2. Setup and notation

We let $\mathbb {N}=\{1,2,3,\ldots \}$ and for $n\in \mathbb {N}$ , set $[n]=\{1,\ldots ,n\}$ . Given a sequence $(\gamma _{n})_{n\in \mathbb {N}}$ taking values in $(-1/2,1/2)$ and $\Omega =\{-1,1\}$ , we define the product measure $\nu $ on $\Omega ^{\mathbb {N}}$ by

$$ \begin{align*} \nu=\prod_{n=1}^{\infty}\nu_{n}\quad\text{where }\nu_{n}(\{1\})=\frac{1}{2}+\gamma_{n}\,\text{and}\,\nu(\{-1\})=\frac{1}{2}-\gamma_{n}. \end{align*} $$

Let $\mu ^k$ denote the uniform product measure on $\Omega ^k$ and consider $\mathbb {P}_k =\nu \times \mu ^k$ defined on $\Omega ^{\mathbb {N}}\times \Omega ^k$ . Denote by ${\mathbb E}_k $ the corresponding expectation.

For $1\leq j\leq 2^{k}$ , define the indicator random variables $I_{j}:\Omega ^{\mathbb {N}}\times \Omega ^k\rightarrow \{0,1\}$ by

(2.1) $$ \begin{align} I_{j}(x,\omega)= \begin{cases} 1 & \text{for }x_{j}\ldots x_{j+k-1}=\omega,\\ 0 & \text{otherwise}, \end{cases} \end{align} $$

and $M_{k}:\Omega ^{\mathbb {N}}\times \Omega ^k\rightarrow \mathbb {Z}_{\ge 0}$ by

(2.2) $$ \begin{align} M_{k}(x,\omega)=\#\{1\leq i\leq2^{k}\mid x_{i}\ldots x_{i+k-1}=\omega\}=\sum_{j\in[2^k]}I_{j}(x,\omega). \end{align} $$

For $\omega \in \Omega ^k$ and $j,k\ge 1$ , we introduce the quantity

(2.3) $$ \begin{align} P_{j,k}(\omega)=\prod_{i=1}^{k}(1+2\omega_{i}\gamma_{i+j-1}). \end{align} $$

Sometimes, we think of $P_{j,k}$ as a random variable on $\Omega ^{\mathbb {N}}\times \Omega ^k$ . By the independence of the random variables $\{\omega _i\gamma _{i+j-1}\}_{1\le i\le k}$ , we point out that

(2.4) $$ \begin{align} {\mathbb E}_k [P_{j,k}]=\prod_{i=1}^k{\mathbb E}_k [1+2\omega_i\gamma_{i+j-1}]=1. \end{align} $$

We also note that for any fixed $\omega \in \Omega ^k$ ,

(2.5) $$ \begin{align} \mathbb{P}_k (I_j=1|\{\omega\})= \prod_{i=1}^{k}\Big(\frac{1}{2}+\omega_{i}\gamma_{i+j-1}\Big) =2^{-k}P_{j,k}(\omega). \end{align} $$

Observe that, for any fixed $x\in \Omega ^{\mathbb {N}}$ , there is a unique $\omega \in \Omega ^k$ such that $I_j(x,\omega )=1$ , and the probability of this $\omega $ , like all others, is $2^{-k}$ ; thus, ${\mathbb E}_k [I_{j}]=2^{-k}$ . When $|i-j|\ge k$ , the variables $I_j$ and $I_i$ are independent conditionally to $\omega \in \Omega ^k$ . However, the independence fails if we do not condition on $\omega $ , since

$$ \begin{align*} {\mathbb E}_k [I_iI_j]=2^{-k}\sum_{\omega\in\Omega^k}\nu(x:I_j(x,\omega)I_i(x,\omega)=1)= 2^{-3k}\sum_{\omega\in\Omega^k}P_{j,k}(\omega)P_{i,k}(\omega) \end{align*} $$

is different from ${\mathbb E}_k [I_j]{\mathbb E}_k [I_i]=2^{-2k}$ .

