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Ray–Knight representation of flows of branching processes with competition by pruning of Lévy trees

Abstract

We introduce flows of branching processes with competition, which describe the evolution of general continuous state branching populations in which interactions between individuals give rise to a negative density dependence term. This generalizes the logistic branching processes studied by Lambert (Ann Appl Probab 15(2):1506–1535, 2005). Following the approach developed by Dawson and Li (Ann Probab 40(2):813–857, 2012), we first construct such processes as the solutions of certain flow of stochastic differential equations. We then propose a novel genealogical description for branching processes with competition based on interactive pruning of Lévy-trees, and establish a Ray–Knight representation result for these processes in terms of the local times of suitably pruned forests.

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Fig. 1

Notes

  1. The reader is referred to [13] for further topological background.

  2. The full result is actually that the Poisson point process can be constructed to obtain a strong solution with respect to a given noise to Eq. (1.15).

  3. Notice that \(d_0\) corresponds to the metric on \(\mathscr {M}(\mathbb {R}_+)\).

  4. Notice that results in Section 4.1 in [12] are stated in terms of the process \(((\rho _t, W_t): t\ge 0)\) with \(W_t\) corresponding here to the increasing \({{\mathcal {M}}}_{at}(\mathbb {R}_+)\)-valued path \(W_t =(h\mapsto {{\mathcal {N}}}_t([0,h], d\nu ): h\in [0,H_t) \, )\), which was denoted \(\xi ^{(t)}\) in Sect. 3.

  5. In the notation of Section 4.1 of [12], \({{\mathcal {N}}}^{(i)} \) would correspond to the “increment” of an excursion of the path-valued process \(W_t\).

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Acknowledgements

J.B. was partially supported by ANR grants ANR-14-CE25-0014 (ANR GRAAL) and ANR-14-CE25-0013 (ANR NONLOCAL). M.C.F.’s research was supported by a doctoral grant of BASAL-Conicyt Center for Mathematical Modeling (CMM) at University of Chile and a postdoctoral grant from the National Agency for Science and Technology Promotion, Argentina. She thanks also ICM Millennium Nucleus NC120062 (Chile) for travel support, a doctoral mobility grant of the French Embassy in Chile for a three months visit to Laboratoire de Probabilités et Modèles Aléatoires of University Paris 6 in 2012, and hospitality of the latter institution during her stay. J.F. acknowledges partial support of ICM Millennium Nucleus NC120062 and BASAL-Conicyt CMM. The authors thank Jean-François Delmas for enlightening discussions at the beginning of this work and Victor Rivero and Juan Carlos Pardo for kindly providing some useful mathematical insights. We also thank two anonymous referees for their careful reading of the manuscript, for valuable remarks and comments which allowed us to clarify the presentation of our results, and for pointing out to us references [7, 23].

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Appendix

Appendix

1.1 Proof of Proposition 1.2

Recall from [3, Theorem 0.3] that \((\rho ^\theta _t: t\ge 0) : = (\rho _{C^\theta _t}: t\ge 0)\), with \(C^\theta _t \) the right-continuous inverse of \(A^\theta _t : = \int _0^t \mathbf {1}_{\{m^{\theta }_s =0\}} \mathrm {d}s\), has the same law as the exploration process associated with a Lévy process of Laplace exponent (1.11). Denote by \(\left( {\bar{L}}_s^a :t\ge 0 \right) \) its local time at level a. Applying (3.4) to \(\rho ^{\theta }\) and performing the change of variable \(C^{\theta }_r\mapsto u\), we deduce that

$$\begin{aligned} {\bar{L}}_t^a= \lim \limits _{\epsilon \rightarrow 0} \epsilon ^{-1}\int _0^{C^{\theta }_t} \mathbf {1}_{\{a<H_u\le a+\epsilon , \, m^{\theta }_u=0\}}\mathrm {d}u \quad \text{ for } \text{ all } t\ge 0, \end{aligned}$$

