2.1. Fragmentation chains

Let $\varepsilon>0$ . As in [Reference Hoffmann and Krell16], we start with the space

\begin{align*}\mathcal{S}^{\downarrow}=\Bigg\{ \mathbf{s}=(s_{1},s_{2},\dots),\,s_{1}\geq s_{2}\geq\dots\geq0,\,\sum_{i=1}^{+\infty}s_{i}\leq1\Bigg\} .\end{align*}

A fragmentation chain is a process in $\mathcal{S}^{\downarrow}$ characterized by

• a dislocation measure $\nu$ , which is a finite measure on $\mathcal{S}^{\downarrow}$ ;

• a description of the law of the times between fragmentations.

A fragmentation chain with dislocation measure $\nu$ is a Markov process $X=(X(t),t\geq0)$ with values in $\mathcal{S}^{\downarrow}$ . Its evolution can be described as follows: a fragment with size x lives for some time (which may or may not be random) then splits and gives rise to a family of smaller fragments distributed as $x\xi$ , where $\xi$ is distributed according to $\nu(\!\cdot\!)/\nu(\mathcal{S}^{\downarrow})$ . We suppose the lifetime of a fragment of size x is an exponential time of parameter $x^{\alpha}\nu(\mathcal{S}^{\downarrow})$ , for some $\alpha$ . We could make different assumptions here on the lifetime of fragments, but this would not change our results. Indeed, as we are interested in the sizes of the fragments frozen as soon as they are smaller than $\varepsilon$ , the time they need to become this small is not important.

We denote by $\mathbb{P}_{m}$ the law of X started from the initial configuration $(m,0,0,\dots\!)$ with m in (0, 1]. The law of X is entirely determined by $\alpha$ and $\nu(\!\cdot\!)$ [Reference Bertoin2, Theorem 3].

We make the same assumption as in [Reference Hoffmann and Krell16].

Let $\mathcal{U}\;:\!=\;\{0\}\cup\bigcup_{n=1}^{+\infty}(\mathbb{N}^{*})^{n}$ denote the infinite genealogical tree. For u in $\mathcal{U}$ , we use the conventional notation $u=()$ if $u=\{0\}$ and $u=(u_{1},\dots,u_{n})$ if $u\in(\mathbb{N}^{*})^{n}$ with $n\in\mathbb{N}^{*}$ . This way, any u in $\mathcal{U}$ can be denoted by $u=(u_{1},\dots,u_{n})$ for some $u_{1},\dots,u_{n}$ and with n in $\mathbb{N}$ . Now, for $u=(u_{1},\dots,u_{n})\in\mathcal{U}$ and $i\in\mathbb{N}^{*}$ , we say that u is in the nth generation and we write $|u|=n$ ; we write $ui=(u_{1},\dots,u_{n},i)$ , $u(k)=(u_{1},\dots,u_{k})$ for all $k\in[n]$ . For any $u=(u_{1},\dots,u_{n})$ and $v=ui$ ( $i\in\mathbb{N}^{*}$ ), we say that u is the mother of v. For any u in $\mathcal{U}\backslash\{0\}$ ( $\mathcal{U}$ deprived of its root), u has exactly one mother and we denote it by ${\boldsymbol{{m}}}(u)$ . The set $\mathcal{U}$ is ordered alphanumerically:

• If u and v are in $\mathcal{U}$ and $|u|<|v|$ then $u<v$ .

• If u and v are in $\mathcal{U}$ and $|u|=|v|=n$ , $u=(u_{1},\dots,u_{n})$ , and $v=(v_{1},\dots,v_{n})$ with $u_{1}=v_{1}, \dots , u_{k}=v_{k},u_{k+1}<v_{k+1}$ then $u<v$ .

Suppose we have a process X which has the law $\mathbb{P}_{m}$ . For all $\omega$ , we can index the fragments that are formed by the process X with elements of $\mathcal{U}$ in a recursive way.

• We start with a fragment of size m indexed by $u=()$ .

• If a fragment x, with a birth time $t_{1}$ and a split time $t_{2}$ , is indexed by u in $\mathcal{U}$ , at time $t_{2}$ this fragment splits into smaller fragments of sizes $(xs_{1},xs_{2},\dots)$ with $(s_{1},s_{2},\dots)$ of law $\nu(\!\cdot\!)/\nu(\mathcal{S}^{\downarrow})$ . We index the fragment of size $xs_{1}$ by u1, the fragment of size $xs_{2}$ by u2, and so on.

A mark is an application from $\mathcal{U}$ to some other set. We associate a mark $\xi_{\dots}$ on the tree $\mathcal{U}$ to each path of the process X. The mark at node u is $\xi_{u}$ , where $\xi_{u}$ is the size of the fragment indexed by u. The distribution of this random mark can be described recursively as follows.

2.2. Tagged fragments

From now on, we suppose that we start with a block of size $m=1$ . We assume that the total mass of the fragments remains constant through time, as follows.

This assumption was already present in [Reference Hoffmann and Krell16]. We observe that the Malthusian exponent of [Reference Bertoin3, p. 27] is equal to 1 under our assumptions. Without this assumption, the link between the empirical measure $\gamma_{-\log(\varepsilon)}$ and the tagged fragments, (5.1), vanishes and our proofs of Proposition 5.1 and Theorem 5.1 fail.

We can now define tagged fragments. We use the representation of fragmentation chains as random infinite marked trees to define a fragmentation chain with q tags. Suppose we have a fragmentation process X of law $\mathbb{P}_{1}$ . We take $(Y_{1},Y_{2},\dots,Y_{q})$ to be q i.i.d. variables of law $\mathcal{U}([0,1])$ . We set, for all u in $\mathcal{U}$ , $(\xi_{u},A_{u},I_{u})$ with $\xi_{u}$ defined as above. The random variables $A_{u}$ take values in the subsets of [q]. The random variables $I_{u}$ are intervals. These variables are defined as follows.

• We set $A_{\{0\}}=[q]$ , $I_{\{0\}}=(0,1]$ ( $I_{\{0\}}$ is of length $\xi_{\{0\}}=1$ ).

• For all $n\in\mathbb{N}$ , suppose we are given the marks of the first n generations. Suppose that, for u in the nth generation, $I_{u}=(a_{u},a_{u}+\xi_{u}]$ for some $a_{u}\in\mathbb{R}$ (it is of length $\xi_{u}$ ). Then the marks at generation $n+1$ are given by Proposition 2.1 (concerning $\xi_{\cdot}$ ) and, for all u such that $|u|=n$ and for all i in $\mathbb{N}^{*}$ ,

  \begin{align*}I_{ui}=\big(a_{u}+\xi_{u}\big(\widetilde{\xi}_{u1}+\dots+\widetilde{\xi}_{u(i-1)}\big),a_{u}+\xi_{u}\big(\widetilde{\xi}_{u1}+\dots+\widetilde{\xi}_{ui}\big)\big],\end{align*}
  $k\in A_{ui}$ if and only if $Y_{k}\in I_{ui}$ ( $I_{ui}$ is then of length $\xi_{ui}$ ). We observe that for all $j\in[q]$ , $u\in\mathcal{U}$ , $i\in\mathbb{N}^{*}$ ,
  (2.1) \begin{equation} \mathbb{P}\big(j\in A_{ui}\mid j\in A_{u},\widetilde{\xi}_{ui}\big)=\widetilde{\xi}_{ui}. \end{equation}

In this definition, we imagine having q dots on the interval [0, 1], and we impose that dot j has the position $Y_{j}$ (for all j in [q]). During the fragmentation process, if we know that dot j is in the interval $I_{u}$ of length $\xi_{u}$ , then the probability that this dot is on $I_{ui}$ (for some i in $\mathbb{N}^{*}$ , $I_{ui}$ of length $\xi_{ui}$ ) is equal to $\xi_{ui}/\xi_{u}=\widetilde{\xi}_{ui}$ .

In the case $q=1$ , the branch $\{u\in\mathcal{U}\;:\; A_{u}\neq\emptyset\}$ has the same law as the randomly tagged branch of [Reference Bertoin3, Section 1.2.3]. The presentation is simpler in our case because the Malthusian exponent is 1 under Assumption 2.2.

2.3. Observation scheme

We freeze the process when the fragments become smaller than a given threshold $\varepsilon>0$ . That is, we have the data $(\xi_{u})_{u\in\mathcal{U}_{\varepsilon}}$ , where $\mathcal{U}_{\varepsilon}=\{u\in\mathcal{U},\,\xi_{{\boldsymbol{{m}}}(u)}\geq\varepsilon,\,\xi_{u}<\varepsilon\}$ .

We now look at q tagged fragments ( $q\in\mathbb{N}^{*}$ ). For each i in [q], we call $L_{0}^{(i)}=1, L_{1}^{(i)}, L_{2}^{(i)},\dots{}$ the successive sizes of the fragment having the tag i. More precisely, for each $n\in\mathbb{N}^{*}$ , there is almost surely exactly one $u\in\mathcal{U}$ such that $|u|=n$ and $i\in A_{u}$ ; and so, $L_{n}^{(i)}=\xi_{u}$ . For each i, the process $S_{0}^{(i)}=-\log\big(L_{0}^{(i)}\big)=0\leq S_{1}^{(i)}=-\log\big(L_{1}^{(i)}\big)\leq\cdots$ is a renewal process without delay, with waiting time following a law $\pi$ (see [Reference Asmussen1, Chapter V] for an introduction to renewal processes). The waiting times (for i in [q]) are $S_{0}^{(i)}, S_{1}^{(i)}-S_{0}^{(i)}, S_{2}^{(i)}-S_{1}^{(i)},\dots{}$ The renewal times (for i in [q]) are $S_{0}^{(i)},S_{1}^{(i)}, S_{2}^{(i)}, \dots{}$ The law $\pi$ is defined by the following:

For all bounded measurable $f\colon[0,1]\rightarrow[0,+\infty),$

(2.2) \begin{align} \qquad\qquad\qquad\qquad\qquad \int_{\mathcal{S}^{\downarrow}}\sum_{i=1}^{+\infty}s_{i}f(s_{i})\nu(\textrm{d}{\boldsymbol{{s}}})= \int_{0}^{+\infty}f(\textrm{e}^{-x})\pi(\textrm{d} x)\end{align}

(see [Reference Bertoin3, Proposition 1.6, p. 34], or [Reference Hoffmann and Krell16, (3) and (4), p. 398]). Under Assumptions 2.1 and 2.2, [Reference Bertoin3, Proposition 1.6] is true, even without the Malthusian hypothesis of [Reference Bertoin3].

We make the following assumption on $\pi$ .

We have added a comment about Assumption 2.3 in Remark 4.1. We believe that we could replace it by the following.

However, this would lead to difficult computations.

We set

(2.3) \begin{equation}T=-\log(\varepsilon).\end{equation}

We set, for all $i\in[q]$ , $t\geq0$ ,

(2.4) \begin{align} N_{t}^{(i)} & = \inf\big\{\,j\;:\; S_{j}^{(i)}>t\big\}, \nonumber \\[5pt] B_{t}^{(i)} & =S_{N_{t}^{(i)}}^{(i)}-t, \\[5pt] C_{t}^{(i)} & = t-S_{N_{t}^{(i)}-1}^{(i)} \nonumber \end{align}

(see Fig. 2 for an illustration). The processes $B^{(i)}$ , $C^{(i)}$ , and $N^{(i)}$ are time-homogeneous Markov processes [Reference Asmussen1, Proposition 1.5, p. 141]. All of them are càdlàg (i.e. right-continuous with a left-hand-side limit). We call $B^{(i)}$ the residual lifetime of the fragment tagged by i. We call $C^{(i)}$ the age of the fragment tagged by i. We call $N^{(i)}$ the number of renewals up to time t. In the following, we treat t as a time parameter. This has nothing to do with the time in which the fragmentation process X evolves.

Figure 2.

Process $B^{(1)}$ and $C^{(1)}$ .

We observe that, for all t, $\big(B_{t}^{(1)},\dots,B_{t}^{(q)}\big)$ is exchangeable, meaning that for all $\sigma$ in the symmetric group of order q, $\big(B_{t}^{(\sigma(1))},\dots,B_{t}^{(\sigma(q))}\big)$ has the same law as $\big(B_{t}^{(1)},\dots,B_{t}^{(q)}\big)$ . When we look at the fragments of sizes $(\xi_{u},\,u\in\mathcal{U}_{\varepsilon}\;:\; A_{u}\neq\emptyset)$ , we have almost the same information as when we look at $\big(B_{T}^{(1)},B_{T}^{(2)},\dots,B_{T}^{(q)}\big)$ . We say almost because knowing $\big(B_{T}^{(1)},B_{T}^{(2)},\dots,B_{T}^{(q)}\big)$ does not give exactly the number of u in $\mathcal{U}_{\varepsilon}$ such that $A_{u}$ is not empty.

In the remainder of Section 2 we define processes that will be useful when we describe the asymptotics of our model (in Section 4).

2.4. Stationary renewal processes $\big(\overline{B}^{(1)}$ , $\overline{B}^{(1),v}\big)$

We define $\widetilde{X}$ to be an independent copy of X. We suppose it has q tagged fragments. Therefore it has a mark $\big(\widetilde{\xi},\widetilde{A}\big)$ and renewal processes $\big(\widetilde{S}_{k}^{(i)}\big)_{k\geq0}$ (for all i in [q]) defined in the same way as for X. We let $\big(\widetilde{B}^{(1)},\widetilde{B}^{(2)}\big)$ be the residual lifetimes of the fragments tagged by 1 and 2.