3. Convergence to Poisson

Let $\gamma _n\in (-1/2,1/2)$ be such that $\gamma _n= O (\log ^{-(1/2+\delta )}n)$ for some $\delta>0$ . Without loss of generality (decreasing $\delta $ if necessary), we assume that there is $n_0\ge 1$ such that

$$ \begin{align*} |\gamma_n|\le\log^{-(1/2+\delta)}n\quad \text{for all } n\ge n_0. \end{align*} $$

We consider $M_k$ defined in (2.2) on the probability space $(\Omega ^{\mathbb {N}}\times \Omega ^k,\mathbb {P}_k )$ ; the main result of this section is that $M_{k}$ converges in distribution to a Poisson random variable with mean one.

This is commonly referred to as the annealed case, because it involves a coupled probability space. By contrast, the quenched scenario refers to an almost sure result on the probability space $\Omega ^{\mathbb {N}}$ , corresponding precisely to the convergence statement of Theorem 1.1. The following proposition establishes the connection between annealed and quenched results.

By Proposition 3.2, the convergence result in Theorem 1.1 follows from Proposition 3.1; hence, the remainder of this section is dedicated to proving Proposition 3.1.

3.1. A general convergence theorem applied to our setting

To prove Proposition 3.1, we rely on a general result on Poisson approximation, [Reference Barbour, Holst and Janson3, Theorem 1.A], derived by the Chen–Stein method, which provides a bound on the total variation distance $d_{\text {TV}}$ (see the reference above for a definition). We note that convergence in total variation implies convergence in distribution. Given a family of random variables $\{X_j\}_{j\in J}$ , we denote by $\sigma (X_j:j\in J)$ the $\sigma $ -algebra generated by such a family, that is, the smallest $\sigma $ -algebra for which each of the $X_j$ is measurable.

Theorem 3.3. Let $I_{1},\ldots ,I_{n}$ be indicator random variables and $S=\sum _{j\in [n]}I_{j}$ . For every $j\in [n]$ , let there be given a partition $\Gamma _{j}^{s},\Gamma _{j}^{w}\subseteq [n]$ of $[n]\setminus \{j\}$ , let

$$ \begin{align*} \unicode{x3bb}=\sum_{j\in[n]}{\mathbb E} [I_{j}], \end{align*} $$

and let

$$ \begin{align*} \eta_{j}={\mathbb E} |{\mathbb E} [I_{j}|\sigma(I_{i}:i\in\Gamma_{j}^{w})]-{\mathbb E} [I_{j}]|. \end{align*} $$

Then,

$$ \begin{align*} \begin{aligned} d_\mathrm{TV}(S,\operatorname*{\mathrm{Po}}(\unicode{x3bb}))\le & \min\{1,\unicode{x3bb}^{-1}\}\bigg(\!\sum_{j\in[n]}({\mathbb E} [I_{j}]^{2}+\sum_{i\in\Gamma_{j}^{s}}({\mathbb E} [I_{j}]{\mathbb E} [I_{i}]+{\mathbb E} [I_{j}I_{i}])\bigg)\\ & +\min\{1,\unicode{x3bb}^{-1/2}\}\sum_{j\in[n]}\eta_{j}. \end{aligned} \end{align*} $$

The sets $\Gamma _j^s,\Gamma _j^w$ partition the variables into those that are strongly and weakly correlated with $I_j$ , respectively. This is the meaning of the superscripts: ‘s’ for strong and ‘w’ for weak.

For a fixed k, we apply this with $n=2^{k}$ , the indicators $I_{1},\ldots ,I_{2^{k}}$ from (2.1), and $S=M_{k}=\sum _{j\in [2^k]}I_{j}$ . Recall from §2 that ${\mathbb E}_k [I_{j}]=2^{-k}$ and so $\unicode{x3bb} =\sum _{j\in [2^{k}]}{\mathbb E}_k [I_{j}]=1.$ For $j\in [2^{k}]$ , we let

(3.1) $$ \begin{align} \Gamma_{j}^{s}=\{n\in[2^{k}]\setminus\{j\}:|n-j|<k\} \quad\text{and}\quad \Gamma_{j}^{w}=[2^{k}]\setminus(\Gamma_{j}^{s}\cup\{j\}). \end{align} $$