in the \(L^1(\mathbb {P})\) sense. Let us check that this limit is equal to \(L_{C^{\theta }_t}^a(m^\theta )\). Since \(C^{\theta }_t<\infty \) a.s. (see [3]), we deduce from (3.4) applied to the original exploration process \((\rho _t: t\ge 0)\) that, on the interval \([0,C^{\theta }_t ]\), the finite measures \(\epsilon ^{-1} \mathbf {1}_{\{a-\epsilon <H_s\le a\}}\mathrm {d}s\) converge weakly in probability towards \(\mathrm {d}L_s^a\) as \(\epsilon \rightarrow 0\). By Lemma 3.3 the function \(s\mapsto \mathbf {1}_{\{ m_s^{\theta } ([0,H_s))=0\}}\) has a.s. at most countably many discontinuities, therefore a.s. it is continuous a.e. with respect to the measure \(\mathbf {1}_{[0,C^{\theta }_t]}(s)\mathrm {d}L_s^a\). We thus deduce that:

$$\begin{aligned} L_{C^{\theta }_t}^a(m^{\theta }) = \mathbb {P}-\lim \limits _{\epsilon \rightarrow 0} \epsilon ^{-1}\int _0^{C^{\theta }_t} \mathbf {1}_{\{a<H_u\le a +\epsilon , \, m^{\theta }_u=0\}}\mathrm {d}u = {\bar{L}}_t^a . \end{aligned}$$
(A.1)

By right-continuity in \(t\ge 0\) of the first and third expressions, \(\left( {\bar{L}}_t^a :t\ge 0 \right) \) and \(\left( L_{C^{\theta }_t}^a(m^{\theta }) :t\ge 0 \right) \) are indistinguishable for each \(a\ge 0\). In particular, if we set \({\bar{T}}_x=\inf \{ s>0 : \, {\bar{L}}_t^0 >x\}\), for each \(a\ge 0\) we a.s. have

$$\begin{aligned} {\bar{L}}_{{\bar{T}}_x}^a = L_{C^{\theta }_{{\bar{T}}_x}}^a(m^\theta )= L_{T_x}^a(m^\theta ) \end{aligned}$$

since \({\bar{L}}^0_{\cdot }=L_{C^{\theta }_{\cdot }}^0(m^\theta )\) (by (A.1) with \(a=0\)) and \(L^0_{\cdot }(m^{\theta })=L^0_{\cdot }\). The result follows from the above identities and Theorem 1.1 applied to the local times of \(\rho ^{\theta }\).

1.2 Proof of Lemma 2.1

Write \(M_t= (M^i_t)_{ i\le n}\), \(N_t= (N^k_t)_{ k\le m} \), let \(\lambda =(\lambda _1,\ldots ,\lambda _n)\in \mathbb {R}^n\), \(\mu =(\mu _1,\ldots ,\mu _m)\in \mathbb {R}^m\) and set \({\hat{\lambda }} =(\lambda _1 a_1^{-1},\ldots ,\lambda _n a_n^{-1})\). It is enough to show for all \(t,s\ge 0\) that

$$\begin{aligned} \mathbb {E}(\exp \{i {\hat{\lambda }} \cdot (M_{t+s}-M_t)+ i\mu \cdot (N_{t+s}-N_t)\}\vert {{\mathcal {S}}}_t)= \exp \left\{ -s \frac{|\lambda |^2}{2}+ s\sum _{k=1}^m b_k(e^{i\mu _k}-1)\right\} \end{aligned}$$

(where “\(\cdot \)” is the Euclidean inner product and \(i=\sqrt{-1}\)). Equivalently, we need to show that

$$\begin{aligned} \exp \left\{ i {\hat{\lambda }} \cdot M_{t} +t\frac{|\lambda |^2}{2} + i\mu \cdot N_t -t\sum _{k=1}^m b_k(e^{i\mu _k}-1) \right\} \end{aligned}$$

is a \(({{\mathcal {S}}}_t)\)- martingale. By i), \( {{\mathcal {E}}}_t^{(M)}:=\exp \left\{ i {\hat{\lambda }} \cdot M_{t} +t\frac{|\lambda |^2}{2} \right\} = \exp \left\{ i {\hat{\lambda }} \cdot M_{t} - \frac{i^2}{2}\right. \left. [ {\hat{\lambda }} \cdot M]_t \right\} \) clearly is a \(({{\mathcal {S}}}_t)\)-martingale. On the other hand, writing \({{\mathcal {E}}}_t^{(N)}= \exp \left\{ i\mu \cdot N_t -t\sum _{k=1}^m b_k(e^{i\mu _k}-1) \right\} \) we have