Let $\mu=\int_{0}^{+\infty}x\pi(\textrm{d} x)$ , and let $\pi_{1}$ be the distribution with density $x\mapsto x/\mu$ with respect to $\pi$ . We set $\overline{C}$ to be a random variable of law $\pi_{1}$ ; U to be independent of $\overline{C}$ and uniform on (0, 1); and $\widetilde{S}_{-1}=\overline{C}(1-U)$ . The process $\overline{S}_{0}=\widetilde{S}_{-1}$ , $\overline{S}_{1}=\widetilde{S}_{-1}+\widetilde{S}_{0}^{(1)}$ , $\overline{S}_{2}=\widetilde{S}_{-1}+\widetilde{S}_{1}^{(1)}$ , $\overline{S}_{2}=\widetilde{S}_{-1}+\widetilde{S}_{2}^{(1)}$ , … is a renewal process with delay $\pi_{1}$ (with waiting times $\overline{S}_{0},\overline{S}_{1}-\overline{S}_{0}, \dots{}$ all smaller than b by Assumption 2.3). The renewal times are $\overline{S}_{0}, \overline{S}_{1}, \overline{S}_{2},\dots{}$ We set $(\overline{B}_{t}^{(1)})_{t\geq0}$ to be the residual lifetime process of this renewal process,

\begin{equation*} \overline{B}_{t}^{(1)}=\left\{\begin{array}{l@{\quad}l} \overline{C}(1-U)-t & \mbox{ if }t<\overline{S}_{0},\\[5pt] \inf_{n\geq0}\big\{\overline{S}_{n}\;:\;\overline{S}_{n}>t\big\}-t & \mbox{ if }t\geq\overline{S}_{0}; \end{array} \right. \end{equation*}

we define $(\overline{C}_{t}^{(1)})_{t\geq0}$ as

\begin{equation*} \overline{C}_{t}^{(1)}=\left\{\begin{array}{l@{\quad}l} \overline{C}U+t & \mbox{ if }t<\overline{S}_{0},\\[5pt] t-\sup_{n\geq0}\big\{\overline{S}_{n}\;:\;\overline{S}_{n}\leq t\big\} & \mbox{ if }t\geq\overline{S}_{0} \end{array} \right. \end{equation*}

(we call it the age process of our renewal process); and we set $\overline{N}_{t}^{(1)}=\inf\big\{\,j\;:\;\overline{S}_{j}>t\big\}$ .

Figure 3 provides a graphic representation of $\overline{B}_{\cdot}^{(1)}$ . It might be counter-intuitive to start with $\overline{B}_{0}^{(1)}$ having a law which is not $\pi$ in order to get a stationary process, but [Reference Asmussen1, Corollary 3.6, p. 153] is clear on this point: a delayed renewal process (with waiting time of law $\pi$ ) is stationary if and only if the distribution of the initial delay is $\eta$ (defined below).

Figure 3.

Renewal process with delay.

We define a measure $\eta$ on $\mathbb{R}^{+}$ by its action on bounded measurable functions:

For all bounded measurable f:

(2.5) \begin{align} \mathbb{R}^{+}\rightarrow\mathbb{R},\quad \eta(f)=\frac{1}{\mu}\int_{\mathbb{R}^{+}}\mathbb{E}(f(Y-s)\mathbf{1}_{\{Y-s>0\}})\,\textrm{d} s\quad (Y\sim\pi). \end{align}

Proof. We show the proof for $\overline{B}_{t}^{(1)}$ only. Let $\xi\geq0$ . We set $f(y)=\mathbf{1}_{y\geq\xi}$ , for all y in $\mathbb{R}$ . We have, with Y of law $\pi$ ,

\begin{align*} \frac{1}{\mu}\int_{\mathbb{R}^{+}}\mathbb{E}(f(Y-s)\mathbf{1}_{Y-s>0})\,\textrm{d} s &= \frac{1}{\mu}\int_{\mathbb{R}^{+}}\bigg(\int_{0}^{y}\mathbf{1}_{y-s\geq\xi}\,\textrm{d} s\bigg)\pi(\textrm{d} y) \\[5pt] & = \frac{1}{\mu}\int_{\mathbb{R}^{+}}(y-\xi)_{+}\pi(\textrm{d} y) \\[5pt] &= \int_{\xi}^{+\infty}\bigg(1-\frac{\xi}{y}\bigg)\frac{y}{\mu}\pi(\textrm{d} y) = \mathbb{P}(\overline{C}(1-U)\geq\xi). \end{align*}

We set $\eta_{2}$ to be the law of $\big(\overline{C}_{0}^{(1)},\overline{B}_{0}^{(1)}\big)=\big(\overline{C}U,\overline{C}(1-U)\big)$ . The support of $\eta_{2}$ is $\mathcal{C}\;:\!=\;\{(u,v)\in[0,b]^{2}\;:\; a\leq u+v\leq b\}$ .

For v in $\mathbb{R}$ , we now want to define a process

(2.6) \begin{align} \big(\overline{C}_{t}^{(1),v},\overline{B}_{t}^{(1),v}\big)_{t\geq v-2b}\text{ having the same} &\;\text{transition as } \nonumber \\[5pt] & \quad \big(C_{t}^{(1)},B_{t}^{(1)}\big)\text{ and being stationary.}\end{align}

We set $\big(\overline{C}_{v-2b}^{(1),v},\overline{B}_{v-2b}^{(1),v}\big)$ such that it has the law $\eta_{2}$ . As we have given its transition, the process $\big(\overline{C}_{t}^{(1),v},\overline{B}_{t}^{(1),v}\big)_{t\geq v-2b}$ is well defined in law. In addition, we suppose that it is independent of all the other processes. By Fact 2.1, the process $\big(\overline{C}_{t}^{(1),v},\overline{B}_{t}^{(1),v}\big)_{t\geq v-2b}$ is stationary.

We define the renewal times of $\overline{B}^{(1),v}$ by $\overline{S}_{1}^{(1),v}=\inf\big\{t\geq v-2b\;:\;\overline{B}_{t+}^{(1),v}\neq\overline{B}_{t-}^{(1),v}\big\}$ , and, by recurrence, $\overline{S}_{k}^{(1),v}=\inf\big\{t>\overline{S}_{k-1}^{(1),v}\;:\;\overline{B}_{t+}^{(1),v}\neq\overline{B}_{t-}^{(1),v}\big\}$ . We also define, for all t, $\overline{N}_{t}^{(1),v}=\inf\big\{j\;:\;\overline{S}_{j}^{(1),v}>t\big\}$ . As will be seen later, the processes $\overline{B}^{(1),v}$ and $\overline{B}^{(2),v}$ are used to define asymptotic quantities (see, for example, Proposition 4.1) and we need them to be defined on an interval $[v,+\infty)$ with v possibly in $\mathbb{R}^{-}$ . The process $\overline{B}^{(2),v}$ is defined below (Section 2.6).

Figure 4.

Processes $\widehat{B}^{(1),v}$ , $\widehat{B}^{(2),v}$ .

2.5. Tagged fragments conditioned to split up $\big(\widehat{B}^{(1),v},\widehat{B}^{(2),v}\big)$

For v in $[0,+\infty)$ , we define a process $\big(\widehat{B}_{t}^{(1),v},\widehat{B}_{t}^{(2),v}\big)_{t\geq0}$ such that

(2.7) \begin{align} \;\widehat{B}^{(1),v}=B^{(1)}\text{ and, with }B^{(1)}\text{ fixed, }& \widehat{B}^{(2),v}\text{ has the law of }B^{(2)}\text{ conditioned on } \nonumber \\[5pt] & \text{for all } u\in\mathcal{U},\ 1\in A_{u}\Rightarrow[2\in A_{u}\Leftrightarrow-\log(\xi_{u})\leq v], \end{align}

which reads as follows: the tag 2 remains on the fragment bearing the tag 1 until the size of the fragment is smaller than $\textrm{e}^{-v}$ . We observe that, conditionally on $\widehat{B}_{v}^{(1),v}$ and $\widehat{B}_{v}^{(2),v}$ , $\big(\widehat{B}_{v+\widehat{B}_{v}^{(1),v}+t}^{(1),v}\big)_{t\geq0}$ and $\big(\widehat{B}_{v+\widehat{B}_{v}^{(2),v}+t}^{(2),v}\big)_{t\geq0}$ are independent. We also define $\widehat{C}^{(1),v}=C^{(1)}$ . There is an algorithmic way to define $\widehat{B}^{(1),v}$ and $\widehat{B}^{(2),v}$ , which is illustrated in Fig. 4. Remember that $\widehat{B}^{(1),v}=B^{(1)}$ , and the definition of the mark $(\xi_{u},A_{u},I_{u})_{u\in\mathcal{U}}$ in Section 2.2. We call $\big(\widehat{S}_{j}^{(i)}\big)_{i=1,2;j\geq1}$ the renewal times of these processes (as before, they can be defined as the times when the right-hand-side and left-hand-side limits are not the same). If $\widehat{S}_{j}^{(1)}\leq v$ then $\widehat{S}_{j}^{(2)}=\widehat{S}_{j}^{(1)}$ . If k is such that $\widehat{S}_{k-1}^{(1)}\leq v$ and $\widehat{S}_{k}^{(1)}>v$ , we remember that

(2.8) \begin{equation} \exp\big(\widehat{S}_{k}^{(1)}-\widehat{S}_{k-1}^{(1)}\big)=\widetilde{\xi}_{ui}\end{equation}

for some u in $\mathcal{U}$ with $|u|=k-1$ and some i in $\mathbb{N}^{*}$ (because $\widehat{B}^{(1),v}=B^{(1)}$ ). We have points $Y_{1}, Y_{2}\in[0,1]$ such that $Y_{1}$ and $Y_{2}$ are in $I_{u}$ of length $\xi_{u}$ . Conditionally on $\{Y_{1},Y_{2}\in I_{u}\}$ , $Y_{1}$ and $Y_{2}$ are independent and uniformly distributed on $I_{u}$ . The interval $I_{ui}$ , of length $\xi_{u}\widetilde{\xi}_{ui}$ , is a sub-interval of $I_{u}$ such that $Y_{1}\in I_{ui}$ , because of (2.8). Then, for $r\in\mathbb{N}^{*}\backslash\{i\}$ , we want $Y_{2}$ to be in $I_{ur}$ with probability $\widetilde{\xi}_{ur}/\big(1-\widetilde{\xi}_{ui}\big)$ (because we want $2\notin A_{ui}$ ). So we take $\widehat{S}_{k}^{(2)}=\widehat{S}_{k-1}^{(1)}-\log\widetilde{\xi}_{ur}$ with probability $\widetilde{\xi}_{ur}/\big(1-\widetilde{\xi}_{ui}\big)$ ( $r\in\mathbb{N}^{*}\backslash\{i\}$ ).

The subsequent waiting times $\widehat{S}_{k+1}^{(1)}-\widehat{S}_{k}^{(1)},\widehat{S}_{k+1}^{(2)}-\widehat{S}_{k}^{(2)}, \dots{}$ are chosen independently of each other, each of them having the law $\pi$ . For j equal to 1 or 2 and t in $[0,+\infty)$ , we define $\widehat{N}_{t}^{(j)}=\inf\{i\;:\;\widehat{S}_{i}^{(j)}>t\}$ . We observe that, for $t\geq2b$ , $\widehat{N}_{t}^{(1)}$ is bigger than 2 (because of Assumption 2.3).

2.6. Two stationary processes after a split-up ( $\overline{B}^{(1),v},\overline{B}^{(2),v}$ )

Let k be an integer, $k\geq2$ , such that

(2.9) \begin{equation} k\times(b-a)\geq b.\end{equation}

Now we state a small lemma that will be useful in what follows. Remember that the process $\big(\overline{C}_{t}^{(1),v},\overline{B}_{t}^{(1),v}\big)_{t\geq v-2b}$ is defined in (2.6). The process $\big(\widehat{C}_{t}^{(1),v},\widehat{B}_{t}^{(1),v}\big)_{t\geq0}$ is defined in the previous section.

Proof. The law $\eta_{2}$ is the law of $\big(\overline{C}_{0}^{(1)},\overline{B}_{0}^{(1)}\big)$ ( $\eta_{2}$ is defined below Lemma 2.1). As previously stated, the support of $\eta_{2}$ is $\mathcal{C}$ ; so, by stationarity, the support of $\big(\overline{C}_{v}^{(1),v},\overline{B}_{v}^{(1),v}\big)$ is $\mathcal{C}$ .

Keep in mind that $\widehat{B}^{(1),v}=B^{(1)}$ , $\widehat{C}^{(1),v}=C^{(1)}$ . By Assumption 2.3, the support of $S_{k}^{(1)}$ is [ka, kb] and the support of $S_{k+1}^{(1)}-S_{k}^{(1)}$ is [a, b]. If $S_{k+1}^{(1)}>kb$ then $B_{kb}^{(1)}=S_{k+1}^{(1)}-S_{k}^{(1)}-\big(kb-S_{k}^{(1)}\big)$ and $C_{kb}^{(1)}=kb-S_{k}^{(1)}$ (see Fig. 5).

Figure 5.

$B_{kb}^{(1)}$ and $C_{kb}^{(1)}$ .

