn-n^{1/2+\beta}$ . Let us assume the former. It will be apparent below that the other case can be treated similarly. From (16), we have one of

\begin{equation*} M_s > \frac{1}{4}n^{1/2 + \beta} , \qquad \frac{2}{n}\sum_{i=1}^{s-1}W_{\tau_1+i} -2sp \leq -\frac{1}{4}n^{1/2 + \beta}. \end{equation*}

If the latter is the case, by (18) and (19), there must exist a time s′ such that

\begin{align*} s^{\prime} = \max \left\{ 0 \leq t > s \colon W_{\tau_1+t} \leq pn+2n^{1/2+\beta/2} \right\}. \end{align*}

Then observe that

\begin{align*} \frac{2}{n}\sum_{i=1}^{s^{\prime}-1}W_{\tau_1+i} -2s^{\prime}p < \frac{2}{n}\sum_{i=1}^{s-1}W_{\tau_1+i} -2sp,\end{align*}

since $pn < W_{\tau_1+t}$ for $s^{\prime} \leq t <s$ . Therefore,

\begin{equation*} M_{s^{\prime}} = W_{\tau_1+s^{\prime}}-W_{\tau_1}+\frac{2}{n}\sum_{i=1}^{s^{\prime}-1}W_{\tau_1+i} -2s^{\prime}p \leq 3 n^{1/2+\beta/2} -\frac{1}{4}n^{1/2 + \beta} \leq - n^{1/2+\beta/2}. \end{equation*}

In either case, we have $\max_{1 \leq t \leq s}|M_t|\geq n^{1/2+\beta/2}$ . Therefore, by (17),

\begin{align*} \mathbb{P}\Big(\max\limits_{\tau_1 \leq t \leq \tau_1+s }|W_t -pn| > n^{1/2+\beta} \mid W_{\tau_1}=W^* \Big) & < \mathbb{P}\Big(\max\limits_{1 \leq t \leq s}|M_t|\geq n^{1/2+\beta/2} \mid W_{\tau_1}=W^* \Big) \\[5pt] & < C(p)kn^{-\beta}. \end{align*}

Now let us take $s=kn$ . Since the established upper bound is uniform for $W^*$ in A, it follows via a conditioning argument that

\begin{align*} \mathbb{P}\Big(\max\limits_{\tau_1 \leq t \leq \tau_1+kn }|W_t -pn| > n^{1/2+\beta}\Big) & < \mathbb{P}\Big(\max\limits_{1 \leq t \leq kn}|M_t|\geq n^{1/2+\beta/2} \mid A\Big)\mathbb{P}(A) + \mathbb{P}(A^\mathrm{c}) \\[5pt] & < C(p)kn^{-\beta}\mathbb{P}(A)+ \mathbb{P}(A^\mathrm{c}) \\[5pt] & \leq C(p)kn^{-\beta}(1-4n^{-\beta})+4n^{-\beta} < 4n^{-\alpha}, \end{align*}

where n is chosen large enough to satisfy the last inequality.

We return to (11). We use the lemmas above to bound the transition probabilities of $\Delta_t$ uniformly. We first need the following corollary.

Corollary 1. Let $t \in [\tau_1, \tau_1+kn]$ for some $k>0$ independent of n. For every $\varepsilon > 0$ , there exist $\alpha \in \big(0,\frac12\big)$ and $\beta \in \big(\alpha,\frac12\big)$ such that

(20) \begin{align} p_t & \;:\!=\; \frac{\sqrt{2} W_t}{n} \in [p-\varepsilon, p+\varepsilon], \\[5pt] \delta & \;:\!=\; \sqrt{2}\frac{Z_t-W_t}{n} = \mathcal{O}(n^{\beta-1/2}), \nonumber \end{align}

with probability $1-8n^{-\alpha}$ .

Therefore, under the assumptions of Corollary 1 we can rewrite (11) as

\begin{align*}\Delta_{t+1} = \Delta_t + \begin{cases} -2 & \mathrm{with\ probability}\ p_t^2q_t^2 +2p_tq_t^2\delta + o(n^{2\beta-1}), \\[5pt] -1 & \mathrm{with\ probability}\ 2p_tq_t(p_t^2+q_t^2)+q_t((q_t-p_t)^2+p_t)\delta +o(n^{2\beta-1}), \\[5pt] \hphantom{-}1 & \mathrm{with\ probability}\ 2p_tq_t(p_t^2+q_t^2)-p_t((p_t-q_t)^2+q_t)\delta + o(n^{2\beta-1}), \\[5pt] \hphantom{-}2 & \mathrm{with\ probability}\ p_t^2q_t^2 -2p_t^2q_t\delta + o(n^{2\beta-1}), \\[5pt] \hphantom{-}0 & \mathrm{otherwise}. \\[5pt] \end{cases} \end{align*}