Theorem 3.3 yields that

$$ \begin{align*} d_\mathrm{TV}(M_{k},\operatorname*{\mathrm{Po}}(1))\le\underset{A_{k}}{\underbrace{2^{-2k}\sum_{j\in[2^{k}]}(1+|\Gamma_{j}^{s}|)}}+\underset{B_{k}}{\underbrace{\sum_{j\in[2^{k}]}\sum_{i\in\Gamma_{j}^{s}}{\mathbb E}_k [I_{j}I_{i}]}}+\underset{C_{k}}{\underbrace{\sum_{j\in[2^{k}]}\eta_{j}}}. \end{align*} $$

To conclude that $M_{k}\xrightarrow {d}\operatorname *{\mathrm {Po}}(1)$ , we will show that each of the positive terms $A_{k},B_{k},C_{k}$ tend to zero as $k\to \infty $ .

3.3. $B_{k}\rightarrow 0$

Lemma 3.4. There exists $j_{0}\in \mathbb {N}$ such that ${\mathbb E}_k [I_{i}I_{j}]<2^{-3k/2}$ for all $j_{0}\leq i<j\leq 2^{k}$ satisfying $0<j-i< k$ .

Proof. Since $(\gamma _n)$ is a null sequence, we let $j_{0}$ be such that $1+2\gamma _{n}<2^{1/4}$ for all $n\ge j_0$ , and let i, j be as in the statement. Arguing as in the proof of [Reference Álvarez, Becher and Mereb2, Lemma 1], let

$$ \begin{align*} \Omega^k_{i,j}=\{\omega\in\Omega^k\mid(\omega_{1},\ldots,\omega_{k-(j-i)})=(\omega_{j-i},\ldots,\omega_{k})\}, \end{align*} $$

and note that a word $\omega \in \Omega ^k$ can satisfy $I_{i}(x,\omega )I_{j}(x,\omega )=1$ for some $x\in \Omega ^{\mathbb {N}}$ , only if ${\omega \in \Omega ^k_{i,j}}$ . The elements of $\Omega ^k_{i,j}$ are in bijection with their prefix of length $j-i$ , so $\mu ^k(\Omega ^k_{i,j})=2^{-k+(j-i)}$ .

For a fixed $\omega \in \Omega ^k_{i,j}$ , we define $\widetilde {\omega }\in \Omega ^{k+(j-i)}$ as the juxtaposition of two copies of $\omega $ , namely $\widetilde {\omega }_h=\omega _h$ if $h\in \{1,\ldots ,k\}$ and $\widetilde {\omega }_h=\omega _{h-(j-i)}$ if $h\in \{k+1,\ldots ,k+(j-i)\}$ . By $i\ge j_0$ ,

$$ \begin{align*} \begin{aligned} \nu(x:I_{i}(x,\omega)I_{j}(x,\omega)=1)&= \prod_{h=i}^{j+k-1}(1/2+\widetilde{\omega}_{h-i+1}\gamma_{h})\\ &\le 2^{-(k+j-i)}\prod_{h=i}^{j+k-1}(1+2\gamma_{h})\\ &\le 2^{-(k+j-i)}2^{1/4(k+j-i)}. \end{aligned} \end{align*} $$

Using that $k+j-i<2k$ , it follows that

$$ \begin{align*} \nu(x:I_{i}(x,\omega)I_{j}(x,\omega)=1)\le 2^{-(k/2+j-i)}. \end{align*} $$

So,

$$ \begin{align*} {\mathbb E}_k [I_{i} I_{j}] e &= \int_{\Omega^k_{i,j}}\nu(x:I_{i}(x,\omega)I_{j}(x,\omega)=1)\mathop{}\!\mathrm{d}\mu^k(\omega)\\& \leq\mu^k(\Omega^k_{i,j})2^{-(k/2+j-i)}=2^{-3k/2}, \end{align*} $$

which completes the proof.