The support of $S_{k}^{(1)}$ is [ka, kb] and $kb-ka\geq b$ (see (2.9)), so, as $S_{k}^{(1)}$ and $S_{k+1}^{(1)}-S_{k}^{(1)}$ are independent, we get that the support of $\big(C_{kb}^{(1)},S_{k+1}^{(1)}-S_{k}^{(1)}\big)$ includes $\{(u,w)\in[0;\;b]^{2}\;:\; w\geq\sup(a,u)\}$ . Hence, the support of $\big(C_{kb}^{(1)},B_{kb}^{(1)}\big)=\big(C_{kb}^{(1)},S_{k+1}^{(1)}-S_{k}^{(1)}-C_{kb}^{(1)}\big)$ includes $\mathcal{C}$ . As this support is included in $\mathcal{C}$ , we have proved the desired result.

For v in $\mathbb{R}$ , we define a process $\big(\overline{B}_{t}^{(2),v}\big)_{t\geq v}$ . We start with:

(2.10) \begin{equation} \overline{B}_{v}^{(2),v}\text{ has the law } \eta'\big(\dots\mid\overline{C}_{v}^{(1),v},\overline{B}_{v}^{(1),v}\big)\end{equation}

(remember that $\eta'$ is defined in Fact 3.1). This conditioning is correct because the law of $\big(\overline{C}_{v}^{(1),v},\overline{B}_{v}^{(1),v}\big)$ is the law $\eta_{2}$ , whose support is included in the support of the law of $\big(\widehat{C}_{kb}^{(1),kb},\widehat{B}_{kb}^{(1),kb}\big)$ , which is $\eta_{2}$ (see the lemma above, and below (2.6)). We then let the process $\big(\overline{B}_{t}^{(1),v},\overline{B}_{t}^{(2),v}\big)_{t\geq v}$ run its course as a Markov process having the same transition as $\big(\widehat{B}_{t-v+kb}^{(1),kb},\widehat{B}_{t-v+kb}^{(2),kb}\big)_{t\geq v}$ . This means that, after time v, $\overline{B}_{t}^{(1),v}$ and $\overline{B}_{t}^{(2),v}$ decrease linearly (with slope $-1$ ) until they reach 0. When they reach 0, each of these two processes makes a jump of law $\pi$ , independently of the other one. After that, they decrease linearly, and so on.

4.1. Notation

We fix $q\in\mathbb{N}^{*}$ , and set $\mathcal{S}_{q}$ to be the symmetric group of order q. A function $F\;:\;\mathbb{R}^{q}\rightarrow\mathbb{R}$ is symmetric if, for all $\sigma\in\mathcal{S}_{q}$ and all $(x_{1},\dots,x_{q})\in\mathbb{R}^{q}$ ,

\begin{align*}F(x_{\sigma(1)},x_{\sigma(2)},\dots,x_{\sigma(q)})=F(x_{1},x_{2},\dots,x_{q}).\end{align*}

For $F\;:\;\mathbb{R}^{q}\rightarrow\mathbb{R}$ , we define a symmetric version of F by

(4.1) \begin{equation} F_{\text{sym}}(x_{1},\dots,x_{q}) = \frac{1}{q!}\sum_{\sigma\in\mathcal{S}_{q}}F(x_{\sigma(1)},\dots,x_{\sigma(q)})\qquad \text{for all }(x_{1},\dots,x_{q})\in\mathbb{R}^{q}.\end{equation}

We set $\mathcal{B}_{\text{sym}}(q)$ to be the set of bounded, measurable, symmetric functions F on $\mathbb{R}^{q}$ , and we set $\mathcal{B}_{\text{sym}}^{0}(q)$ to be the F of $\mathcal{B}_{\text{sym}}(q)$ such that

\begin{align*} \int_{x_{1}}F(x_{1},x_{2},\dots,x_{q})\eta(\textrm{d} x_{1}) = 0 \qquad \text{for all } (x_{2},\dots,x_{q})\in\mathbb{R}^{q-1}.\end{align*}

Suppose that k is in [q] and $l\geq1$ . For t in [0, T], we consider the following collections of nodes of $\mathcal{U}$ (remember that $T=-\log\varepsilon$ , and $\mathcal{U}$ and $\mathbf{m}(\!\cdot\!)$ are defined in Section 2.1):

(4.2) \begin{align} \mathcal{T}_{1} & = \{u\in\mathcal{U}\backslash\{0\}\;:\; A_{u}\neq\emptyset,\,\xi_{{\boldsymbol{{m}}}(u)}\geq\varepsilon\}\cup\{0\}, \nonumber \\[5pt] S(t) & =\{u\in\mathcal{T}_{1}\;:\;-\log(\xi_{{\boldsymbol{{m}}}(u)})\leq t,\,-\log(\xi_{u})>t\}=\mathcal{U}_{\textrm{e}^{-t}},\end{align}

(4.3) \begin{align} L_{t} & = \sum_{u\in S(t)\,:\,A_{u}\neq\emptyset}(\#A_{u}-1).\end{align}

We set $\mathcal{L}_{1}$ to be the set of leaves in the tree $\mathcal{T}_{1}$ . For t in [0, T] and i in [q], there exists one and only one u in S(t) such that $i\in A_{u}$ . We call it $u\{t,i\}$ . Under Assumption 2.3, there exists a constant bounding the numbers of vertices of $\mathcal{T}_{1}$ almost surely.

Figure 6.

Example tree and marks.

Let us consider the example in Fig. 6. Here, we have a graphical representation of a realization of $\mathcal{T}_{1}$ . Each node u of $\mathcal{T}_{1}$ is written above a rectangular box in which we read $A_{u}$ ; the right side of the box has the coordinate $-\log(\xi_{u})$ on the x-axis. For simplicity, the node (1,1) is designated by 11, the node (1,2) by 12, and so on. In this example:

\begin{align*} \mathcal{T}_{1} & = \{(0),(1),(2),(1,1),(2,1),(1,2),(1,1,1),(2,1,1),(1,1,1,1),(1,2,1)\}, \\[5pt] \mathcal{L}_{1} & = \{(2,1,1),(1,1,1,1),(1,2,1)\}, \\[5pt] A_{(1)} &= \{1,2,3\},\ A_{(1,2)}=\{1,2\},\ \dots{}, \\[5pt] S(t) &= \{(1,2),(1,1),(2,1)\}, \\[5pt] u\{t,1\} &=(1,2),\ u\{t,2\}=(1,2),\ u\{t,3\}=(1,1),\ u\{t,4\}=(2,1).\end{align*}

For $k, l \in \mathbb{N}$ and $t \in [0,T]$ , we define the event

\begin{align*} C_{k,l}(t)=\Bigg\{\sum_{u\in S(t)}\mathbf{1}_{\#A_{u}=1}=k,\,\sum_{u\in S(t)}(\#A_{u}-1)=l\Bigg\}.\end{align*}

For example, in Fig. 6, we are in the event $C_{2,1}(t)$ .

We define $\mathcal{T}_{2}=\{u\in\mathcal{T}_{1}\backslash\{0\}\;:\;\#A_{{\boldsymbol{{m}}}(u)}\geq2\}\cup\{0\}$ , $m_{2}\;:\; u\in\mathcal{T}_{2}\mapsto(\xi_{u},\inf\{i,i\in A_{u}\})$ . For example, in Fig. 6, $\mathcal{T}_{2}=\{(0),(1),(2),(1,1),(1,2),(1,2,1)\}$ . Let $\alpha$ be in (0, 1).

For any t, we can compute $\sum_{u\in S(t)}(\#A_{u}-1)$ if we know $\sum_{u\in S(t)}\mathbf{1}_{\#A_{u}=1}$ and $\#S(t)$ . As $T-\alpha T>b$ , any u in $S(\alpha T)$ satisfies $\#A_{u}\geq2$ if and only if u is the mother of some v in $\mathcal{T}_{2}$ . So we deduce that $C_{k,l}(\alpha T)$ is measurable with respect to $(\mathcal{T}_{2},m_{2})$ . We set, for all u in $\mathcal{T}_{2}$ ,

(4.4) \begin{equation} T_{u}=-\log(\xi_{u}).\end{equation}

For any i in [q], $t\mapsto u\{t,i\}$ is piecewise constant and the ordered sequence of its jump times is $S_{_{1}}^{(i)}<S_{2}^{(i)}<\cdots$ (the $S_{\dots}^{(i)}$ are defined in Section 2.3). We simply have that $1, \textrm{e}^{-S_{1}^{(i)}}, \textrm{e}^{-S_{2}^{i)}}, \dots$ are the successive sizes of the fragment supporting the tag i. For example, in Fig. 6, we have

(4.5) \begin{equation} S_{1}^{(1)}=-\log(\xi_{1}), \quad S_{2}^{(1)}=-\log(\xi_{(1,2)}), \quad S_{3}^{(1)}=-\log(\xi_{(1,2,1)}), \dots\end{equation}

Let $\mathcal{L}_{2}$ be the set of leaves u in the tree $\mathcal{T}_{2}$ such that the set $A_{u}$ has a single element $n_{u}$ . For example, in Fig. 6, $\mathcal{L}_{2}=\{(2),(1,1)\}$ . We observe that $\# \mathcal{L}_{1}=q\Leftrightarrow\# \mathcal{L}_{2}=q$ , and thus

(4.6) \begin{equation} \{\# \mathcal{L}_{1}=q\}\in\sigma(\mathcal{L}_{2}).\end{equation}

We summarize the definition of $n_{u}$ :

(4.7) \begin{equation} \#A_{u}=1\Rightarrow A_{u}=\{n_{u}\}.\end{equation}

For q even ( $q=2p$ ) and for all t in [0, T], we define the events

\begin{align*} & G_{t} =\{\text{for all } i\in[p],\text{ there exists } u_{i}\in\mathcal{U}\;:\;\xi_{u_{i}} \lt \textrm{e}^{-t},\,\xi_{{\boldsymbol{{m}}}(u_{i})}\geq \textrm{e}^{-t},\,A_{u_{i}}=\{2i-1,2i\}\}, \\[5pt] & \text{for all } i\in[p],\ G_{i,i+1}(t) = \{\text{there exists } u\in S(t)\;:\;\{2i-1,2i\}\subset A_{u}\}.\end{align*}

We set, for all t in [0, T], $\mathcal{F}_{S(t)}=\sigma(S(t),(\xi_{u},A_{u})_{u\in S(t)})$ .

4.2. Intermediate results

The reader must keep in mind that $T=-\log(\varepsilon)$ , (2.3), and that $\delta$ is defined in Assumption 2.3. The set $\mathcal{B}_{\textrm{sym}}^0(q)$ is defined in Section 4.1.

Lemma 4.1. We suppose that F is in $\mathcal{B}_{\textrm{sym}}^0(q)$ and that F is of the form $F=(f_{1}\otimes f_{2}\otimes\dots\otimes f_{q})_{\textrm{sym}}$ , with $f_{1}, f_{2}, \dots, f_{q} \in \mathcal{B}_{\textrm{sym}}^0(1)$ . Let A be in $\sigma(\mathcal{L}_{2})$ . For any $\alpha$ in ]0, 1[, k in [q], and l in $\{0,1,\dots,(q-k-1)_{+}\}$ , we have

\begin{align*} \big|\mathbb{E}(\mathbf{1}_{C_{k,l}(\alpha T)}\mathbf{1}_{A}F\big(B_{T}^{(1)},B_{T}^{(2)},\dots,B_{T}^{(q)}\big))\big| \leq \Vert F\Vert_{\infty}\Gamma_{1}^{q}C_\textrm{tree}(q)\bigg(\frac{1}{\delta}\bigg)^{q}\varepsilon^{q/2} \end{align*}

(for a constant $C_\textrm{tree}(q)$ defined in the proof, and $\Gamma_{1}$ defined in Corollary 3.1), and

\begin{align*} \varepsilon^{-q/2}\mathbb{E}\big(\mathbf{1}_{C_{k,l}(\alpha T)}\mathbf{1}_{A}F\big(B_{T}^{(1)},B_{T}^{(2)},\dots,B_{T}^{(q)}\big)\big) \underset{\varepsilon\rightarrow0}{\longrightarrow}0. \end{align*}

Proof. Let A be in $\sigma(\mathcal{L}_{2})$ . We have

\begin{align*} & \mathbb{E}\big(\mathbf{1}_{C_{k,l}(\alpha T)}\mathbf{1}_{A}F\big(B_{T}^{(1)},B_{T}^{(2)},\dots,B_{T}^{(q)}\big)\big) \\[5pt] & = \mathbb{E}\Bigg(\mathbf{1}_{A}\sum_{f : \mathcal{T}_{2}\rightarrow\mathcal{P}([q])\text{s.t.} \dots} \mathbb{E}\big(F\big(B_{T}^{(1)},B_{T}^{(2)},\dots,B_{T}^{(q)}\big) \mathbf{1}_{A_{u}=f(u),\,\text{for all }u\in\mathcal{T}_{2}}\mid\mathcal{L}_{2},\mathcal{T}_{2},m_{2}\big)\Bigg) \end{align*}

( $\mathcal{P}$ defined in Section 1.5), where we sum on the $f\;:\;\mathcal{T}_{2}\rightarrow\mathcal{P}([q])$ such that

(4.8) \begin{equation} \begin{cases} f(u)=\sqcup_{v\;:\;{\boldsymbol{{m}}}(v)=u}f(v)\qquad \text{for all}\; u \;\text{in} {\mathcal{T}_{2}},\\[5pt] \sum_{u\in S(\alpha T)}\mathbf{1}_{\#f(u)=1}=k\text{ and }\sum_{u\in S(\alpha T)}(\#f(u)-1)=l. \end{cases} \end{equation}

We remind the reader that $\sqcup$ is defined in Section 1.5 (disjoint union), ${\boldsymbol{{m}}}(\!\cdot\!)$ is defined in Section 2.1 (mother), and $S(\dots\!)$ is defined in (4.2). Here, we mean that we sum over the f compatible with a description of tagged fragments.

