On the Coupling Time of the Heat-Bath Process for the Fortuin–Kasteleyn Random–Cluster Model

Abstract

We consider the coupling from the past implementation of the random–cluster heat-bath process, and study its random running time, or coupling time. We focus on hypercubic lattices embedded on tori, in dimensions one to three, with cluster fugacity at least one. We make a number of conjectures regarding the asymptotic behaviour of the coupling time, motivated by rigorous results in one dimension and Monte Carlo simulations in dimensions two and three. Amongst our findings, we observe that, for generic parameter values, the distribution of the appropriately standardized coupling time converges to a Gumbel distribution, and that the standard deviation of the coupling time is asymptotic to an explicit universal constant multiple of the relaxation time. Perhaps surprisingly, we observe these results to hold both off criticality, where the coupling time closely mimics the coupon collector’s problem, and also at the critical point, provided the cluster fugacity is below the value at which the transition becomes discontinuous. Finally, we consider analogous questions for the single-spin Ising heat-bath process.

Appendices

Appendix A: Autocorrelation Functions of Strictly Increasing Observables

Let P denote the transition matrix of the FK heat-bath process on a finite graph $G=(V,E)$ with parameters $p\in (0,1)$ and $q\ge 1$, and let $k=2^{|E|}$. To avoid trivialities, we assume $|E|>1$. We regard elements of $\mathbb {R}^k$ as functions from $2^E$ to $\mathbb {R}$, and we endow $\mathbb {R}^k$ with the inner product $\langle \cdot ,\cdot \rangle$ defined by

$$\begin{aligned} \langle g,h\rangle := \sum _{A\subseteq E} g(A)\,h(A)\,\phi (A). \end{aligned}$$

Denote the eigenvalues of P by $1=\lambda _1 > \lambda _2\ge \ldots \ge \lambda _k$. As mentioned in Sect. 2.2, general results for heat-bath chains [13] imply that all $\lambda _i$ are non-negative. Let $\{\psi _i\}_{i=1}^k$ be an orthonormal basis for $\mathbb {R}^k$ such that $\psi _i$ is an eigenfunction of P corresponding to $\lambda _i$. The Perron-Frobenius theorem implies that the eigenspace of $\lambda _1$ is one-dimensional, and that we can take $\psi _1(A)=1$ for all $A\subseteq E$. Let W denote the eigenspace of $\lambda _2$. For $g\in \mathbb {R}^k$, we let $g_W$ denote its projection onto W.

We say $g\in \mathbb {R}^k$ is increasing if $A\subset B$ implies $g(A)\le g(B)$, and strictly increasing if $A\subset B$ implies $g(A) < g(B)$.

Proposition A.1

Let $(X_t)_{t\in \mathbb {N}}$ be a stationary FK heat-bath process, and for $g\in \mathbb {R}^k$ define $(g_t)_{t\in \mathbb {N}}$ via $g_t\,:=g(X_t)$. If g is strictly increasing, then its autocorrelation function satisfies

$$\begin{aligned} \rho _{g}(t) := \frac{{\text {cov}}(g_0,g_t)}{{\text {var}}(g_0)} \sim C e^{-t/t_{\exp }},\qquad t\rightarrow \infty , \end{aligned}$$

for constant $C>0$.

Proof

Let $\varPi$ denote the projection matrix onto the space of constant functions. General arguments (see e.g. [45] or [36, Chap. 9]) imply

$$\begin{aligned} {\text {cov}}(g_0,g_t) = \langle g,(P^t-\varPi )g\rangle = \sum _{l=2}^k \langle g,\psi _l\rangle ^2\lambda _l^t = \Vert g_W\Vert ^2\lambda _2^t \,+\, \sum _{l=\dim (W)+2}^k\langle g,\psi _l\rangle ^2\lambda _l^t. \end{aligned}$$

Since g is strictly increasing, Lemma A.2 implies that $\Vert g_W\Vert ^2>0$, and therefore

$$\begin{aligned} {\text {cov}}(g_0,g_t) \sim \Vert g_W\Vert ^2 e^{-t/t_{\exp }}, \qquad t\rightarrow \infty . \end{aligned}$$

It follows that

$$\begin{aligned} \rho _g(t) \sim \frac{\Vert g_W\Vert ^2}{{\text {var}}(g)} e^{-t/t_{\exp }}, \qquad t\rightarrow \infty . \end{aligned}$$

$\square$

Lemma A.2

If g is strictly increasing, then its projection onto W is non-zero.