To conclude the proof that $B_{k}\rightarrow 0$ , we use that ${\mathbb E}_k [I_{j}]=2^{-k}$ to get

$$ \begin{align*} {\mathbb E}_k [I_{i}I_{j}]\leq{\mathbb E}_k [I_{j}]=2^{-k}. \end{align*} $$

Therefore, with $j_{0}$ as in Lemma 3.4,

$$ \begin{align*} B_{k} & =\sum_{j\in[2^{k}]}\sum_{i\in\Gamma_{j}^{s}}{\mathbb E}_k [I_{i}I_{j}]\\ & \leq\sum_{j=1}^{j_{0}-1}k{\mathbb E}_k [I_{i}I_{j}]+\sum_{j=j_{0}}^{2^{k}-1}k2^{-3k/2}\\ & \leq j_{0}k2^{-k}+k2^{-k/2}, \end{align*} $$

and $B_{k}\rightarrow 0$ follows.

Remark 3.5. The arguments used so far do not rely on the specific rate at which $(\gamma _n)$ decays to zero. This property becomes crucial in the next subsection.

3.4. $C_{k}\to 0$

Let $P_{j,k}$ be as in (2.3). The main step to prove $C_k\to 0$ is the following proposition.

Proposition 3.6. Let $\varepsilon>0$ . Then, ${\mathbb E}_k[|P_{j,k}-1|]\to 0$ uniformly in $j\ge 2^{\varepsilon k}$ as $k\to \infty $ .

Proof. As $k\to \infty $ , the decay rate for $(\gamma _n)$ yields uniformly in $j\ge 2^{\varepsilon k}$ that

$$ \begin{align*} 0\le \gamma_{j}^2\le\log^{-(1+2\delta)}j \le \log^{-(1+2\delta)}(2^{k\varepsilon}) = O (k^{-(1+2\delta)}). \end{align*} $$

So,

(3.2) $$ \begin{align} \sum_{i=1}^k\gamma_{i+j-1}^2= O (k^{-2\delta}). \end{align} $$

By a first-order expansion around $x=0$ of $f(x)=\log (1+x)$ and (3.2), for $j>2^{\varepsilon k}$ , we have

(3.3) $$ \begin{align} & P_{j,k}(\omega) = \exp\bigg\{\sum_{i=1}^k\log(1+2\omega_i\gamma_{i+j-1})\bigg\} =\exp\bigg\{2\sum_{i=1}^k\omega_i\gamma_{i+j-1} + O (k^{-2\delta})\bigg\}. \end{align} $$

For a fixed $\theta \in (0,1/2)$ , we define

$$ \begin{align*} A_{k,j}^\theta=\bigg\{\omega\in \Omega^k: \bigg|\sum_{i=1}^k\omega_i\gamma_{i+j-1}\bigg|\le \bigg(\sum_{i=1}^k\gamma_{i+j-1}^2\bigg)^{1/2-\theta} \bigg\}. \end{align*} $$

By (3.2), we have uniformly in $\omega \in A_{k,j}^\theta $ and $j\ge 2^{\varepsilon k}$ that $\sum _{i=1}^k\omega _i\gamma _{i+j-1}=O(k^{-\delta (1-2\theta )})$ . So, identity (3.3) yields that $P_{j,k}(\omega )=1+o(1)$ uniformly on $A^\theta _{k,j}$ and j. It follows that

(3.4)

In particular, this implies

Since by (2.4), it follows that

(3.5)

Under the measure $\mu ^k$ , the random variables $(\omega _i\gamma _{i+j-1})_{1\le i\le k}$ are independent with mean zero and variance $\gamma _{i+j-1}^2$ . Hence, by Chebyshev’s inequality and (3.2), we get

$$ \begin{align*} \mu^k(\Omega^k\setminus A_{k,j}^\theta)\le \bigg(\sum_{i=1}^k\gamma_{i+j-1}^2\bigg)^{2\theta}= O (k^{-4\delta\theta})\longrightarrow 0, \end{align*} $$

uniformly in $j\ge 2^{\varepsilon k}$ , as $k\to \infty $ . Applying (3.5),

Combining the latter with (3.4),

which finishes the proof.

Remark 3.7. The exponent $1/2+\delta $ in the decay of $(\gamma _n)$ is heuristically explained by applying of the central limit theorem to (3.3). The sum of independent random variables $\sum _{i=1}^k\omega _i\gamma _{i+j-1}$ typically grows proportionally to $(\sum _{i=1}^k{\gamma _{i+j-1}^2})^{1/2}$ . Thus, the elements of $A^\theta _{k,j}$ characterize the asymptotics of $P_{j,k}$ .