If $u \in \mathcal{L}_{2}$ and $T_{u}<T$ , then, conditionally on $\mathcal{T}_{2}$ and $m_{2}$ , $B_{T}^{(n_{u})}$ is independent of all the other variables and has the same law as $B_{T-T_{u}}^{(1)}$ ( $T_{u}$ is defined in (4.4), $n_{u}$ in (4.7)). Thus, using Theorem 3.1 and Corollary 3.1, we get, for any $\varepsilon'\in(0,\theta-1)$ , $u\in\mathcal{L}_{2}$ ,

\begin{align*} \big|\mathbb{E}\big(f_{n_{u}}\big(B_{T}^{(n_{u})}\big)|\mathcal{L}_{2},\mathcal{T}_{2},m_{2}\big)\big| & \leq \Gamma_{1}\Vert f_{n_{u}}\Vert_{\infty}\textrm{e}^{-(\theta-\varepsilon')(T-T_{u})_{+}} \\[5pt] & \text{for }T-T_{u}\notin Z_{0},\text{ where} \, {Z_{0}} \text{is of Lebesgue measure zero.} \end{align*}

Thus, we get

For a fixed $\omega$ and a fixed f, we have

\begin{equation*} \prod_{u\in\mathcal{L}_{2}}\textrm{e}^{-(T-T_{{\boldsymbol{{m}}}(u)})-\log(\delta)} \prod_{u\in\mathcal{T}_{2}\backslash\{0\}}\textrm{e}^{-(\#f(u)-1)(T_{u}-T_{{\boldsymbol{{m}}}(u)})} = \bigg(\frac{1}{\delta}\bigg)^{\#\mathcal{L}_{2}}\exp\bigg(-\int_{0}^{T}a(s)\,\textrm{d} s\bigg), \end{equation*}

where, for all s,

\begin{align*} a(s) &= \sum_{u\in\mathcal{L}_{2}\backslash\{0\}\;:\; T_{{\boldsymbol{{m}}}(u)}\leq s<T}\mathbf{1}_{\#f(u)=1} + \sum_{u\in\mathcal{T}_{2}\backslash\{0\}\;:\; T_{{\boldsymbol{{m}}}(u)}\leq s\leq T_{u}}(\#f(u)-1) \\[5pt] &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad \text{(if $u\in\mathcal{T}_{2}\backslash\mathcal{L}_{2}$, $\mathbf{1}_{\#f(u)=1}=0$)} \\[5pt] &= \sum_{u\in\mathcal{T}_{2}\backslash\{0\}\;:\; T_{{\boldsymbol{{m}}}(u)}\leq s<T}\mathbf{1}_{\#f(u)=1} + \sum_{u\in\mathcal{T}_{2}\backslash\{0\}\;:\; T_{{\boldsymbol{{m}}}(u)}\leq s\leq T_{u}}(\#f(u)-1) \\[5pt] &\;\,\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad \text{($S(\!\cdot\!)$ defined in (4.2)} \\[5pt] & \geq \sum_{u\in S(s)}\mathbf{1}_{\#f(u)=1}+\sum_{u\in S(s)}(\#f(u)-1). \end{align*}

We observe that, under (4.8),

\begin{align*} a(t) \geq \left\lceil \frac{q}{2}\right\rceil \quad\text{for all }t, \qquad a(\alpha T)\geq k+l, \end{align*}

and, if t is such that $\sum_{u\in S(t)}\mathbf{1}_{\#f(u)=1}=k'$ and $\sum_{u\in S(t)}(\#f(u)-1)=l'$ for some integers k ^′, l ^′, then, for all $s\geq t$ ,

\begin{align*} a(s)\geq k'+\left\lceil \frac{q-k'}{2}\right\rceil . \end{align*}

We observe that, under Assumption 2.3, there exists a constant which bounds $\#\mathcal{T}_{2}$ almost surely (because, for all u in $\mathcal{U}\backslash\{0\}$ , $-\log(\xi_{u})+\log(\xi_{\mathbf{m}(u)})\geq a$ ), and so there exists a constant $C_{tree}(q)$ which bounds $\#\{f\;:\;\mathcal{T}_{2}\rightarrow\mathcal{P}([q])\}$ almost surely. So, we have

(4.9) \begin{align} & \big|\mathbb{E}\big(\mathbf{1}_{A}F(B_{T}^{(1)},B_{T}^{(2)},\dots,B_{T}^{(q)})\big)\big| \nonumber \\[5pt] &\leq \Vert F\Vert_{\infty}\Gamma_{1}^{q} \mathbb{E}\Bigg(\sum_{f\;:\;\mathcal{T}_{2}\rightarrow\mathcal{P}([q])\,\text{s.t.} \dots} \mathbf{1}_{A} \bigg(\frac{1}{\delta}\bigg)^{\#\mathcal{L}_{2}}\textrm{e}^{-\lceil q/2\rceil\alpha T} \exp\bigg\{{-}\bigg(k+\left\lceil \frac{q-k}{2}\right\rceil \bigg)(T-\alpha T)\bigg\}\Bigg) \nonumber \\[5pt] &\leq \Vert F\Vert_{\infty}\Gamma_{1}^{q}C_\textrm{tree}(q) \bigg(\frac{1}{\delta}\bigg)^{q}\textrm{e}^{-\lceil q/2\rceil\alpha T} \exp\bigg\{{-}\bigg(k+\left\lceil \frac{q-k}{2}\right\rceil \bigg)(1-\alpha)T\bigg\}. \end{align}

Since $k\geq1$ , $k+\left\lceil ({q-k})/{2}\right\rceil \gt{q}/{2}$ , and so we have proved the desired result (remember that $T=-\log\varepsilon$ ).

Remember that $L_{t}$ ( $t\geq0$ ) is defined in (4.3).

Lemma 4.2. Let k be an integer, $k\geq q/2$ , and let $\alpha\in[q/(2k),1]$ . Then we have $\mathbb{P}(L_{\alpha T}\geq k)\leq K_{1}(q)\varepsilon^{q/2}$ , where

\begin{align*}K_{1}(q)=\sum_{i\in[q]}\frac{q!}{(q-i)!}\times i^{q-i}.\end{align*}

Let k be an integer, $k>q/2$ , and let $\alpha\in(q/(2k),1)$ . Then

\begin{align*} \varepsilon^{-q/2}\mathbb{P}(L_{\alpha T}\geq k)\underset{\varepsilon\rightarrow0}{\longrightarrow}0. \end{align*}

(We remind the reader that $T=-\log(\varepsilon)$ .)

Proof. Let k be an integer, $k\geq q/2$ , and let $\alpha\in[q/(2k),1]$ . Remember that $S(\!\cdot\!)$ is defined in (4.2). Observe that $\#S(\alpha T)=i$ if and only if $L_{\alpha T}=q-i$ (see (4.3)). We use the decomposition

\begin{align*} \{L_{\alpha T}\geq k\} &= \{L_{\alpha T}\in\{k,k+1,\dots,q-1\}\} \\[5pt] &= \cup_{i\in[q-k]}\{\#S(\alpha T)=i\} \\[5pt] &= \cup_{i\in[q-k]}\cup_{m:[i]\hookrightarrow[q]}(F(i,m)\cap\{\#S(\alpha T)=i\}) \end{align*}

(remember that ‘ $\hookrightarrow$ ’ means we are summing on injections; see Section 1.5), where

\begin{align*} F(i,m)=\{i_{1},i_{2}\in&[i]\text{ with }i_{1}\ne i_{2}\Rightarrow \\[5pt] & \text{there exists } u_{1},u_{2}\in S(\alpha T),u_{1}\ne u_{2},m(i_{1})\in A_{u_{1}},m(i_{2})\in A_{u_{2}}\}. \end{align*}

(To make the above equations easier to understand, observe that if $\#S(\alpha T)=i$ , we have, for each $j\in[i]$ , an index m(j) in $A_{u}$ for some $u\in S(\alpha T)$ , and we can choose m such that we are in the event F(i, m)). Suppose we are in the event F(i, m). For $u\in S(\alpha T)$ and for all j in [i] such that $m(j)\in A_{u}$ , we define (remember $|u|$ and $\mathbf{m}$ are defined in Section 2.1)

\begin{align*} T_{|u|}^{(j)}=-\log(\xi_{u}),\,T_{|u|-1}^{(j)}=-\log(\xi_{{\boldsymbol{{m}}}(u)}),\,\dots,\, T_{1}^{(j)}=-\log(\xi_{{\boldsymbol{{m}}}^{\circ(|u|-1)}(u)}),T_{0}^{(j)}=0, \end{align*}

with $l(j)=|u|$ , $v(j)=u$ . We have

\begin{align*} &\mathbb{P}(L_{\alpha T}\geq k)\leq\sum_{i\in[q-k]}\sum_{m:[i]\hookrightarrow[q]}\mathbb{P}(F(i,m)\cap\{\#S(\alpha T)=i\}) \\[5pt] &=\sum_{i\in[q-k]}\sum_{m\;:\;[i]\hookrightarrow[q]}\mathbb{E}\big(\mathbf{1}_{F(i,m)} \mathbb{E}\big(\mathbf{1}_{\#S(\alpha T)=i}\mid F(i,m),\big(T_{p}^{(j)}\big)_{j\in[i],p\in[l(j)]},(v(j))_{j\in[i]}\big)\big) \\[5pt] & \!\!\!\quad \qquad \quad \qquad\text{(below, we sum over the partitions $\mathcal{B}$ of $[q]\backslash m([i])$ into i subsets ${\mathcal{B}_{1},\mathcal{B}_{2},\dots,\mathcal{B}_{i}}$)} \\[5pt] & = \sum_{i\in[q-k]}\sum_{m\;:\;[i]\hookrightarrow[q]}\sum_{\mathcal{B}}\mathbb{E}\Bigg(\mathbf{1}_{F(i,m)} \mathbb{E}\Bigg(\prod_{j\in[i]}\prod_{r\in\mathcal{B}_{j}}\mathbf{1}_{r\in A_{v(j)}}\mid F(i,m), \big(T_{p}^{(j)}\big)_{j\in[i],p\in[l(j)]},(v(j))_{j\in[i]}\Bigg)\Bigg) \\[5pt] & \;\,\quad\qquad\qquad \qquad \qquad \qquad \qquad \qquad \text{ (as $Y_{1},\dots,Y_{q}$ defined in Section 2.2 are independent)}\end{align*}

\begin{align*} &= \sum_{i\in[q-k]}\sum_{m\;:\;[i]\hookrightarrow[q]}\sum_{\mathcal{B}} \mathbb{E}\Bigg(\mathbf{1}_{F(i,m)}\prod_{j\in[i]}\prod_{r\in\mathcal{B}_{j}}\mathbb{E}\big(\mathbf{1}_{r\in A_{v(j)}}\mid F(i,m), \big(T_{p}^{(j)}\big)_{j\in[i],p\in[l(j)]},(v(j))_{j\in[i]}\big)\Bigg) \\[5pt] &\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad\qquad\qquad\;\, \text{ (because of (2.1) and (4.4))} \\[5pt] &= \sum_{i\in[q-k]}\sum_{m\;:\;[i]\hookrightarrow[q]}\sum_{\mathcal{B}} \mathbb{E}\Bigg(\mathbf{1}_{F(i,m)}\prod_{j\in[i]}\prod_{r\in\mathcal{B}_{j}}\prod_{s=1}^{l(j)} \exp\big({-}T_{s}^{(j)}+T_{s-1}^{(j)}\big)\Bigg) \text{ (as $v(j)\in S(\alpha T)$)} \\[5pt] &\leq \sum_{i\in[q-k]}\sum_{m\;:\;[i]\hookrightarrow[q]}\sum_{\mathcal{B}} \prod_{j\in[i]}\prod_{r\in\mathcal{B}_{j}}\textrm{e}^{-\alpha T} \\[5pt] &= \sum_{i\in[q-k]}\sum_{m\;:\;[i]\hookrightarrow[q]}\sum_{\mathcal{B}}\textrm{e}^{-\alpha(q-i)T} \\[5pt] &\leq \sum_{i\in[q-k]}\sum_{m\;:\;[i]\hookrightarrow[q]}\sum_{\mathcal{B}}\textrm{e}^{-k\alpha T} \leq \textrm{e}^{-k\alpha T}\sum_{i\in[q]}\frac{q!}{(q-i)!}i^{q-i}. \end{align*}

If we suppose that $k>q/2$ and $\alpha\in(q/(2k),1)$ , then

\begin{equation*} \exp\bigg(\frac{qT}{2}\bigg)\exp({-}k\alpha T)\underset{T\rightarrow+\infty}{\longrightarrow}0. \end{equation*}

Immediate consequences of the two lemmas above are the following corollaries.