Proof

Lemma A.3 implies there exists $\psi \in W$ which is non-zero and increasing. Positive association (see e.g. [24, Theorem 3.8 (b)]) then implies that for any other increasing g we have

$$\begin{aligned} \langle g,\psi \rangle \ge \mathbb {E}(g)\,\mathbb {E}(\psi ) = 0, \end{aligned}$$

(A.1)

since $\mathbb {E}(\psi )=\langle \psi _1,\psi \rangle =0$. In particular, suppose that g is strictly increasing. Choosing $\alpha >0$ so that

$$\begin{aligned} g(B)-g(A) > \alpha [\psi (B)-\psi (A)], \qquad \text { for all } A \subset B\subseteq E, \end{aligned}$$

implies that $g-\alpha \psi$ is also strictly increasing. Applying (A.1) to $g-\alpha \psi$ then yields

$$\begin{aligned} \langle g-\alpha \psi ,\psi \rangle \ge 0. \end{aligned}$$

Rearranging, and using the fact that $\psi$ is non-zero then implies

$$\begin{aligned} \langle g,\psi \rangle \ge \alpha \langle \psi ,\psi \rangle >0. \end{aligned}$$

Therefore, g has a non-zero projection onto $\psi \in W$, and the stated result follows. $\square$

The following lemma is the natural analogue, in the FK setting, of the result [39, Lemma 3] established for the Ising heat-bath process.

Lemma A.3

There exists $\psi \in W$ which is non-zero and increasing.

Proof

Let $g=\psi _2+ C (\mathscr {N}-\mathbb {E}(\mathscr {N}))$, where $\mathscr {N}\in \mathbb {R}^k$ is defined so that $\mathscr {N}(A)=|A|$ for each $A\subseteq E$, and $C>0$ is a constant. We have

$$\begin{aligned} g = [1+C\langle \mathscr {N},\psi _2\rangle ]\psi _2 + C \sum _{j=3}^k \langle \mathscr {N},\psi _j\rangle \psi _j. \end{aligned}$$

If $\langle \mathscr {N},\psi _2\rangle =0$, then g has a non-zero projection onto $\psi _2$, for any choice of $C>0$. If $\langle \mathscr {N},\psi _2\rangle \ne 0$, then choosing $C>|\langle \mathscr {N},\psi _2\rangle |^{-1}$ suffices to guarantee that g again has a non-zero projection onto $\psi _2$. In either case, assume C is so chosen. It follows that $g_W$ is non-zero.

If $A\subset B$, then

$$\begin{aligned} g(B) - g(A) = \psi _2(B)-\psi _2(A) + C[\mathscr {N}(B)-\mathscr {N}(A)] \ge \min _{A\subset B\subseteq E} [\psi _2(B) - \psi _2(A)] + C. \end{aligned}$$

Therefore, by choosing $C>\left| \min \limits _{A\subset B\subseteq E} [\psi _2(B) - \psi _2(A)]\right|$ we guarantee that g is increasing. Lemma A.4 then implies that $g_W$ is increasing. Therefore, $\psi =g_W$ is an increasing, non-zero element of W. $\square$

Lemma A.4

If g is increasing and has zero-mean, then its projection onto W is also increasing.