$$\begin{aligned} \begin{aligned} {{\mathcal {E}}}_t^{(N)}- 1=\,&\sum _{0<s\le t} {{\mathcal {E}}}_{s-}^{(N)} \left( e^{ i\mu \cdot \Delta N_s }-1\right) - \int _0^t \sum _{k=1}^m b_k(e^{i\mu _k}-1) {{\mathcal {E}}}_{s}^{(N)}\mathrm {d}s \\ =\,&\sum _{0<s\le t} \sum _{k=1}^m {{\mathcal {E}}}_{s-}^{(N)} \left( e^{ i\mu _k }-1\right) \Delta N^k_s - \int _0^t \sum _{k=1}^m b_k(e^{i\mu _k}-1) {{\mathcal {E}}}_{s-}^{(N)}\mathrm {d}s \end{aligned} \end{aligned}$$

by the first property in ii) and since \(\Delta N^k_s=0\) or 1 for each \(k=1,\dots ,m\). The second property in ii) then grants that \( {{\mathcal {E}}}_t^{(N)}- 1=\int _0^t \sum _{k=1}^m b_k(e^{i\mu _k}-1) {{\mathcal {E}}}_{s-}^{(N)} \mathrm {d}{\tilde{N}}^k_s\), with \({\tilde{N}}^k_t=N^k_t-b_k t\), is a \(({{\mathcal {S}}}_t)\)- martingale too. Since the martingales \({{\mathcal {E}}}^{(M)}\) and \({{\mathcal {E}}}^{(N)}\) respectively are continuous and pure jump, \([{{\mathcal {E}}}^{(M)},{{\mathcal {E}}}^{(N)}]=0\) and \({{\mathcal {E}}}^{(M)}{{\mathcal {E}}}^{(N)}\) is a \(({{\mathcal {S}}}_t)\)- martingale as required.

1.3 Proof of estimate (5.7)

We follow ideas in the proofs of [11, Theorem 2.1] and [14, Proposition 3.1]. Define a function \(\varrho \) on \(\mathbb {R}_+\) by \(\varrho (x):= [\sigma + \int _{0}^{1} r^{2}\Pi (\mathrm {d}r)]\sqrt{x}\). Notice that for all \(y\le x\) one has

$$\begin{aligned} \int _0^{\infty } \mathrm {d}\nu \int _0^1 \Pi (\mathrm {d}r) \int _0^1 \dfrac{r^2\mathbf {1}_{\{y<\nu<x\}} (1-t)}{\varrho (|x-y +tr\mathbf {1}_{\{y<\nu <x\}}|)^2}\mathrm {d}t \le c := \dfrac{\int _{0}^{1}r^{2}\Pi (\mathrm {d}r)}{\left( \sigma + \int _{0}^{1} r^{2}\Pi (\mathrm {d}r)\right) ^2}.\nonumber \\ \end{aligned}$$
(A.2)

Consider a real sequence \(\{a_{j}\}_{j\ge 1}\) defined by \(a_{0} = 1\) and \(a_{j}= a_{j-1} \,e^{-j[\sigma + \int _{0}^{1} r^2\Pi (\mathrm {d}r)]^{2}}\), so that \(a_{j}\searrow 0\) and \(\int _{a_{j}}^{a_{j-1}} \varrho (z)^{-2 }\mathrm {d}z = j\). For each \(j\ge 1\), let \(\psi _{j}\) be a non-negative continuous function on \(\mathbb {R}\) supported in \((a_{j} , a_{j-1})\) such that \(0 \le \psi _{j}(z)\le 2j^{-1}\varrho (z)^{-2}\) and \(\int _{a_{j}}^{a_{j-1}} \psi _{j}(z)\mathrm {d}z = 1\). Define also non-negative twice continuously differentiable functions \(\phi _{j}\) on \(\mathbb {R}\) by

$$\begin{aligned} \phi _{j}(x) =\int _{0}^{|x|}\mathrm {d}y\int _{0}^{y}\psi _{j}(z)\mathrm {d}z, \quad x \in \mathbb {R}. \end{aligned}$$

Notice that for each \(x\in \mathbb {R}\) it holds for all \(j\ge 1\) that \( 0 \le \phi '_{j}(x)\text{ sign }(x) \le 1 \), \(\phi _{j}''(x)\ge 0\) and \( \phi ''_{j}(x)\sigma ^2|x| \le 2j^{-1}\). Moreover, we have \(\phi _{j}(x) \nearrow |x|\) as \(j \rightarrow \infty \).