Corollary 4.1. If q is odd and if $F\in\mathcal{B}_{\textrm{sym}}^{0}(q)$ is of the form $F=(f_{1}\otimes\dots\otimes f_{q})_{\textrm{sym}}$ , then

\begin{align*} \varepsilon^{-q/2}\mathbb{E}\big(F\big(B_{T}^{(1)},\dots,B_{T}^{(q)}\big)\mathbf{1}_{\#\mathcal{L}_{1}=q}\big) \underset{\varepsilon\rightarrow0}{\longrightarrow}0. \end{align*}

( $\mathcal{B}_{\textrm{sym}}^{0}$ and $\mathcal{L}_{1}$ are defined in Section 4.1 .)

Proof. We take $\alpha\in(({q}/{2})\lceil{q}/{2}\rceil^{-1},1)$ . We observe that, for k in [q], t in (0, T),

\begin{align*} \sum_{u\in S(t)}\mathbf{1}_{\#A_{u}=1} = k \Rightarrow \sum_{u\in S(t)}(\#A_{u}-1)\in\{0,1,\dots,(q-k-1)_{+}\}, \end{align*}

and ( $L_{t}$ is defined in (4.3)) $\sum_{u\in S(t)}\mathbf{1}_{\#A_{u}=1}=0\Rightarrow L_{t}\geq\lceil{q}/{2}\rceil$ . So, we can use the decomposition

(4.10) \begin{align} & \varepsilon^{-q/2}\big|\mathbb{E}\big(F\big(B_{T}^{(1)},\dots,B_{T}^{(q)}\big)\mathbf{1}_{\#\mathcal{L}_{1}=q}\big)\big| \nonumber \\[5pt] & = \Bigg|\varepsilon^{-q/2}\sum_{k\in[q]}\sum_{l\in\{0,1,\dots,(q-k-1)_{+}\}} \mathbb{E}\big(\mathbf{1}_{C_{k,l}(\alpha T)}\mathbf{1}_{\#\mathcal{L}_{1}=q}F\big(B_{T}^{(1)},\dots,B_{T}^{(q)}\big)\big) \nonumber \\[5pt] &\qquad + \varepsilon^{-q/2}\mathbb{E}\big(\mathbf{1}_{L_{\alpha T}\geq\left\lceil q/2\right\rceil }\mathbf{1}_{\#\mathcal{L}_{1}=q} F\big(B_{T}^{(1)},\dots,B_{T}^{(q)}\big)\big)\Bigg| \underset{\varepsilon\rightarrow0}{\longrightarrow}0 \end{align}

(by (4.6) and Lemmas 4.1 and 4.2).

( $\mathcal{L}_{1}$ and $\mathcal{L}_{2}$ are defined in Section 4.1.)

Corollary 4.2. Suppose $F\in\mathcal{B}_\textrm{sym}^{0}(q)$ is of the form $F=(f_{1}\otimes\dots\otimes f_{q})_{\textrm{sym}}$ . Let $A \in \sigma(\mathcal{L}_{2})$ . Then

\begin{align*} \big|\mathbb{E}\big(F\big(B_{T}^{(1)},\dots,B_{T}^{(q)}\big)\mathbf{1}_{A}\big)\big| \leq \Vert F\Vert_{\infty}\varepsilon^{q/2}\bigg\{ K_{1}(q)+\Gamma_{1}^{q}C_{\textrm{tree}}(q) \bigg(\frac{1}{\delta}\bigg)^{q}(q+1)^{2}\bigg\}. \end{align*}

Proof. We get, as in (4.10),

\begin{align*} &\big|\mathbb{E}\big( F\big(B_{T}^{(1)},\dots,B_{T}^{(q)}\big)\mathbf{1}_{A}\big)\big| \\[5pt] & \qquad = \Bigg|\mathbb{E}\Bigg(F\big(B_{T}^{(1)},\dots,B_{T}^{(q)}\big)\mathbf{1}_{A}\Bigg( \mathbf{1}_{L_{\alpha T}\geq q/2}+\sum_{k'\in[q]}\sum_{0\leq l\leq(q-k'-1)_{+}}\mathbf{1}_{C_{k',l}(\alpha T)}\Bigg)\Bigg)\Bigg|\\[5pt] & \;\,\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \text{(from Lemmas 4.1 and 4.2)} \\[5pt] & \qquad \leq\Vert F\Vert_{\infty}\varepsilon^{q/2} \Bigg\{K_{1}(q)+\Gamma_{1}^{q}C_{\textrm{tree}}(q)\bigg(\frac{1}{\delta}\bigg)^{q} \sum_{k'\in[q]}1+(q-k'-1)_{+}\Bigg\}, \end{align*}

and $\sum_{k'\in[q]}1+(q-k'-1)_{+}\leq(q+1)^{2}$ (see Section B for a detailed proof).

We now want to find the limit of $\varepsilon^{-q/2}\mathbb{E}\big(\mathbf{1}_{L_{T}\leq q/2}\mathbf{1}_{\#\mathcal{L}_{1}=q}F\big(B_{T}^{(1)},\dots,B_{T}^{(q)}\big)\big)$ when $\varepsilon$ goes to 0, for q even. First we need a technical lemma.

For any i, the process $\big(B_{t}^{(i)}\big)$ has a stationary law (see [Reference Asmussen1, Theorem 3.3 p. 151]). Let $B_{\infty}$ be a random variable having this stationary law $\eta$ (it has already appeared in Section 3). We can always suppose that it is independent of all the other variables.

Lemma 4.3. Let $f_{1} , f_{2}$ be in $\mathcal{B}_\textrm{sym}^{0}(1)$ . Let $\alpha$ belong to (0, 1), and $\varepsilon'$ belong to $(0,\theta-1)$ ( $\theta$ is defined in Fact 3.1). We have

(4.11) \begin{equation} \int_{-\infty}^{-\log(\delta)}\textrm{e}^{-v} \big|\mathbb{E}\big(f_{1}\big(\overline{B}_{0}^{(1),v}\big)f_{2}\big(\overline{B}_{0}^{(2),v}\big)\big)\big|\,\textrm{d} v \lt \infty , \end{equation}

and, almost surely, for T large enough,

\begin{align*} & \bigg|\textrm{e}^{T-\alpha T-B_{\alpha T}^{(1)}}\mathbb{E}\big(f_{1}\otimes f_{2}\big(B_{T,}^{(1)}B_{T}^{(2)}\big) \mathbf{1}_{G_{1,2}(T)^{\textrm{c}}}\mid\mathcal{F}_{S(\alpha T)},G_{\alpha T}\big) \\[5pt] & \qquad\qquad\qquad -\int_{-\infty}^{-\log(\delta)}\textrm{e}^{-v}\mathbb{E}\big(\mathbf{1}_{v\leq\overline{B}_{0}^{(1),v}} f_{1}\big(\overline{B}_{0}^{(1),v}\big)f_{2}\big(\overline{B}_{0}^{(2),v}\big)\big)\,\textrm{d} v\bigg| \\[5pt] &\qquad\qquad\qquad\qquad\qquad\qquad \leq \Gamma_{2}\Vert f_{1}\Vert_{\infty}\Vert f_{2}\Vert_{\infty} \exp\bigg({-}(T-\alpha T)\bigg(\frac{\theta-\varepsilon'-1}{2}\bigg)\bigg), \end{align*}

where

\begin{align*} \Gamma_{2}=\frac{\Gamma_{1}^{2}}{\delta^{2+2(\theta-\varepsilon')}(2(\theta-\varepsilon')-1)} + \frac{\Gamma_{1}}{\delta^{\theta-\varepsilon'}} + \frac{\Gamma_{1}^{2}}{\delta^{2(\theta-\varepsilon')}(2(\theta-\varepsilon')-1)}. \end{align*}

(The processes $B^{(1)}$ , $B^{(2)}$ , $\overline{B}^{(1),v}$ , and $\overline{B}^{(2),v}$ are defined in Sections 2.3, 2.4, and 2.6 .)

Proof. We have, for all s in $\big[\alpha T+B_{\alpha T}^{(1)},T\big]$ (because of (2.1) and (4.4)),

\begin{align*} \mathbb{P}\big(u\{s,2\}=u\{s,1\}\mid\mathcal{F}_{S(\alpha T)},G_{\alpha T},\big(S_{j}^{(1)}\big)_{j\geq1}\big) = \exp\big[{-}\big(s+B_{s}^{(1)}-\big(\alpha T+B_{\alpha T}^{(1)}\big)\big)\big] \end{align*}

(we remind the reader that $u\{s,1\}$ , $G_{1,2}$ , are defined in Section 4.1, below (4.3)). Let us introduce the breaking time $\tau_{1,2}$ between 1 and 2 as a random variable having the following property: conditionally on $\mathcal{F}_{S(\alpha T)}$ , $G_{\alpha T}$ , and $\big(S_{j}^{(1)}\big)_{j\geq1}$ , $\tau_{1,2}$ has the density

\begin{align*} s\in\mathbb{R}\mapsto\mathbf{1}_{[\alpha T+B_{\alpha T}^{(1)},+\infty)}(s)\textrm{e}^{-(s-(\alpha T+B_{T}^{(1)}))} \end{align*}

(this is a translation of an exponential law). We have the equalities $\alpha T+B_{\alpha T}^{(1)}=S_{j_{0}}^{(1)}$ for some $j_{0}$ , and $T+B_{T}^{(1)}=S_{i_{0}}^{(1)}$ for some $i_{0}$ . Here, we need to comment on the definitions of Section 4.1. In Fig. 6 we have $-\log(\xi_{(1,2)})=S_{2}^{(1)}$ (as in (4.5)), $S(S_{2}^{(1)})=\{(1,2,1),(1,1,1),(2,1)\}$ , and $u\{-\!\log(\xi_{(1,2)}),1\}=\{(1,2,1)\}$ . It is important to understand this example before reading what follows. The breaking time $\tau_{1,2}$ has the following interesting property (for all $k\geq j_{0}$ ):

\begin{align*} \mathbb{P}\big(u\big\{S_{k}^{(1)},2\big\} \neq u\big\{S_{k}^{(1)},1\big\} & \mid \mathcal{F}_{S(\alpha T)},G_{\alpha T},\big(S_{j}^{(1)}\big)_{j\geq1}\big) \\[5pt] & = \mathbb{P}\big(\tau_{1,2}\in\big[\alpha T+B_{\alpha T}^{(1)},S_{k}^{(1)}\big] \mid \mathcal{F}_{S(\alpha T)},G_{\alpha T},\big(S_{j}^{(1)}\big)_{j\geq1}\big). \end{align*}

Just because we can, we impose, for all $k\geq j_{0}$ , conditionally on $\mathcal{F}_{S(\alpha T)}$ , $G_{\alpha T}$ , and $\big(S_{j}^{(1)}\big)_{j\geq1}$ ,

\begin{align*} \big\{u\big\{S_{k}^{(1)},2\big\} \neq u\big\{S_{k}^{(2)},1\big\}\big\} = \big\{\tau_{1,2}\in\big[\alpha T+B_{\alpha T}^{(1)},S_{k}^{(1)}\big]\big\}. \end{align*}

Now, let v be in $\big[\alpha T+B_{\alpha T}^{(1)},T+B_{T}^{(1)}\big]$ . We observe that, for all v in $\big[\alpha T+B_{\alpha T}^{(1)},T+B_{T}^{(1)}\big]$ ,

\begin{align*} \qquad\qquad\qquad \mathbb{E}\big(f_{1}\otimes f_{2}\big(B_{T,}^{(1)}&B_{T}^{(2)}\big)\mathbf{1}_{G_{1,2}(T)^{\textrm{c}}} \mid \mathcal{F}_{S(\alpha T)},G_{\alpha T},\big(S_{j}^{(1)}\big)_{j\geq1},\tau_{1,2}=v\big) \\[5pt] &= \mathbb{E}\big(f_{1}\otimes f_{2}\big(\widehat{B}_{T,}^{(1),v}\widehat{B}_{T}^{(2),v}\big) \mid \mathcal{F}_{S(\alpha T)},G_{\alpha T},\big(S_{j}^{(1)}\big)_{j\geq1}\big) \\[5pt] &\;\,\qquad\quad\qquad\qquad\qquad\quad\qquad\qquad\qquad\qquad \text{(because of (2.7)).} \end{align*}