Proof

Let $g\in \mathbb {R}^k$ be any increasing observable with mean zero, and let $t\in \mathbb {N}^+$. Since Lemma A.6 implies $\lambda _2>0$, we can write

$$\begin{aligned} \frac{P^t g}{\lambda _2^t} = g_W + \sum _{l=\dim (W)+2}^k \langle g,\psi _l\rangle \psi _l\,\left( \frac{\lambda _l}{\lambda _2}\right) ^t. \end{aligned}$$

It follows that

$$\begin{aligned} \lim _{t\rightarrow \infty } \frac{P^t g}{\lambda _2^t} = g_W. \end{aligned}$$

(A.2)

Now, for any given $t\ge 1$, Lemma A.5 implies that $P^t g(A)$ is an increasing function of A, and so $\lambda _2^{-t}P^t g (A)$ is also an increasing function of A. It then follows, as an elementary consequence of (A.2), that $g_W$ is also increasing. We have therefore established that if g is an increasing zero-mean function, then its projection $g_W$ is also increasing. $\square$

Lemma A.5

If $g\in \mathbb {R}^k$ is increasing, then $P^tg$ is also increasing, for every $t\ge 1$.

Proof

Let $(f,\mathscr {E},U)$ be the random mapping representation for P given in Sect. 2.1; see (2.3). Let $A_1\subset A_2\subseteq E$, and let $B_i=f(A_i,\mathscr {E},U)$ for $i=1,2$. Clearly, $(B_1,B_2)$ is a coupling of the distributions $P(A_1,\cdot )$ and $P(A_2,\cdot )$, and the monotonicity of f implies $B_1\subseteq B_2$. Strassen’s theorem (see e.g. [25, Theorem 4.2]) then implies that

$$\begin{aligned} \mathbb {E}_{P(A_1,\cdot )}(g) \le \mathbb {E}_{P(A_2,\cdot )}(g) \end{aligned}$$

for any increasing $g\in \mathbb {R}^k$. It follows that

$$\begin{aligned} (Pg)(A_1) = \sum _{B\subseteq E} P(A_1,B) g(B)= & {} \mathbb {E}_{P(A_1,\cdot )}(g)\\\le & {} \mathbb {E}_{P(A_2,\cdot )}(g) = \sum _{B\subseteq E} P(A_2,B) g(B) = (Pg)(A_2). \end{aligned}$$

Since this holds for any $A_1\subset A_2\subseteq E$, it follows that Pg is increasing. It then follows by a simple induction that $P^tg$ is increasing for any $t\ge 1$. $\square$

Lemma A.6

The second-largest eigenvalue of P is positive.

Proof

Since P is reversible and irreducible we have the spectral decomposition (see e.g. [35, Lemma 12.2])

$$\begin{aligned} \frac{P(A,B)}{\phi (B)} = 1 + \sum _{j=2}^k\psi _i(A)\psi _i(B)\lambda _i. \end{aligned}$$

Since $\lambda _2\ge \lambda _j\ge 0$ for all $j>2$, it follows that if $\lambda _2=0$, then $P(A,B)=\phi (B)$ for all $A,B\subseteq E$. But since, by assumption, we have $|E|>1$, we can choose $A,B\subseteq E$ with $|A\triangle B| > 1$, where $\triangle$ denotes symmetric difference, and (2.2) then implies

$$\begin{aligned} P(A,B) = 0 \ne \phi (B). \end{aligned}$$

We have therefore reached a contradiction, and we conclude that $\lambda _2>0$. $\square$

Appendix B: Coupon Collecting

Let $n\in \mathbb {N}^+$, and let $C_1,C_2,\ldots$ be an iid sequence of uniformly random elements of $[n]\,:=\{1,2,\ldots ,n\}$. For $t\in \mathbb {N}^+$, we think of $C_t$ as the coupon collected at time t. For $i\in [n]$, let $D_i\in [n]$ denote the ith distinct type of coupon collected; i.e. the ith distinct element of the sequence $C_1,C_2,\ldots$. Let $S_i(t)\,:=\#\{s\le t : C_s = D_i\}$, the number of copies of $D_i$ collected by time t. Define $R_t\,:=\{c\in [n]: C_s = c \text { for some } s\le t\}$, the set of distinct coupon types collected up to time t. For any $1\le k \le n$, let $W_k=\inf \{t\in \mathbb {N}^+: |R_t| = k\}$, and note that $W_k$ is simply the hitting time of $D_{k}$. The coupon collector’s time is then defined as $W\,:=W_n$.