For any \(z,h, r, \nu ,x,y\ge 0\) and a differentiable function f, set now \(l(r,\nu ; ,x, y): = r\left[ \mathbf {1}_{\{\nu<x\}} - \mathbf {1}_{\{\nu <y\}}\right] \), \(\Delta _{h}f(z):= f(z+h)-f(z)\) and \(D_h f (\zeta ) :=\Delta _h f(\zeta ) - f' (\zeta )h\). Using (5.6) and Itô’s formula, we get

$$\begin{aligned} \phi _{j}(\zeta ^{\varepsilon ,\delta }_{t\wedge \tau _{m}}(x))= & {} M_{t\wedge \tau _{m}}+ \alpha _0 \int _{0}^{t\wedge \tau _{m}} \phi '_{j}(\zeta ^{\varepsilon ,\delta }_{s}(x)) \zeta ^{\varepsilon ,\delta }_{s}(x) \mathrm {d}s \nonumber \\&+ \int _{0}^{t\wedge \tau _{m}}\int _{0}^{\infty }\int _1^{\infty } \Delta _{l(r,\nu ;Z_{s}(x),Z^{\varepsilon ,\delta }_{s}(x))}\phi _{k}(\zeta ^{\varepsilon ,\delta }_{s}(x))\mathrm {d}\nu \Pi (\mathrm {d}r) \mathrm {d}s\nonumber \\&+ \frac{ \sigma ^2}{2}\int _{0}^{t\wedge \tau _{m}}\phi ''_{j}(\zeta ^{\varepsilon ,\delta }_{s}(x)) \zeta ^{\varepsilon ,\delta }_{s}(x) \mathrm {d}s\nonumber \\&+ \int _{0}^{t\wedge \tau _{m}} \int _0^{\infty } \int _0^1 D_{l(r,\nu ; Z_{s}(x),Z^{\varepsilon ,\delta }_{s}(x))}\phi _{k}(\zeta ^{\varepsilon ,\delta }_{s})\mathrm {d}\nu \Pi (\mathrm {d}r) \mathrm {d}s \nonumber \\&- \int _0^{t\wedge \tau _{m}} \phi '_{j}(\zeta ^{\varepsilon ,\delta }_{s}(x)) \left[ G(Z_s(x)) - G(Z^{\varepsilon ,\delta }_s(x)) \right] \mathrm {d}s\nonumber \\&- \int _0^{t\wedge \tau _m} \sum \limits _{n =0}^{ n^{k_s\varepsilon }_x} \int _0^{Z^{\varepsilon ,\delta }_{k_s\varepsilon }(x)}\phi '_{j}(\zeta ^{\varepsilon ,\delta }_{s}(x)) \mathbf {1}_{\{n\delta <u\le (n+1)\delta \}}\nonumber \\&\qquad \times \left[ g(u) - g(n\delta ) \right] \mathrm {d}_uZ_{k_s\varepsilon ,s}(u)\mathrm {d}s, \end{aligned}$$
(A.3)

where \((M_{t \wedge \tau _{m}})_{t\ge 0}\) is a martingale. By properties of \(\phi '_j\), the integrands in the first and second lines are respectively bounded by \(|\alpha _0||Z_{s}(x) - Z^{\varepsilon ,\delta }_{s}(x)|\) and \( \int _1^{\infty } r \Pi (\mathrm {d}r) |Z_{s}(x) - Z^{\varepsilon ,\delta }_{s}(x)|\). The fact that

$$\begin{aligned} 0 \le \int _{0}^{\infty }\int _0^{1} D_{l(r,\nu ; x,y)}\phi _{j}(x-y)\mathrm {d}\nu \Pi (\mathrm {d}r)\le \dfrac{2c}{j} \end{aligned}$$

following from estimate (A.2), together with the mentioned properties of \(\phi ''_{j}\) ensure that the terms on the third and fourth lines vanish when \(j\rightarrow \infty \). Taking expectation in (A.3) and letting \(j\rightarrow \infty \), we get the desired bound, noting that g(m) is a Lipschitz constant for the function G in [0, m].