And so,

\begin{align*} & \mathbb{E}\big(f_{1}\otimes f_{2}\big(B_{T,}^{(1)}B_{T}^{(2)}\big)\mathbf{1}_{G_{1,2}(T)^{\textrm{c}}} \mid \mathcal{F}_{S(\alpha T)},G_{\alpha T}\big) \\[5pt] & =\mathbb{E}\big(\mathbb{E}\big(f_{1}\otimes f_{2}\big(B_{T,}^{(1)}B_{T}^{(2)}\big)\mathbf{1}_{G_{1,2}(T)^{\textrm{c}}} \mid \mathcal{F}_{S(\alpha T)},G_{\alpha T},\big(S_{j}^{(1)}\big)_{j\geq1}\big) \mid \mathcal{F}_{S(\alpha T)},G_{\alpha T}\big) \\[5pt] & = \mathbb{E}\big(\mathbb{E}\big(\mathbb{E}\big(f_{1}\otimes f_{2}\big(B_{T,}^{(1)}B_{T}^{(2)}\big)\mathbf{1}_{G_{1,2}(T)^{\textrm{c}}} \mid \mathcal{F}_{S(\alpha T)},G_{\alpha T},\big(S_{j}^{(1)}\big)_{j\geq1},\tau_{1,2}\big) \mid \mathcal{F}_{S(\alpha T)},G_{\alpha T},\big(S_{j}^{(1)}\big)_{j\geq1}\big) \\[5pt] &\qquad \qquad \qquad \mid \mathcal{F}_{S(\alpha T)},G_{\alpha T}\big) \\[5pt] & \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \!\!\!\qquad\qquad \text{(keep in mind that $\widehat{B}^{(1),v}=B^{(1)}$ for all v)}\end{align*}

\begin{align*}& = \mathbb{E}\bigg(\!\mathbb{E}\bigg(\!\int_{\alpha T+B_{\alpha T}^{(1)}}^{T+B_{T}^{(1)}} \!\textrm{e}^{-\big(v-\alpha T-\widehat{B}_{\alpha T}^{(1),v}\big)} \mathbb{E}\big(f_{1}\!\otimes f_{2}\big(B_{T,}^{(1)}B_{T}^{(2)}\big)\mathbf{1}_{G_{1,2}(T)^{\textrm{c}}}\! \mid \mathcal{F}_{S(\alpha T)},\!G_{\alpha T},\!\big(S_{j}^{(1)}\big)_{j\geq1},\!\tau_{1,2}=v\big)\,\textrm{d} v \\[5pt] & \qquad \qquad \qquad \mid \mathcal{F}_{S(\alpha T)},G_{\alpha T},\big(S_{j}^{(1)}\big)_{j\geq1}\bigg) \mid \mathcal{F}_{S(\alpha T)},G_{\alpha T}\bigg) \\[5pt] & = \mathbb{E}\bigg(\mathbb{E}\bigg(\int_{\alpha T+B_{\alpha T}^{(1)}}^{T+B_{T}^{(1)}} \textrm{e}^{-(v-\alpha T-\widehat{B}_{\alpha T}^{(1),v})}\mathbb{E}\bigg(f_{1}\big(\widehat{B}_{T}^{(1),v}\big) f_{2}\big(\widehat{B}_{T}^{(2),v}\big) \mid \mathcal{F}_{S(\alpha T)},G_{\alpha T},\big(S_{j}^{(1)}\big)_{j\geq1}\bigg)\,\textrm{d} v \\[5pt] & \qquad \qquad \qquad \mid \mathcal{F}_{S(\alpha T)},G_{\alpha T},\big(S_{j}^{(1)}\big)_{j\geq1}\bigg) \mid \mathcal{F}_{S(\alpha T)},G_{\alpha T}\bigg) \\[5pt] & = \mathbb{E}\bigg(\int_{\alpha T+B_{\alpha T}^{(1)}}^{T+B_{T}^{(1)}}\textrm{e}^{-(v-\alpha T-\widehat{B}_{\alpha T}^{(1),v})} f_{1}\big(\widehat{B}_{T}^{(1),v}\big)f_{2}\big(\widehat{B}_{T}^{(2),v}\big)\,\textrm{d} v \mid \mathcal{F}_{S(\alpha T)},G_{\alpha T}\bigg) . \end{align*}

Let us split the above integral into two parts and multiply them by $\textrm{e}^{T-\alpha T-B_{\alpha T}^{(1)}}$ . For the first part:

(4.12) \begin{align} & \bigg|\textrm{e}^{T-\alpha T-B_{\alpha T}^{(1)}}\mathbb{E}\bigg(\int_{\alpha T+B_{\alpha T}^{(1)}}^{(T+\alpha T)/2} \textrm{e}^{-\big(v-\alpha T-\widehat{B}_{\alpha T}^{(1),v}\big)}f_{1}\big(\widehat{B}_{T}^{(1),v}\big) f_{2}\big(\widehat{B}_{T}^{(2),v}\big)\,\textrm{d} v \mid \mathcal{F}_{S(\alpha T)},G_{\alpha T}\bigg)\bigg| \nonumber \\[5pt] &= \textrm{e}^{T-\alpha T-B_{\alpha T}^{(1)}}\bigg|\mathbb{E}\bigg(\int_{\alpha T+B_{\alpha T}^{(1)}}^{(T+\alpha T)/2} \!\textrm{e}^{-\big(v-\alpha T-\widehat{B}_{\alpha T}^{(1),v}\big)} \mathbb{E}\big(f_{1}\big(\widehat{B}_{T}^{(1),v}\big)f_{2}\big(\widehat{B}_{T}^{(2),v}\big)\! \mid\! \widehat{B}_{v}^{(1),v},\widehat{B}_{v}^{(2),v},\mathcal{F}_{S(\alpha T)},\!G_{\alpha T}\big)\textrm{d} v\nonumber \\[5pt] &\qquad \qquad \qquad \qquad \mid\mathcal{F}_{S(\alpha T)},G_{\alpha T}\bigg)\bigg| \nonumber \\[5pt] & \qquad\qquad\qquad\qquad \qquad\;\, \text{(using the fact that $\widehat{B}_{T}^{(1),v}$ and $\widehat{B}_{T}^{(2),v}$ are independent conditionally on} \nonumber \\[5pt] & \qquad\qquad\qquad\qquad\qquad \qquad\qquad\qquad\text{$\big\{\widehat{B}_{v}^{(1),v},\widehat{B}_{v}^{(2),v},\mathcal{F}_{S(\alpha T)},G_{\alpha T}\big\}$ if $T\geq v-\log(\delta)$, we get,} \nonumber \\[5pt] & \;\;\quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \!\!\!\!\qquad\quad\qquad \text{by Theorem 3.1, Corollary 3.1, and Fact 4.2)} \nonumber \\[5pt] & \leq \textrm{e}^{T-\alpha T-B_{\alpha T}^{(1)}} \mathbb{E}\bigg(\int_{\alpha T+B_{\alpha T}^{(1)}}^{(T+\alpha T)/2} \textrm{e}^{-(v-\alpha T-\widehat{B}_{\alpha T}^{(1)})} \nonumber \\[5pt] & \qquad \qquad \qquad \big(\Gamma_{1}\Vert f_{1}\Vert_{\infty} \textrm{e}^{-(\theta-\varepsilon')(T-v-\widehat{B}_{v}^{(1),v})_{+}} \Gamma_{1}\Vert f_{2}\Vert_{\infty} \textrm{e}^{-(\theta-\varepsilon')(T-v-\widehat{B}_{v}^{(2),v})_{+}}\big)\,\textrm{d} v \mid \mathcal{F}_{S(\alpha T)},G_{\alpha T}\bigg) \nonumber \\[5pt] & \,\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\text{(using Assumption 2.3)} \nonumber \\[5pt] &\leq\Gamma_{1}^{2}\Vert f_{1}\Vert_{\infty}\Vert f_{2}\Vert_{\infty}\textrm{e}^{(T-\alpha T-\log(\delta))} \int_{\alpha T}^{(T+\alpha T)/2}\textrm{e}^{-(v-\alpha T+\log(\delta))} \textrm{e}^{-2(\theta-\varepsilon')(T-v+\log(\delta))}\,\textrm{d} v \nonumber \\[5pt] & = \frac{\Gamma_{1}^{2}\Vert f_{1}\Vert_{\infty}\Vert f_{2}\Vert_{\infty}}{\delta^{2+2(\theta-\varepsilon')}} \textrm{e}^{(T-2(\theta-\varepsilon')T)}\bigg[ \frac{\textrm{e}^{(2(\theta-\varepsilon')-1)v}}{2(\theta-\varepsilon')-1}\bigg]_{\alpha T}^{(T+\alpha T)/2} \nonumber \\[5pt] & \leq \frac{\Gamma_{1}^{2}\Vert f_{1}\Vert_{\infty}\Vert f_{2}\Vert_{\infty}}{\delta^{2+2(\theta-\varepsilon')}} \frac{\exp(\!-\!(2(\theta-\varepsilon')-1)T+(2(\theta-\varepsilon')-1){(T+\alpha T)}/{2})}{2(\theta-\varepsilon')-1} \nonumber \\[9pt] & = \frac{\Gamma_{1}^{2}\Vert f_{1}\Vert_{\infty}\Vert f_{2}\Vert_{\infty}}{\delta^{2+2(\theta-\varepsilon')}} \frac{\exp(\!-\!(2(\theta-\varepsilon')-1)({(T-\alpha T)}/{2}))}{2(\theta-\varepsilon')-1}. \end{align}

For the second part, minus some other terms:

(4.13) \begin{align} & \left| \begin{array}{c} \underbrace{\textrm{e}^{T-\alpha T-B_{\alpha T}^{(1)}}\mathbb{E}\bigg(\int_{(T+\alpha T)/2}^{T+B_{T}^{(1)}} \textrm{e}^{-(v-\alpha T-B_{\alpha T}^{(1)})}f_{1}\big(\widehat{B}_{T}^{(1),v}\big) f_{2}\big(\widehat{B}_{T}^{(2),v}\big) \, \textrm{d} v \mid \mathcal{F}_{S(\alpha T)},G_{\alpha T}\bigg)} \\[7pt] \text{ second part} \end{array}\right. \nonumber \\[7pt] & \left. \qquad \qquad \qquad \qquad -\,\begin{array}{c} \underbrace{\int_{(T+\alpha T)/2}^{T-\log(\delta)}\textrm{e}^{-(v-T)}\mathbb{E}\bigg(\mathbf{1}_{v\leq T+\overline{B}_{T}^{(1),v}} f_{1}\big(\overline{B}_{T}^{(1),v}\big)f_{2}\big(\overline{B}_{T}^{(2),v}\big)\bigg)\,\textrm{d} v} \\[7pt] ({ \heartsuit}) \end{array}\right| \nonumber \\[7pt] & = \bigg|\textrm{e}^{T-\alpha T-B_{\alpha T}^{(1)}} \mathbb{E}\bigg(\int_{(T+\alpha T)/2}^{T-\log(\delta)} \textrm{e}^{-(v-\alpha T-B_{\alpha T}^{(1)})}\mathbf{1}_{v\leq T+B_{T}^{(1)}}f_{1}\big(\widehat{B}_{T}^{(1),v}\big) f_{2}\big(\widehat{B}_{T}^{(2),v}\big) \, \textrm{d} v \mid \mathcal{F}_{S(\alpha T)},G_{\alpha T}\bigg) \nonumber \\[7pt] & \qquad - \textrm{e}^{T-\alpha T-B_{\alpha T}^{(1)}}\mathbb{E}\bigg(\int_{(T+\alpha T)/2}^{T-\log(\delta)} \textrm{e}^{-(v-\alpha T-B_{\alpha T}^{(1)})}\mathbf{1}_{v\leq T+\overline{B}_{T}^{(1),v}}f_{1}\big(\overline{B}_{T}^{(1),v}\big) f_{2}\big(\overline{B}_{T}^{(2),v}\big)\,\textrm{d} v\bigg)\bigg| \nonumber \\[7pt] & = \textrm{e}^{T-\alpha T-B_{\alpha T}^{(1)}}\bigg|\int_{(T+\alpha T)/2}^{T-\log(\delta)} \textrm{e}^{-(v-\alpha T-B_{\alpha T}^{(1)})}\mathbb{E}\big(\mathbb{E}\big(\mathbf{1}_{v\leq T+B_{T}^{(1)}} f_{1}\big(\widehat{B}_{T}^{(1),v}\big)f_{2}(\widehat{B}_{T}^{(2),v}) \nonumber \\[7pt] & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \mid \widehat{C}_{v}^{(1),v},\mathcal{F}_{S(\alpha T)},G_{\alpha T}\big) \mid \mathcal{F}_{S(\alpha T)},G_{\alpha T}\big)\,\textrm{d} v \nonumber \\[7pt] & \qquad \qquad -\int_{(T+\alpha T)/2}^{T-\log(\delta)}\textrm{e}^{-(v-\alpha T-B_{\alpha T}^{(1)})} \mathbb{E}\big(\mathbb{E}\big(\mathbf{1}_{v\leq T+\overline{B}_{T}^{(1),v}}f_{1}\big(\overline{B}_{T}^{(1),v}\big) f_{2}\big(\overline{B}_{T}^{(2),v}\big) \mid \overline{C}_{v}^{(1),v}\big)\big)\,\textrm{d} v\bigg|. \end{align}

We observe that, for all v in $[(T+\alpha T)/2,T-\log(\delta)]$ , once $\widehat{C}_{v}^{(1),v}$ is fixed, we can make a simulation of $\widehat{B}_{T}^{(1),v}=B_{T}^{(1)}$ and $\widehat{B}_{T}^{(2),v}$ (these processes are independent of $\mathcal{F}_{S(\alpha T)},G_{\alpha T}$ conditionally on $\widehat{C}_{v}^{(1),v}$ ). Indeed, we draw $\widehat{B}_{v}^{(1),v}$ conditionally on $\widehat{C}_{v}^{(1),v}$ (with law $\eta_{1}\big(\,\cdot\mid\widehat{C}_{v}^{(1),v}\big)$ defined in Fact 2.2), then we draw $\widehat{B}_{v}^{(2),v}$ conditionally on $\widehat{B}_{v}^{(1),v}$ and $\widehat{C}_{v}^{(1),v}$ (with law $\eta'\big(\,\cdot\mid\widehat{B}_{v}^{(1),v},\widehat{C}_{v}^{(1),v}\big)$ , see Fact 2.2). Then, $\big(\widehat{B}_{t}^{(1),v}\big)_{t\geq v}$ and $\big(\widehat{B}_{t}^{(2),v}\big)_{t\geq v}$ run their courses as independent Markov processes, until we get $\widehat{B}_{T}^{(1),v}$ , $\widehat{B}_{T}^{(2),v}$ .