Then, examining the probabilities, we see that it is more likely for $\{\Delta_{t}\}_{t > \tau_1}$ to go to the left if $\Delta_t >0$ . In fact,

(21) \begin{align} \begin{split} \mathbb{P}(\Delta_{t+1}-\Delta_t =-2)& \geq p_t^2q_t^2\geq \mathbb{P}(\Delta_{t+1}-\Delta_t =2), \\[5pt] \mathbb{P}(\Delta_{t+1}-\Delta_t =-1) &\geq 2p_tq_t(p_t^2+q_t^2) \geq \mathbb{P}(\Delta_{t+1}-\Delta_t =1), \end{split} \end{align}

provided that $Z_t\geq W_t$ . Thus, we can compare it to the following random walk, which is equally likely to go in either direction at any time:

\begin{align*}S_{t+1} = S_t + \begin{cases} -2 & \mathrm{with\ probability}\ p_t^2q_t^2, \\[5pt] -1 & \mathrm{with\ probability}\ 2p_tq_t(p_t^2+q_t^2), \\[5pt] \hphantom{-}0 & \mathrm{with\ probability}\ (p_t^2+q_t^2)^2+2p_t^2q_t^2, \\[5pt] \hphantom{-}1 & \mathrm{with\ probability}\ 2p_tq_t(p_t^2+q_t^2), \\[5pt] \hphantom{-}2 & \mathrm{with\ probability}\ p_t^2q_t^2. \end{cases} \end{align*}

Let us take $S_0=\Delta_{\tau_1}=\big\lceil \sqrt{n} \big\rceil$ . We define the stopping times $\tau_2=\min_t \{\Delta_{\tau_1+t} \in I\}$ and $\tau_S=\min_t \{S_t \in I\}$ , where $I=\{-1,0,1\}$ . Comparing the transition probabilities of the two random walks by (21), we have $\Delta_{\tau_1+t}\leq S_t$ since $Z_t \geq W_t$ . Therefore, we can find a coupling for two processes such that

(22) \begin{equation} \tau_2 \leq \tau_S. \end{equation}

Furthermore, observe that $S_{t+1}-S_t$ is a sum of independent and identically distributed random variables which take values $-1$ or 1 with probability $p_tq_t$ and 0 with probability $p_t^2+q_t^2$ . But then $S_t=S'_{\!\!2t}$ , where $S'_{\!\!t}$ is none other than a lazy simple random walk on the integers, so we may use the results on the lazy simple random walk to bound $\mathbb{P}(\tau_S > s)$ , such as [Reference Levin, Peres and Wilmer11, Corollary 2.28]. The only difficulty is that $S_t$ is not time-homogeneous as the parameter $p_t$ varies, which is dealt with next.

Given $r \in [0,1]$ , we let $\tau_S(r)=\min_t\{S_t(r) \in I\}$ where $S_t(r)$ is the lazy simple random walk with holding probability $r^2+(1-r)^2$ . Considering the bound in (20), we define $p^{*}= (p-\varepsilon)\textbf{1}_{\{p\leq 1/2\}}+ (p+\varepsilon)\textbf{1}_{\{p > 1/2\}}$ . The choice of $p^*$ ensures that $S_t(p^*)$ has the maximal holding probability among all values in $[p-\varepsilon,p+\varepsilon]$ .

Proof. Let us take $q^*=1-p^*$ . We define a Bernoulli random variable for each t as follows:

\begin{align*} \xi_t = \begin{cases} 0 & \mathrm{with\ probability}\ [{1-(p^{*2}+q^{*2})}]/[{1-(p_t^2+q_t^2)}], \\[5pt] 1 & \mathrm{otherwise}. \\[5pt] \end{cases}\end{align*}

$\xi$ is well-defined by the choice of $p^*$ . Now observe that $S_t(p^{*})$ has the same distribution as the chain which moves according to the law of $S_t$ (which is the same as $S_t(p_t)$ ) if $\xi_t=1$ , and stays at the same state if $\xi_t=0$ . Then, it is obvious that we can find a coupling satisfying $\tau_S(p^*) \geq \tau_S$ . The result follows.