For each $c\in [n]$, define

$$\begin{aligned} H(c) = \sup \{t\le W: C_t=c\}. \end{aligned}$$

We refer to the time $H(c)$ as the last visit to c. Let $(H_i)_{i=1}^n$ denote the sequence of the $H(c)$, arranged in increasing order. In particular, $H_1$ is the first time that a last visit occurs.

Lemma B.1

There exists $\varphi >0$ such that $\mathbb {P}(|R_{H_{1}}| \le \left\lfloor \ln n \right\rfloor ) = O(n^{-\varphi })$.

Proof

Inserting $a_n=\lfloor \ln (n)\rfloor$ and $c_n=\lfloor \ln (n)/4\rfloor$ into Lemma B.2 and applying the union bound, implies

$$\begin{aligned} \mathbb {P}\left( \bigcup _{i=1}^{a_{n}} \left\{ S_{i}(W) \le c_n \right\} \right)&\le \ln (n) \exp \left( -\frac{1}{2}\ln (n-a_n) + \frac{\ln (n)}{4} + 1\right) \\&= e\,\ln (n) \exp \left( -\frac{1}{4}\ln (n) - \frac{1}{2}\ln \left( 1-\frac{a_n}{n}\right) \right) \\&= \frac{e}{\sqrt{1-\lfloor \ln (n)\rfloor /n}}\,\ln (n)\, n^{-1/4}. \end{aligned}$$

Therefore, for any $0<\rho <1/4$, we have

$$\begin{aligned} \mathbb {P}\left( \bigcup _{i=1}^{a_{n}} \left\{ S_{i}(W) \le c_n \right\} \right) = O(n^{-\rho }), \qquad n\rightarrow \infty . \end{aligned}$$

It follows that,

$$\begin{aligned} \mathbb {P}(|R_{H_1}| \le a_n) = \mathbb {P}\left( |R_{H_1}|\le a_n, \bigcap _{i=1}^{a_n}\{S_i(W)>c_n\}\right) + O(n^{-\rho }) \end{aligned}$$

(B.1)

Let $I\,:=\inf \{t\in \mathbb {N}^+: S_i(t) = c_n \text { for some } i\in [n] \}$, the first time that there exists a coupon type for which exactly $c_n$ copies have been collected, and define the random variable $K\in [n]$ via $C_{H_1}=D_K$. If $|R_{H_1}|\le a_n$, then $1\le K \le a_n$. Therefore, observing that $S_{K}(W)=S_{K}(H_1)$, we find

$$\begin{aligned} \mathbb {P}\left( |R_{H_1}|\le a_n, \bigcap _{i=1}^{a_n}\{S_i(W)>c_n\}\right)&\le \mathbb {P}(|R_{H_1}|\le a_n, S_{K}(W)> c_n) \nonumber \\&= \mathbb {P}(|R_{H_1}|\le a_n, S_{K}(H_1) > c_n) \nonumber \\&\le \mathbb {P}(|R_I|\le a_n) \end{aligned}$$

(B.2)

since if $|R_{H_1}|\le a_n$ and $S_{K}(H_1) > c_n$ then $|R_I|\le a_n$. Combining (B.1) and (B.2) then implies

$$\begin{aligned} \mathbb {P}(|R_{H_1}| \le a_n) \le \mathbb {P}(|R_I|\le a_n) + O(n^{-\rho }). \end{aligned}$$

However, Lemma B.3 implies that there exists $\delta >0$ such that $\mathbb {P}(|R_I|\le a_n) = O(n^{-\delta })$. We therefore conclude that, if $\varphi =\min \{\rho ,\delta \}$, then

$$\begin{aligned} \mathbb {P}(|R_{H_{1}}| \le \left\lfloor \ln n \right\rfloor ) = O(n^{-\varphi }). \end{aligned}$$