In the same way (for all v in $[(T+\alpha T)/2,T-\log(\delta)]$ ), we observe that the process $\big(\overline{C}^{(1),v},\overline{B}^{(1),v}\big)$ starts at time $v-2b$ and has the same transition as $\big(C^{(1)},B^{(1)}\big)$ (see (2.6)). By Assumption 2.1, the following time exists: $S=\sup\big\{t\;:\; v-b\leq t\leq v\,,\,\overline{C}_{t}^{(1),v}=0\big\}$ . We then have $v-S=\overline{C}_{v}^{(1),v}$ . When $\overline{C}_{v}^{(1),v}$ is fixed, this entails that $\overline{B}_{v}^{(1),v}$ has the law $\eta_{1}\big(\,\cdot\mid\overline{C}_{v}^{(1),v}\big)$ . We have $\overline{B}_{v}^{(2),v}$ of law $\eta'\big(\,\cdot\mid\overline{C}_{v}^{(1),v},\overline{B}_{v}^{(1),v}\big)$ (by (2.10)). As before, we then let the process $\big(\overline{B}_{t}^{(1),v},\overline{B}_{t}^{(2),v}\big)_{t\geq v}$ run its course as a Markov process having the same transition as $\big(\widehat{B}_{t-v+kb}^{(1),kb},\widehat{B}_{t-v+kb}^{(2),kb}\big)_{t\geq v}$ until we get $\overline{B}_{T}^{(1),v}$ , $\overline{B}_{T}^{(2),v}$ .

So we get that (for all v in $[(T+\alpha T)/2,T-\log(\delta)]$ )

\begin{align*} & \mathbb{E}\big(\mathbf{1}_{v\leq T+B_{T}^{(1)}}f_{1}\big(\widehat{B}_{T}^{(1),v}\big) f_{2}\big(\widehat{B}_{T}^{(2),v}\big) \mid \widehat{C}_{v}^{(1),v},\mathcal{F}_{S(\alpha T)},G_{\alpha T}\big) = \Psi\big(\widehat{C}_{v}^{(1),v}\big), \\[5pt] & \mathbb{E}\big(\mathbf{1}_{v\leq T+\overline{B}_{T}^{(1),v}}f_{1}\big(\overline{B}_{T}^{(1),v}\big) f_{2}\big(\overline{B}_{T}^{(2),v}\big)\mid\overline{C}_{v}^{(1),v}\big) = \Psi\big(\overline{C}_{v}^{(1),v}\big) \overset{\text{law}}{=}\Psi(C_{\infty}) \end{align*}

for some function $\Psi$ , the same in both lines, such that $\Vert\Psi\Vert_{\infty}\leq\Vert f_{1}\Vert_{\infty}\Vert f_{2}\Vert_{\infty}$ (where $C_{\infty}$ is defined in Section 3). So, by Theorem 3.1 and Corollary 3.1 applied on the time interval $\big[\alpha T+B_{\alpha T}^{(1)},v\big]$ , the quantity in (4.14) can be bounded (remember that $\widehat{C}^{(1),v}=C^{(1)}$ , Section 2.5) by

\begin{align*} \textrm{e}^{T-\alpha T-B_{\alpha T}^{(1)}}\int_{(T+\alpha T)/2}^{T-\log(\delta)} \textrm{e}^{-(v-\alpha T-B_{\alpha T}^{(1)})}\Gamma_{1}\Vert f_{1}\Vert_{\infty}\Vert f_{2}\Vert_{\infty} \textrm{e}^{-(\theta-\varepsilon')(v-\alpha T-B_{\alpha T}^{(1)})}\,\textrm{d} v . \end{align*}

(Coming from Corollary 3.1 there is an integral over a set of Lebesgue measure zero in the above bound, but this term vanishes.) The above bound can in turn be bounded by

(4.14) \begin{align} \qquad\qquad& \Gamma_{1}\Vert f_{1}\Vert_{\infty}\Vert f_{2}\Vert_{\infty}\delta^{-(\theta-\varepsilon')}\textrm{e}^{T} \int_{(T+\alpha T)/2}^{T-\log(\delta)}\textrm{e}^{(\theta-\varepsilon')\alpha T}\textrm{e}^{-(\theta-\varepsilon'+1)v}\,\textrm{d} v \nonumber \\[5pt] & \qquad\qquad\qquad\qquad\qquad\qquad\quad\qquad\qquad\qquad\qquad\qquad\qquad\quad \text{(as $\theta-\varepsilon'+1>1$)} \nonumber \\[5pt] & \leq\Gamma_{1}\Vert f_{1}\Vert_{\infty}\Vert f_{2}\Vert_{\infty}\delta^{-(\theta-\varepsilon')} \textrm{e}^{T+\alpha T(\theta-\varepsilon')} \exp\bigg[{-}(\theta-\varepsilon'+1)\bigg(\frac{T+\alpha T}{2}\bigg)\bigg] \nonumber \\[5pt] & = \Gamma_{1}\Vert f_{1}\Vert_{\infty}\Vert f_{2}\Vert_{\infty}\delta^{-(\theta-\varepsilon')} \exp\bigg[{-}(\theta-\varepsilon'-1)\bigg(\frac{T-\alpha T}{2}\bigg)\bigg]. \end{align}

We have

(4.15) \begin{align} &\int_{\frac{T+\alpha T}{2}}^{T-\log(\delta)}\textrm{e}^{-(v-T)} \mathbb{E}\big(\mathbf{1}_{v\leq T+\overline{B}_{T}^{(1),v}}f_{1}\big(\overline{B}_{T}^{(1),v}\big) f_{2}\big(\overline{B}_{T}^{(2),v}\big)\big) \, \textrm{d} v \nonumber \\[5pt] &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\;\, \text{(as $\big(\overline{B}_{T}^{(1),v},\overline{B}_{T}^{(2),v}\big)$ and $\big(\overline{B}_{0}^{(1),v-T},\overline{B}_{0}^{(2),v-T}\big)$ have same law)} \nonumber \\[5pt] &= \int_{\frac{T+\alpha T}{2}}^{T-\log(\delta)}e^{-(v-T)} \mathbb{E}\big(\mathbf{1}_{v-T\leq\overline{B}_{0}^{(1),v-T}}f_{1}\big(\overline{B}_{0}^{(1),v-T}\big) f_{2}\big(\overline{B}_{0}^{(2),v-T}\big)\big) \, \textrm{d} v \nonumber \\[5pt] & \qquad\qquad\qquad\qquad\qquad\qquad\qquad\;\quad\qquad\qquad\qquad\qquad\qquad \text{(change of variable $v'=v-T$)} \nonumber \\[5pt] &= \mathbb{E}\bigg(\int_{-\left(\frac{T-\alpha T}{2}\right)}^{-\log(\delta)}\textrm{e}^{-v'} \mathbf{1}_{v'\leq\overline{B}_{0}^{(1),v'}}f_{1}\big(\overline{B}_{0}^{(1),v'}\big) f_{2}\big(\overline{B}_{0}^{(1),v'}\big)\,\textrm{d} v'\bigg) \end{align}

and

(4.16) \begin{align} & \int_{-\infty}^{-\frac{(T-\alpha T)}{2}}\textrm{e}^{-v} \big|\mathbb{E}\big(f_{1}\big(\overline{B}_{0}^{(1),v}\big) f_{2}\big(\overline{B}_{0}^{(2),v}\big)\big)\big|\,\textrm{d} v \nonumber \\[5pt] & \qquad\quad\text{(since $\overline{B}_{0}^{(1),v}$ and $\overline{B}_{0}^{(2),v}$ are independent conditionally on $\overline{B}_{v}^{(1),v}$, $\overline{B}_{v}^{(2),v}$ if $v-\log(\delta)\leq0$;} \nonumber \\[5pt] & \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\,\qquad\qquad\text{ using Theorem 3.1 and Corollary 3.1)} \nonumber \\[5pt] & \leq \int_{-\infty}^{-\frac{(T-\alpha T)}{2}}\textrm{e}^{-v}\Gamma_{1}^{2} \Vert f_{1}\Vert_{\infty}\Vert f_{2}\Vert_{\infty} \mathbb{E}\big(\textrm{e}^{-(\theta-\varepsilon')(-v-\overline{B}_{v}^{(1),v})_{+}} \textrm{e}^{-(\theta-\varepsilon')(-v-\overline{B}_{v}^{(2),v})_{+}}\big)\,\textrm{d} v \nonumber \\[5pt] &\qquad\qquad\qquad\qquad\qquad\!\!\quad\text{ (again, coming from Corollary 3.1 there is an integral over a set of}\nonumber \\ &\qquad\qquad\qquad\qquad\qquad\quad\!\!\text{ Lebesgue measure zero in the above bound, but this term vanishes)} \nonumber \\[5pt] &\leq \int_{-\infty}^{-\frac{(T-\alpha T)}{2}}\textrm{e}^{-v} \Gamma_{1}^{2}\Vert f_{1}\Vert_{\infty}\Vert f_{2}\Vert_{\infty} \textrm{e}^{-2(\theta-\varepsilon')(-v+\log(\delta))} \, \textrm{d} v \nonumber \\[5pt] & = \frac{\Gamma_{1}^{2}\Vert f_{1}\Vert_{\infty}\Vert f_{2}\Vert_{\infty}}{\delta^{2(\theta-\varepsilon')}} \frac{\exp(\!-\!(2(\theta-\varepsilon')-1){(T-\alpha T)}/{2})}{2(\theta-\varepsilon')-1}. \end{align}

Equations (4.16) and (4.17) give us (4.11). Equations (4.13) and (4.15)–(4.17) give us the desired result (see below to understand the puzzle).

Lemma 4.4. Let $k \in \{0,1,2,\dots,p\}$ . We suppose that q is even and $q=2p$ . Let $\alpha\in(q/(q+2),1)$ . We suppose that $F=f_{1}\otimes f_{2}\otimes\dots\otimes f_{q}$ , with $f_{1}, \ldots, f_{q}$ in $\mathcal{B}_\textrm{sym}^{0}(1)$ . Then,

(4.17) \begin{align} \varepsilon^{-q/2}\mathbb{E}\big(F\big(B_{T,}^{(1)}\dots,&B_{T}^{(q)}\big)\mathbf{1}_{G_{\alpha T}}\mathbf{1}_{\#\mathcal{L}_{1}=q}\big) \nonumber \\[5pt] & \underset{\varepsilon\rightarrow0}{\longrightarrow}\prod_{i=1}^{p}\int_{-\infty}^{-\log(\delta)}\textrm{e}^{-v} \mathbb{E}\big(\mathbf{1}_{v\leq\overline{B}_{0}^{(1),v}}f_{2i-1}\big(\overline{B}_{0}^{(1),v}\big) f_{2i}\big(\overline{B}_{0}^{(2),v}\big)\big)\,\textrm{d} v. \end{align}

(Remember that $T=-\log\varepsilon$ .)

Proof. By Fact 4.1, we have $T>\alpha T-\log(\delta)$ . We have (remember the definitions just before Section 4.2)

\begin{align*} G_{\alpha T}\cap\{\#\mathcal{L}_{1}=q\} = G_{\alpha T}\cap\underset{1\leq i\leq p}{\bigcap}G_{2i-1,2i}(T)^{\textrm{c}}. \end{align*}

We have (remember $T=-\log(\varepsilon)$ )

(4.18) \begin{align} & \varepsilon^{-q/2}\mathbb{E}\big(F\big(B_{T,}^{(1)}\dots,B_{T}^{(q)}\big) \mathbf{1}_{G_{\alpha T}}\mathbf{1}_{\#\mathcal{L}_{1}=q}\big) \nonumber \\[5pt] & = \textrm{e}^{pT}\mathbb{E}\Bigg(\mathbf{1}_{G_{\alpha T}}\mathbb{E}\Bigg(\prod_{i=1}^{p}f_{2i-1}\otimes f_{2i}\big(B_{T}^{(2i-1)},B_{T}^{(2i)}\big)\mathbf{1}_{G_{2i-1,2i}(T)^{\textrm{c}}} \mid \mathcal{F}_{S(\alpha T)},G_{\alpha T}\Bigg)\Bigg) \nonumber \\[5pt] & \qquad\qquad\qquad\qquad\qquad\qquad\!\!\! \text{(as $\big(B_{T}^{(1)},B_{T}^{(2)},\mathbf{1}_{G_{1,2}(T)}\big)$, $\big(B_{T}^{(3)},B_{T}^{(4)},\mathbf{1}_{G_{3,4}(T)}\big)$,} \dots \text{are independent} \nonumber \\[5pt] & \;\;\qquad\quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \text{conditionally on $\mathcal{F}_{S(\alpha T)},G_{\alpha T}$ due to Fact 4.2)} \nonumber \\[5pt] &= \mathbb{E}\Bigg(\mathbf{1}_{G_{\alpha T}}\prod_{i=1}^{p}\textrm{e}^{T}\mathbb{E}\big(f_{2i-1}\otimes f_{2i}\big(B_{T}^{(2i-1)},B_{T}^{(2i)}\big)\mathbf{1}_{G_{2i-1,2i}(T)^{\textrm{c}}} \mid \mathcal{F}_{S(\alpha T)},G_{2i-1,2i}(\alpha T)\big)\Bigg) \nonumber \\[5pt] & \qquad\qquad\qquad\qquad\qquad\qquad\quad\qquad\text{(by Lemma 4.3 and as $\big(B^{(1)},\dots,B^{(q)}\big)$ is exchangeable)} \nonumber \\[5pt] &= \mathbb{E}\Bigg(\mathbf{1}_{G_{\alpha T}}\prod_{i=1}^{p}\textrm{e}^{\alpha T+B_{\alpha T}^{(2i-1)}} \bigg(\int_{-\infty}^{-\log(\delta)}\textrm{e}^{-v} \mathbb{E}\big(\mathbf{1}_{v\leq\overline{B}_{0}^{(1),v}}f_{2i-1}\big(\overline{B}_{0}^{(1),v}\big) f_{2i}\big(\overline{B}_{0}^{(2),v}\big)\big) \, \textrm{d} v + R_{2i-1,2i}\bigg)\Bigg), \end{align}

with (a.s.)