We now can complete the proof that $C_{k}\rightarrow 0$ . For fixed $k\ge 1$ and $j\in [2^{k}]$ , we let

where $\Gamma _j^w\subset [2^k]$ is from (3.1). Consider now the random variable $W(x,\omega )=\omega $ and let $\xi _{j}={\mathbb E}_k [I_{j}-2^{k}|\mathcal {F}_{j}]$ , where $\mathcal {F}_{j}=\sigma (\{I_{i}:i\in \Gamma _{j}^{w}\},W)$ . Applying the tower property twice,

$$ \begin{align*} \eta_{j}={\mathbb E}_k |{\mathbb E}_k [\xi_{j}|\sigma(I_{i}:i\in\Gamma_{j}^{w})]|\le{\mathbb E}_k [{\mathbb E}_k [|\xi_{j}||\sigma(I_{i}:i\in\Gamma_{j}^{w})]]={\mathbb E}_k |\xi_{j}|. \end{align*} $$

Since $|j-i|\ge k$ , the variable $I_j$ is independent of $(I_{i}:i\in \Gamma _{j}^{w})$ conditionally to ${\{W=\omega \}}$ . Hence, by (2.5),

$$ \begin{align*} {\mathbb E}_k [I_{j}|\mathcal{F}_{j}](x,\omega)=\mathbb{P}_k (I_{j}=1|W=\omega) =2^{-k}P_{j,k}(\omega). \end{align*} $$

Therefore, $\xi _j=2^{-k}(P_{j,k}-1)$ and

$$ \begin{align*} C_{k} \le \sum_{j\in[2^{k}]}{\mathbb E}_k |\xi_{j}| = 2^{-k}\sum_{j\in[2^{k}]}{\mathbb E}_k |P_{j,k}-1|. \end{align*} $$

By (2.4), we know that ${\mathbb E}_k [P_{j,k}]=1$ , so as $k\to \infty $ ,

$$ \begin{align*} 2^{-k}\sum_{j\le 2^{\varepsilon k}} {\mathbb E}_k |P_{j,k}-1|\le 2\cdot 2^{-k(1-\varepsilon)}=o(1). \end{align*} $$

Hence, Proposition 3.6 yields that

$$ \begin{align*} C_k\le o(1)+2^{-k}\sum_{2^{\varepsilon k}\le j\le 2^k}{\mathbb E}_k |P_{j,k}-1|\to 0. \end{align*} $$

This concludes the estimate for $C_{k}$ and thus our proof of Proposition 3.1.

4. Non-convergence

Without loss of generality, we fix $\delta \in (0,1/2), n_0\ge 1$ , and assume that

$$ \begin{align*} \gamma_{n}=\log^{-(1/2-\delta)}n \quad \text{for all } n\ge n_0. \end{align*} $$

We consider $M_k$ defined in (2.2) on the probability space $(\Omega ^{\mathbb {N}}\times \Omega ^k,\mathbb {P}_k )$ ; we shall show that $M_{k}$ does not converge in distribution to a Poisson random variable with mean one. In the current section, we prove this result in the annealed setting, whereas the second part of Theorem 1.1 addresses the quenched result. However, since quenched convergence implies annealed convergence, this is sufficient.

Before proving the annealed case, we need to establish a few preliminary results. Let $k\in \mathbb {N}$ and let $D_{+},D_{-}\subseteq \{1,\ldots ,k\}$ be sets of equal size. For $j\ge 1$ , write

$$ \begin{align*} \Xi_j=\Xi_j(D_{+},D_{-})=\prod_{i\in D_{+}}(1+\gamma_{i+j-1})\prod_{i\in D_{-}}(1-\gamma_{i+j-1}). \end{align*} $$

Proof. Let $\ell =|D_{+}|=|D_{-}|\leq k$ . Because $\gamma _{n}$ is decreasing, the product defining $\Xi _j$ can only increase if we replace each $1+\gamma _{i+j-1}$ by $1+\gamma _{j}$ , and each $1-\gamma _{i+j-1}$ by $1-\gamma _{j+k}$ . Thus,

$$ \begin{align*} \Xi_j\leq(1+\gamma_{j})^{\ell}(1-\gamma_{j+k})^{\ell}=(1+\gamma_{j}-\gamma_{j+k}-\gamma_{j}\gamma_{j+k})^{\ell}. \end{align*} $$