$\square$

Lemma B.2

Let $(a_n)_{n\in \mathbb {N}^+}$ and $(c_n)_{n\in \mathbb {N}^+}$ be any two sequences of natural numbers. For $n\in \mathbb {N}^+$, if $a_n<n$ then for each $1\le i\le a_n$ we have

$$\begin{aligned} \mathbb {P}\left( S_i(W) \le c_n \right) \le \exp \left( -\ln (\sqrt{n-a_n}) + c_n + 1\right) . \end{aligned}$$

Proof

Fix $n\in \mathbb {N}^+$ and $1\le i \le a_n$, and assume $a_n<n$. Adopting the convention $W_0=0$, for $0\le k \le n-1$ we define

$$\begin{aligned} Y_i(k) \,:= \sum _{j=W_{k}+1}^{W_{k+1}-1} \mathbf {1}_{\{C_{j}= D_{i}\}}. \end{aligned}$$

Since $Y_i(k)=0$ for all $k<i$, and $C_{W_k}=D_i$ iff $k=i$, we then have

$$\begin{aligned} S_{i}(W) = 1 + \sum _{k=i}^{n-1} Y_i(k). \end{aligned}$$

And since the random variables $Y_i(k)$ are independent, for any $\theta <0$, we have

$$\begin{aligned} \mathbb {P}\left( S_i(W) \le c_n \right)&\le \mathbb {P}\left( \sum _{k=a_n}^{n-1} Y_i(k) \le c_n \right) \nonumber \\&= \mathbb {P}\left( \exp \left[ \theta \sum _{k=a_n}^{n-1} Y_i(k)\right] \ge e^{\theta c_n} \right) \nonumber \\&\le \exp \left( -\theta c_n + \sum _{k=a_n}^{n-1} \ln \mathbb {E}[e^{\theta Y_i(k)}]\right) , \end{aligned}$$

(B.3)

where the final step follows from Markov’s inequality.

The moment generating function of $Y_i(k)$ can be calculated explicitly. Let $i \le k \le n-1$. Given $W_{k}$ and $W_{k+1}$, the random variable $Y_i(k)$ has binomial distribution with $W_{k+1} -W_{k}-1$ trials and success probability 1 / k, which implies

$$\begin{aligned} \mathbb {E}(e^{\theta Y_i(k)})&= \mathbb {E}(\mathbb {E}[e^{\theta Y_i(k)}|W_{k},W_{k+1}]) \\&= \mathbb {E}\left[ \left( \frac{e^{\theta }}{k} + 1 - \frac{1}{k}\right) ^{W_{k+1} - W_{k}-1}\right] . \end{aligned}$$

But since $W_{k+1}-W_{k}$ has geometric distribution with parameter $1-k/n$, this becomes

$$\begin{aligned} \mathbb {E}(e^{\theta Y_i(k)}) = \frac{n-k}{n-k+1-e^{\theta }}. \end{aligned}$$

Therefore, setting $\lambda =1-e^\theta$ and $b_n=n-a_n$, it follows from the fact that $\ln (1+\lambda /k)$ is a decreasing function of k that

$$\begin{aligned} -\sum _{k=a_{n}}^{n-1} \ln \mathbb {E}(e^{\theta Y_i(k)})&= \sum _{k=1}^{b_n} \ln \left( 1 + \frac{\lambda }{k} \right) \nonumber \\&\ge \int _1^{b_n} \ln \left( 1+\frac{\lambda }{x}\right) \mathrm {d}\,x \nonumber \\&= \lambda \ln (b_n) + (b_n + \lambda ) \ln (1 + \lambda /b_n) - (1+\lambda )\ln (1+\lambda )\nonumber \\&\ge \lambda \ln (b_n) + \lambda - (1+\lambda )\ln (1+\lambda )\nonumber \\&\ge \lambda \ln (b_n) - 1 \end{aligned}$$