(4.19) \begin{equation} |R_{2i-1,2i}| \leq \Gamma_{2}\Vert f_{2i-1}\Vert_{\infty}\Vert f_{2i}\Vert_{\infty} \textrm{e}^{-(T-\alpha T){(\theta-\varepsilon'-1)}/{2}}. \end{equation}

We introduce the events (for $t\in[0,T]$ , with $u\{\cdot\}$ defined below (4.3))

\begin{align*} O_{t}=\{ \#\{u\{t,2i-1\},1\leq i\leq p\}=p\}, \end{align*}

and the tribes (for i in [q], $t\in[0,T]$ ) $\mathcal{F}_{t,i}=\sigma(u\{t,i\},\xi_{u\{t,i\}})$ . As $G_{\alpha T}=O_{\alpha T}\cap\bigcap_{1\leq i\leq p}\{u\{\alpha T,2i-1\}=u\{\alpha T,2i\}\}$ , we have

(4.20) \begin{align} \qquad\qquad & \mathbb{E}\Bigg(\mathbf{1}_{G_{\alpha T}}\prod_{i=1}^{p}\textrm{e}^{B_{\alpha T}^{(2i-1)}+\alpha T}\Bigg) \nonumber \\[5pt] &= \mathbb{E}\Bigg(\mathbf{1}_{O_{\alpha T}}\prod_{i=1}^{p}\textrm{e}^{B_{\alpha T}^{(2i-1)}+\alpha T} \mathbb{E}\Bigg(\prod_{i=1}^{p}\mathbf{1}_{u\{\alpha T,2i-1\}=u\{\alpha T,2i\}} \mid \vee_{1\leq i\leq p}\mathcal{F}_{\alpha T,2i-1}\Bigg)\Bigg) \nonumber \\[5pt] & \;\;\;\quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \text{(by Proposition 2.1 and (2.1))} \nonumber \\[5pt] &= \mathbb{E}(\mathbf{1}_{O_{\alpha T}}). \end{align}

We then observe that

\begin{align*} O_{\alpha T}^{\textrm{c}}=\cup_{i\in[p]}\cup_{j\in[p],j\neq i}\{u\{\alpha T,2i-1\}=u\{\alpha T,2j-1\}\}, \end{align*}

and, for $i\neq j$ ,

\begin{align*} \qquad \mathbb{P}(u\{\alpha T,2i-1\}=u\{\alpha T,2j-1\}) & = \mathbb{E}\big(\mathbb{E}\big(\mathbf{1}_{u\{\alpha T,2i-1\}=u\{\alpha T,2j-1\}}\mid\mathcal{F}_{\alpha T,2i-1}\big)\big) \\[5pt] &= \mathbb{E}\big(\textrm{e}^{-\alpha T-B_{\alpha T}^{(2i-1)}}\big) \quad\qquad\text{(by Proposition 2.1 and (2.1))} \\[5pt] &\leq \textrm{e}^{-\alpha T-\log(\delta)}. \,\qquad\qquad \text{(because of Assumption (2.3))} \end{align*}

So, $\mathbb{P}(O_{\alpha T})\underset{\varepsilon\rightarrow0}{\longrightarrow}1$ . This gives us enough material to finish the proof of (4.18).

Indeed, starting from (4.19), we have

\begin{align*} & \varepsilon^{-q/2}\mathbb{E}\big(F\big(B_{T,}^{(1)}\dots,B_{T}^{(q)}\big)\mathbf{1}_{G_{\alpha T}}\mathbf{1}_{\#\mathcal{L}_{1}=q}\big) \\[5pt] & = \mathbb{E}\Bigg(\mathbf{1}_{G_{\alpha T}}\prod_{i=1}^{p}\textrm{e}^{\alpha T+B_{\alpha T}^{(2i-1)}}\Bigg) \prod_{i=1}^{p}\bigg(\int_{-\infty}^{-\log(\delta)}\textrm{e}^{-v}\mathbb{E}(\mathbf{1}_{v\leq\overline{B}_{0}^{(1),v}} f_{2i-1}\big(\overline{B}_{0}^{(1),v}\big)f_{2i}\big(\overline{B}_{0}^{(2),v}\big))\,\textrm{d} v\bigg) \\[5pt] & \quad +\mathbb{E}\Bigg(\mathbf{1}_{G_{\alpha T}}\prod_{i=1}^{p}\textrm{e}^{\alpha T+B_{\alpha T}^{(2i-1)}}R_{2i-1,2i}\Bigg) \;=\!:\;(\textrm{I})+(\textrm{II}). \end{align*}

By (4.21),

\begin{align*} (\textrm{I}) &= \mathbb{P}(O_{\alpha t})\prod_{i=1}^{p}\bigg(\int_{-\infty}^{-\log(\delta)}\textrm{e}^{-v} \mathbb{E}\big(\mathbf{1}_{v\leq\overline{B}_{0}^{(1),v}}f_{2i-1}\big(\overline{B}_{0}^{(1),v}\big) f_{2i}\big(\overline{B}_{0}^{(2),v}\big)\big)\,\textrm{d} v\bigg) \\[5pt] & \qquad \underset{\varepsilon\rightarrow0}{\longrightarrow}\prod_{i=1}^{p}\bigg(\int_{-\infty}^{-\log(\delta)}\textrm{e}^{-v} \mathbb{E}\big(\mathbf{1}_{v\leq\overline{B}_{0}^{(1),v}}f_{2i-1}\big(\overline{B}_{0}^{(1),v}\big) f_{2i}\big(\overline{B}_{0}^{(2),v}\big)\big)\,\textrm{d} v\bigg). \end{align*}

And, by (4.20),

\begin{equation*} (\textrm{II}) \leq \mathbb{P}(O_{\alpha t})\prod_{i=1}^{p}\big( \Gamma_{2}\Vert f_{2i-1}\Vert_{\infty}\Vert f_{2i}\Vert_{\infty} \textrm{e}^{-(T-\alpha T){(\theta-\varepsilon'-1)}/{2}}\big)\underset{\varepsilon\rightarrow0}{\longrightarrow}0. \end{equation*}

4.3. Convergence result

For f and g bounded measurable functions, we set

(4.21) \begin{equation} V(f,g) = \int_{-\infty}^{-\log(\delta)}\textrm{e}^{-v}\mathbb{E}\big(\mathbf{1}_{v\leq\overline{B}_{0}^{(1),v}} f\big(\overline{B}_{0}^{(1),v}\big)g\big(\overline{B}_{0}^{(2),v}\big)\big)\,\textrm{d} v.\end{equation}

For q even, we set $\mathcal{I}_{q}$ to be the set of partitions of [q] into subsets of cardinality 2. We have

(4.22) \begin{equation} \#\mathcal{I}_{q}=\frac{q!}{\left(q/2\right)!2^{q/2}}.\end{equation}

For I in $\mathcal{I}_{q}$ and t in [0, T], we introduce

\begin{align*} G_{t,I}=\{\text{for all }\{i,j\}\in I,\text{ there exists }u\in\mathcal{U}\text{ such that } \xi_{u} \lt \textrm{e}^{-t},\,\xi_{{\boldsymbol{{m}}}(u)}\geq \textrm{e}^{-t},\,A_{u}=\{i,j\}\}.\end{align*}

For t in [0, T], we define $\mathcal{P}_{t}=\cup_{I\in\mathcal{I}_{q}}G_{t,I}$ . The above event can be understood as ‘at time t, the dots are paired on different fragments’. As before, the reader has to keep in mind that $T=-\log(\varepsilon)$ , see (2.3).

Proposition 4.1. Let q be in $\mathbb{N}^{*}$ . Let $F=(f_{1}\otimes\dots\otimes f_{q})_{\textrm{sym}}$ with $f_{1}, \dots , f_{q}$ in $\mathcal{B}_{\textrm{sym}}^0(1)$ ( $(\!\cdot\!)_{\textrm{sym}}$ defined in (4.1)). If q is even ( $q=2p$ ) then

(4.23) \begin{equation} \varepsilon^{q/2}\mathbb{E}\big(F\big(B_{T}^{(1)},\dots,B_{T}^{(q)}\big)\mathbf{1}_{\#\mathcal{L}_{1}=q}\big) \underset{\varepsilon\rightarrow0}{\longrightarrow}\sum_{I\in\mathcal{I}_{q}}\prod_{\{a,b\}\in I} V(f_{a},f_{b}). \end{equation}

Proof. Let $\alpha$ be in $(q/(q+2),1)$ . We have

\begin{equation*} \varepsilon^{-q/2}\mathbb{E}\big(F\big(B_{T}^{(1)},\dots,B_{T}^{(q)}\big)\mathbf{1}_{\#\mathcal{L}_{1}=q}\big) = \varepsilon^{-q/2}\mathbb{E}\big(F\big(B_{T}^{(1)},\dots,B_{T}^{(q)}\big)\mathbf{1}_{\#\mathcal{L}_{1}=q} (\mathbf{1}_{\mathcal{P}_{\alpha T}}+\mathbf{1}_{\mathcal{P}_{\alpha T}^{\textrm{c}}})\big). \end{equation*}

Remember that the events of the form $C_{k,l}(t)$ , $L_{t}$ are defined in Section 4.1. The set $\mathcal{P}_{\alpha T}^{\textrm{c}}$ is a disjoint union of sets of the form $C_{k,l}(\alpha T)$ (with $k\geq1$ ) and $\{L_{\alpha T}>q/2\}$ (this can be understood heuristically by: ‘if the dots are not paired on fragments then some of them are alone on their fragment, or none of them is alone on a fragment and some are a group of at least three on a fragment’). As before, the event $\{\#\mathcal{L}_{1}=q\}$ is measurable with respect to $\mathcal{L}_{2}$ (see (4.6)). So, by Lemmas 4.1 and 4.2,

\begin{align*} \lim_{\varepsilon\rightarrow0}\varepsilon^{-q/2}\mathbb{E}\big(F\big(B_{T}^{(1)},\dots,B_{T}^{(q)}\big) \mathbf{1}_{\#\mathcal{L}_{1}=q}\mathbf{1}_{\mathcal{P}_{\alpha T}^{\textrm{c}}}\big)=0. \end{align*}

We compute:

\begin{align*} &\varepsilon^{-q/2}\mathbb{E}\big(F\big(B_{T}^{(1)},\dots,B_{T}^{(q)}\big) \mathbf{1}_{\#\mathcal{L}_{1}=q}\mathbf{1}_{\mathcal{P}_{\alpha T}}\big) = \varepsilon^{-q/2}\mathbb{E}\bigg(F\big(B_{T}^{(1)},\dots,B_{T}^{(q)}\big) \mathbf{1}_{\#\mathcal{L}_{1}=q}\sum_{I_{q}\in\mathcal{I}_{q}}\mathbf{1}_{G_{\alpha T,I_{q}}}\bigg) \\[5pt] &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\;\;\;\;\text{(as} F \text{is symmetric and $\big(B_{T}^{(1)},\dots,B_{T}^{(q)}\big)$ is exchangeable)} \\[5pt] &= \frac{q!}{2^{q/2}({q}/{2})!}\varepsilon^{-q/2}\mathbb{E}\big(F\big(B_{T}^{(1)},\dots,B_{T}^{(q)}\big) \mathbf{1}_{\#\mathcal{L}_{1}=q}\mathbf{1}_{G_{\alpha T}}\big) \\[5pt] &= \frac{q!\varepsilon^{-q/2}}{2^{q/2}({q}/{2})!}\frac{1}{q!}\sum_{\sigma\in\mathcal{S}_{q}} \mathbb{E}\big((f_{\sigma(1)}\otimes\dots\otimes f_{\sigma(q)})\big(B_{T}^{(1)},\dots,B_{T}^{(q)}\big) \mathbf{1}_{\#\mathcal{L}_{1}=q}\mathbf{1}_{G_{\alpha T}}\big) \\[5pt] &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\qquad\qquad \text{(by Lemma 4.4)} \\[5pt] &\underset{\varepsilon\rightarrow0}{\longrightarrow}\frac{1}{2^{q/2}({q}/{2})!} \sum_{\sigma\in\mathcal{S}_{q}}\prod_{i=1}^{p}V(f_{\sigma(2i-1)},f_{\sigma(2i)}) = \sum_{I\in\mathcal{I}_{q}}\prod_{\{a,b\}\in I}V(f_{a},f_{b}). \end{align*}