Finally, by Lemmas 3 and 4, (22), and modifying the numbers in [Reference Levin, Peres and Wilmer11, Corollary 2.28] accordingly, we have

\begin{equation*} \mathbb{P}(\tau_2 > s) \leq 4n^{-2\alpha}+\frac{4\sqrt{n}}{\sqrt{2p^*(1-p^*)s}\,}. \end{equation*}

So, if we take $s=\gamma_2 n$ ,

(23) \begin{equation} \mathbb{P}(\tau_2 > \gamma_2 n) \leq 4n^{-2\alpha}+ \frac{4}{\sqrt{2p^*(1-p^*)\gamma_2}\,} \leq \frac{4 \max\{p^{-1/2}, (1-p)^{-1/2}\}}{\sqrt{\gamma_2}} \end{equation}

by choosing $\varepsilon$ small enough, say $\varepsilon = {\min\{p,1-p\}}/{100}$ , in (20).

Third stage: Number of sign changes For the last stage of the coupling, let us define $\tau_3=\min_t \{ \Delta_{\tau_1 +\tau_2+t} = 0 \}$ , which is to account for the case that $Z_{\tau_2}-W_{\tau_2}=1$ and $\Delta_t$ does not hit zero in the next step. By symmetry, we see that it is more likely for $\{\Delta_{t}\}_{t > \tau_1}$ to go to the right if $\Delta_t <0$ . So, we can again bound the stopping time by the time associated with $S_t$ . In fact, for the simple random walk, we have the probability that ‘the number of sign changes is equal to k up to $t=2n+1$ ’ is equal to the probability that ‘the walk is at position $2k+1$ at $t=2n+1$ ’; see [Reference Feller9, Theorem 1 in Chapter III.6]. This allows us to use the normal approximation for the sign changes [Reference Feller9, Theorem 2 in Chapter III.6]. Let $F_Z$ be the distribution function of the standard normal distribution. Let us use $S_t(p^*)$ again, the laziest simple random walk in the range. We have

\begin{equation*} \mathbb{P} \bigg(|t \in \{1,2,\ldots,K(p^*)n \} \colon S_{\tau_1+\tau_2+t}(p^*) \in I| \leq \frac{\sqrt{n}}{\gamma_3}\bigg) \leq 2F_Z\bigg(\frac{1}{\gamma_3}\bigg) - 1 \leq \frac{1}{\sqrt{2\pi} \gamma_3}, \end{equation*}

where $K(p)=1+[({p^2+q^2})/{2pq}]$ . Since

\begin{equation*} \mathbb{P}(|t \in \{1,2,\ldots,k \} \colon S_{\tau_1+\tau_2+t} \in I| \leq x) \leq \mathbb{P}(|t \in \{1,2,\ldots,k \} \colon S_{\tau_1+\tau_2+t}(p^*) \in I| \leq x) \end{equation*}

for all k and x, $\Delta_{t}$ visits I on an order larger than $\sqrt{n}$ times a probability bounded away from zero. At every visit to I, it has a positive probability to hit zero at the next stage. Let $\kappa=\mathbb{P}(S_{t+1}(p^*)=0 \mid S_t=-1 \text{ or }1).$ Thus,

(24) \begin{equation} \mathbb{P}(\tau_3 > K(p^{*}) n) \leq (1-\kappa)^{{\sqrt{n}}/{\gamma_3}} + \frac{1}{\sqrt{2\pi} \gamma_3} < \gamma_3^{-1} \end{equation}

for any $\gamma_3>0$ independent of n for large n. Let us take $\gamma_3>K(p^*)$ for convenience. Finally, let $\tau=\tau_1+\tau_2+\tau_3$ , which is the stopping time, $\alpha>0$ , and $\gamma=\gamma_1+\gamma_2+\gamma_3$ . By (10), (23), (24), and using the strong Markov property iteratively,

\begin{equation*} \mathbb{P}\big(\tau > \tfrac{1}{4}n \log n + \gamma n\big) < 1 - (1 - \mathrm{e}^{-\gamma_1}) \big(1-4 \max\{p^{-1/2}, (1-p)^{-1/2}\}\gamma_2^{-1/2}\big)\big(1-\gamma_3^{-1}\big). \end{equation*}

Since we chose the opposite monochromatic configurations for the different initial states of the coupled chains, among all choices for the starting pair chambers the coupling time above is clearly the maximum, so is the probability in (8). Therefore,

\begin{equation*} d\left(\frac{1}{4}n \log n + \gamma n\right) \leq \frac{C}{\sqrt{\gamma}} \end{equation*}

by choosing $\gamma_1$ , $\gamma_2$ , and $\gamma_3$ large enough. Combining with the lower bound in Theorem 5, we have $t_n=\frac{1}{4} n \log n$ and $w_n=n$ in Definition 1.