(B.4)

where, in the penultimate step, we used the fact that $\ln (1+x)\ge x/(1+x)$ holds for all $x>-1$, and in the last step we used the fact that $(1+\lambda )-(1+\lambda )\ln (1+\lambda )>0$ for any $\lambda \in (0,1)$. Combining (B.3) and (B.4), we conclude that for all $\lambda \in (0,1)$ we have

$$\begin{aligned} \mathbb {P}\left( S_i(W) \le c_n \right) \le \exp \left[ -\lambda \ln (n-a_n) -\ln (1-\lambda )c_n + 1) \right] . \end{aligned}$$

Choosing $\lambda =1/2$ yields the stated result. $\square$

Lemma B.3

Fix $c\in (0,1)$, and define sequences $(a_n)_{n\in \mathbb {N}^+}$ and $(c_n)_{n\in \mathbb {N}^+}$ such that $a_n=\lfloor \ln (n)\rfloor$ and $c_n=\lfloor c\ln (n)\rfloor$. Let

$$\begin{aligned} I\,:=\inf \{t\in \mathbb {N}^+: S_i(t) = c_n \text { for some } i\in [n] \}, \end{aligned}$$

the first time that there exists a coupon type for which exactly $c_n$ copies have been collected. Then there exists $\delta >0$ such that

$$\begin{aligned} \mathbb {P}(|R_{I}|\le a_n)=O(n^{-\delta }), \qquad n \rightarrow \infty . \end{aligned}$$

Proof

We assume, in all that follows, that n is sufficiently large that $c_n>1$. For $k\in [n]$, let

$$\begin{aligned} I_k=\inf \{t\in \mathbb {N}^+: S_{k}(t) =c_n\} \end{aligned}$$

be the first time that $c_n$ copies of coupon type $D_k$ have been collected. For any sequence of natural numbers $(b_n)_{n\in \mathbb {N}^+}$, we have

$$\begin{aligned} \mathbb {P}(|R_{I_k}| \le a_n)&= \mathbb {P}(|R_{I_k}|\le a_n, I_k\le b_n) + \mathbb {P}(|R_{I_k}|\le a_n, I_k> b_n)\nonumber \\&\le \mathbb {P}(I_k\le b_n) + \mathbb {P}(|R_{I_k}|\le a_n, I_k> b_n)\nonumber \\&\le \mathbb {P}(I_k\le b_n) + \mathbb {P}(W_{a_n+1}>b_n), \end{aligned}$$

(B.5)

where the last inequality follows by observing that if $|R_{I_k}|\le a_n$ and $I_k>b_n$, then $W_{a_n+1}>b_n$.

To find an upper bound for $\mathbb {P}(I_k\le b_n)$, note that, for any $s\ge 1$, the random time between the sth and $(s+1)$th arrival of coupon type $D_k$ is a geometric random variable with success probability 1 / n. It follows that $\varDelta _k\,:=I_k-W_k$ is a sum of $c_n-1$ independent geometric random variables,⁶ each with success probability 1 / n. Lemma B.4 therefore implies that for any $0<\lambda <1$,

$$\begin{aligned} \mathbb {P}(\varDelta _k\le \lambda n(c_n-1)) \le e^{-f(\lambda ) c_n + f(\lambda )} \end{aligned}$$

where $f(\lambda )>0$. But from the trivial lower bound $W_k\ge 1$, it follows that $\varDelta _k\le I_k -1$. Therefore, for any $b_n\le \lambda n(c_n-1)+1$, we have

$$\begin{aligned} \mathbb {P}(I_k \le b_n) \le \mathbb {P}(I_k\le \lambda n (c_n-1) + 1) \le \mathbb {P}(\varDelta _k\le \lambda n (c_n-1)) \le e^{-f(\lambda ) c_n + f(\lambda )}. \end{aligned}$$

(B.6)