Let $f(x)=\log ^{-(1/2-\delta )}x$ , $x>0$ , so that $f(n)=\gamma _n$ , $n\ge 1$ . Since f is deceasing and concave, for $x<y$ , we have $|f(x)-f(y)|\leq |x-y||f'(x)|$ . Applying this with $x=j$ , $y=j+k$ , and using $j\geq 2^{\varepsilon k}$ , $f'(x)=-c(1/2-\delta )(x\log ^{3/2-\delta }x)^{-1}$ , we get

$$ \begin{align*} \gamma_{j}-\gamma_{j+k}\leq k\cdot|f'(j)|= O(2^{-\varepsilon k}\cdot k^{-(1/2-\delta)}). \end{align*} $$

However, using $j,j+k\leq 2^{k}+k<2^{k+1}$ for all k sufficiently large, we have $\gamma _{j}\gamma _{j+k}\geq c^2(\log 2/(k+1))^{1-2\delta }$ . It follows that $1+\gamma _{j}-\gamma _{j+k-1}-\gamma _{j}\gamma _{j+k}<1$ and the same holds after raising to the $\ell $ th power, giving us $\Xi _j\leq 1$ . This proves the statement.

For $\eta>0$ and $k\ge 1$ , define

(4.1) $$ \begin{align} \Omega_k^\eta =\bigg\{\omega\in\Omega^k:\sum_{i=1}^{k}\omega_{i}<-\eta\sqrt{k}\bigg\}. \end{align} $$

When convenient, we identify $\Omega _k^\eta \subseteq \Omega ^k$ with its lift $\{(x,\omega )\mid \omega \in \Omega _k^\eta \}$ to $\Omega ^{\mathbb {N}}\times \Omega ^k$ .

Proof. By Fubini, it suffices to bound $\nu (x:M_{k}(x,\omega )\ge 1)=\mathbb {P}_k (M_{k}\geq 1|\{\omega \})$ uniformly in $\omega \in \Omega _k^\eta $ . Since $M_{k}=\sum _{j\in [2^{k}]}I_{j}$ , we get by (2.5) that for all $\omega \in \Omega ^k$ ,

$$ \begin{align*} \nu(x:M_{k}(x,\omega)\ge1) \leq \sum_{j\in[2^{k}]}\nu(x:I_{j}(x,\omega)=1) = 2^{-k}\sum_{j\in[2^{k}]}P_{j,k}(\omega). \end{align*} $$

Let $\varepsilon \in (0,1)$ . We first claim that the sum on the right changes by $o(1)$ if we sum over $2^{\varepsilon k}\leq j\leq 2^{k}$ instead of $1\leq j\leq 2^{k}$ . Indeed, using $\gamma _{n}\rightarrow 0$ , there is $j_{0}$ such that ${1+2\gamma _{j}<2^{(1-\varepsilon )/2}}$ for any $j\ge j_0$ . By the fact that $\gamma _{n}\rightarrow 0$ , for every fixed $j\in \mathbb {N}$ , we have $\sup _{\omega \in \Omega ^k}2^{-k}P_{j,k}(\omega )=o(1)$ as $k\rightarrow \infty $ , so

$$ \begin{align*} \sum_{1\leq j\leq j_{0}}2^{-k}P_{j,k}(\omega)=j_{0}\cdot o(1)=o(1). \end{align*} $$

Also, for all $j\ge j_0$ ,

$$ \begin{align*} P_{j,k}(\omega) = \prod_{i=1}^{k}(1+2\omega_{i}\gamma_{ i+j-1}) \leq 2^{(1-\varepsilon)k/2}, \end{align*} $$

so

$$ \begin{align*} 2^{-k}\sum_{j_{0}\leq j\leq2^{\varepsilon k}}P_{j,k}(\omega)<2^{-k}\cdot2^{\varepsilon k}\cdot2^{(1-\varepsilon)k/2}=o(1), \end{align*} $$

uniformly in $\omega \in \Omega ^k$ . Thus, we have shown that

$$ \begin{align*} \nu(x:M_{k}(x,\omega)\ge1)=o(1)+2^{-k}\sum_{2^{\varepsilon k}\le j\le 2^k}P_{j,k}(\omega). \end{align*} $$