To find an upper bound for $\mathbb {P}(W_{a_n+1}>b_n)$, we begin with the observation that, with the convention $W_0=0$, we have

$$\begin{aligned} W_{a_n+1} = \sum _{i=0}^{a_n}(W_{i+1}-W_i). \end{aligned}$$

For $0\le i \le a_n$, the random variables $W_{i+1}-W_i$ are independent, and distributed according to a geometric distribution with success probability $1-i/n$. Therefore, Lemma B.4 implies that for any $\zeta >1$

$$\begin{aligned} \mathbb {P}(W_{a_n+1}\ge \zeta \,\mathbb {E}[W_{a_n+1}]) \le e^{-f(\zeta )(1-a_n/n)\mathbb {E}(W_{a_n+1})} \end{aligned}$$

with $f(\zeta )>0$. But explicit calculation shows that

$$\begin{aligned} \mathbb {E}(W_{a_n+1}) = n(H_n - H_{n-a_n-1})\sim a_n, \qquad n\rightarrow \infty , \end{aligned}$$

where $H_i$ is the ith harmonic number, and the asymptotic result follows from $H_n\sim \ln (n)$ and the fact that $a_n=o(n)$. It follows that for any choice of $b_n \ge \zeta \, \mathbb {E}(W_{a_n+1})$ and $\alpha \in (0,f(\zeta ))$, for sufficiently large n, we have

$$\begin{aligned} \mathbb {P}(W_{a_n+1}\ge b_n) \le e^{-\alpha \, a_n}. \end{aligned}$$

(B.7)

Any choice of $b_n$ satisfying $\zeta \,\mathbb {E}(W_{a_n+1}) \le b_n \le \lambda n(c_n-1)+1$, for sufficiently large n, suffices to ensure (B.6) and (B.7) hold simultaneously. It therefore suffices to set $b_n=n$. For simplicity, $\lambda \in (0,1)$ and $\zeta >1$ can be chosen so that $f(\lambda )=1=f(\zeta )$. Combining (B.5), (B.6) and (B.7) then implies that for any $\alpha <1$ we have

$$\begin{aligned} \mathbb {P}(|R_{I_k}| \le a_n) \le e^{-c_n+1} + e^{-\alpha \,a_n} \end{aligned}$$

for sufficiently large n.

Finally, since $|R_I|\le a_n$ implies $|R_{I_k}|\le a_n$ for some $1\le k \le a_n$, it follows from the union bound that, for sufficiently large n,

$$\begin{aligned} \mathbb {P}(|R_I|\le a_n) \le \mathbb {P}\left( \bigcup _{k=1}^{a_n}\{|R_{I_k}|\le a_n\}\right) \le \sum _{k=1}^{a_n} \mathbb {P}(|R_{I_k}|\le a_n)&\le a_n\, e^{-c_n+1} + a_n\, e^{-\alpha \,a_n}\\&\le e^2\,\ln (n)\,n^{-c} + e^{\alpha }\,\ln (n)\,n^{-\alpha }. \end{aligned}$$

Since $c,\alpha >0$, we can choose $0< \delta <\min \{c,\alpha \}$, and we obtain $\mathbb {P}(|R_I|\le a_n) = O(n^{-\delta })$.   $\square$

Lemma B.4

Let $X_1,X_2,\ldots ,X_n$ be independent random variables, such that $X_i$ has geometric distribution with success probability $p_i$, and let $X=\sum _{i=1}^n X_i$. Then

$$\begin{aligned} \mathbb {P}(X\le \lambda \mu )&\le e^{-p_{*}\mu f(\lambda )}, \qquad \forall \,\lambda \le 1,\\ \mathbb {P}(X\ge \zeta \mu )&\le e^{-p_{*}\mu f(\zeta )}, \qquad \forall \,\zeta \ge 1, \end{aligned}$$

where $\mu =\mathbb {E}(X) = \sum _{i=1}^n 1/p_i$, $p_*= \min _{i\in [n]} p_i$ and $f(x)=x -1 -\ln (x)$.

Proof

These results can be established, in the standard way, by applying Markov’s inequality to $\mathbb {E}(e^{t X})$, and using the explicit form for $\mathbb {E}(e^{t X_i})$; see e.g. [31]. $\square$