Let $ N_+(\omega ) =\#\{1\leq i\leq k\mid \omega _{i}=1\}$ and let $D_{+},D_{-}\subseteq [k]$ denote the sets of positions of the first $N_+(\omega )$ occurrences of $+1,-1$ in $\omega $ , respectively. Since $\sum _{i\in D_+\cup D_-}\omega _i=0$ , the set $E(\omega )=[k]\setminus (D_{+}\cup D_{-})$ has cardinality $|E|=|\sum _{i=1}^k\omega _i|$ . Let now $\omega \in \Omega _k^\eta $ . It follows that $\omega _{i}=-1$ for $i\in E$ and $|E|>\eta \sqrt {k}$ . Since $(\gamma _n)$ is decreasing, by Proposition 4.1,

$$ \begin{align*} P_{j,k}(\omega)=\Xi_{j}(D_{+},D_{-})\cdot\prod_{i\in E(\omega)}(1-2\gamma_{ i+j-1})\le (1-2\gamma_{k+2^{k}})^{|E|} \end{align*} $$

for all $k\ge 1$ sufficiently large, uniformly in $2^{\varepsilon k}\leq j\leq 2^{k}$ and $\omega \in \Omega _k^\eta $ . By $|E|>\eta \sqrt {k}$ ,

$$ \begin{align*} 2^{-k}\sum_{2^{\varepsilon k}\le j\le 2^k}P_{j,k}(\omega) & <(1-2\gamma_{k+2^{k}})^{\eta k^{1/2}}\leq\bigg(1-\frac{c'}{k^{1/2-\delta}}\bigg)^{\eta k^{1/2}} \end{align*} $$

for some $c'>0$ . Since the exponent tends to infinity faster than the denominator, the last expression tends to zero as $k\rightarrow \infty $ , as desired.

If Y is Poisson with parameter $1$ , then $\mathbb {P}_k (Y=0)=1/e$ . Thus, the next proposition shows that $M_{k}$ does not converge in distribution to $\operatorname *{\mathrm {Po}}(1)$ .

Proof. Since $M_{k}\geq 0$ is integer-valued, the complement of the event $\{M_{k}=0\}$ is ${\{M_{k}\geq 1\}}$ ; we shall bound the probability of the latter event from above. For a parameter $\eta>0$ that we shall choose later, let $\Omega _k^\eta $ be as in (4.1) and let $\mathcal {N}$ be a standard Gaussian. Since on the space $(\Omega ^k,\mu ^{k})$ the random variables $\{\omega _i\}_{1\le i\le k}$ are independent and identically distributed with unitary second moment, as $k\to \infty $ , the central limit theorem yields that

$$ \begin{align*} \mathbb{P}_k ((\Omega_k^\eta)^{c})=\mathbb{P}_k (\mathcal{N}\ge-\eta)+o(1). \end{align*} $$

Therefore, by Lemma 4.2,

$$ \begin{align*} \begin{aligned} \mathbb{P}_k (M_{k}\ge1) & \leq \mathbb{P}_k (\Omega_k^\eta\cap\{M_{k}\geq1\})+\mathbb{P}_k ((\Omega_k^\eta)^{c}) = o(1)+\mathbb{P}_k (\mathcal{N}\ge-\eta). \end{aligned} \end{align*} $$

Since $\lim _{\eta \rightarrow 0}\mathbb {P}_k (\mathcal {N}\geq -\eta )=\mathbb {P}_k (\mathcal {N}\geq 0)=1/2$ , by choosing $\eta $ small enough, we can ensure that $\mathbb {P}_k (\mathcal {N}\geq -\eta )<1-1/e$ . It then follows that

$$ \begin{align*} \limsup_{k\rightarrow\infty}\mathbb{P}_k (M_{k}\geq1)<1-\frac{1}{e}, \end{align*} $$

as desired.