1.4 Proof of Lemma 6.1

In a similar way as for the process \(\left( (\rho _t,{{\mathcal {N}}}_t)\, : t\ge 0 \right) \), the trajectories of the process \(\left( (\rho ^a_t,{{\mathcal {N}}}_t^a)\, : t\ge 0 \right) \) are determined in a unique (measurable) way from the atoms of the point process (6.3). It is therefore enough to establish the first claim. To do so, one easily adapts first the arguments of the proof of Proposition 4.2.3 in [12] in order to prove that, under the excursion measure \(\mathbb {N}\), the process

$$\begin{aligned} \sum _{i\in I^j} \delta _{(\ell ^{(i)}, \rho ^{(i)},\, {{\mathcal {N}}}^{(i)})} , \end{aligned}$$
(A.4)

with \(I_j:=\{ i \in I: (\alpha ^{(i)},\beta ^{(i)}) \subset (\alpha ^{j},\beta ^{j})\}\) denoting the sub-excursions above level a of the excursion away from 0 labeled j, is conditionally on \({{\mathcal {E}}}_a\) a Poisson point process of intensity \(\mathrm {d}x \mathbf {1}_{[L^a_{\alpha _j}, L^a_{\beta _j} ]} \otimes \mathbb {N}(\mathrm {d}\rho \, , \mathrm {d}{{\mathcal {N}}})\) (our superscripts “(i)” correspond to superscripts “i” in the statement of [12]). The only difference is that, in the computation analogous to the one at end of that proof, we must consider here test functions depending also on the components \(\ell ^{(i)}\) of the atoms, and depending on the excursions of the spatial component above level a only through their increments respect to their values at that a. Since I is equal to the disjoint union \(\bigcup _{j\in J} I_j\), one then concludes applying conditionally on \({{\mathcal {E}}}_a\) the additivity of Poisson point measures.

1.5 Proof of Lemma 6.3

By standard properties of Poisson processes, for any nonnegative predictable process f and stopping time \(\tau \) in the given filtration, it holds that \(\mathbb {E}\left[ e^{-u \int _0^{\tau \wedge t} f(\ell )N(\mathrm {d}\ell ) + \lambda \int _0^{\tau \wedge t } (1- e^{-u f (\ell )})\mathrm {d}\ell }\vert {{\mathcal {K}}}_0 \right] =1\) a.s. for all \(u\ge 0\) and \(t\ge 0\). If moreover \(\tau \) is such that \( \mathbb {E}\left[ e^{ \lambda \int _0^{\tau } (1- e^{-u f (\ell )})\mathrm {d}\ell }\right] <\infty \), by dominated convergence one gets after letting \(t\rightarrow \infty \) that \(\mathbb {E}\left[ e^{-u \int _0^{\tau } f(\ell )N(\mathrm {d}\ell ) + \lambda \int _0^{\tau } (1- e^{-u f (\ell )})\mathrm {d}\ell }\vert {{\mathcal {K}}}_0 \right] =1\) a.s. Now, from our assumptions and by a monotone class argument, we can check that \(\mathbf {1}_{E}\) is a predictable process. Since \(e^{\lambda \int _0^{\phi _x} (1- e^{-u \mathbf {1}_{E}(\ell )})\mathrm {d}\ell }=e^{\lambda \int _0^{\phi _x} (1- e^{-u}) \mathbf {1}_{E}(\ell )\mathrm {d}\ell }= e^{\lambda (1- e^{-u}) (x \wedge \int \mathbf {1}_{E}(\ell )\mathrm {d}\ell )} \le e^{\lambda (1- e^{-u}) x } \), from the previous we conclude that for all \(x\ge 0\),

$$\begin{aligned} \mathbb {E}\left[ e^{-u \int _0^{\phi _x} \mathbf {1}_{E}(\ell )N(\mathrm {d}\ell ) } \vert {{\mathcal {K}}}_0 \right] = e^{-\lambda x (1- e^{-u}) } \text{ a.s. } \end{aligned}$$

Applying conditionally on \( {{\mathcal {K}}}_{\phi _{y}}\) this argument to increments \(\int _{\phi _y} ^{\phi _x} \mathbf {1}_{E}(\ell )N(\mathrm {d}\ell ) \), the proof is complete.

1.6 Proof of Lemma 6.5

(a) The proof is inspired by that of Lemma 1.3.2 in [12]. We have

$$\begin{aligned}&\mathbb {E}\left( \sup _{y\in [0,x]} \left| y - \frac{1}{\varepsilon } \int _0^{T_y} \mathbf {1}_{\{ 0<H_s \le \varepsilon , \, m_s =0 \}} \mathrm {d}s \right| \mathbf {1}_{\{\tau ^K> T_x\}} \right) \nonumber \\&\quad \le \mathbb {E}\left( \sup _{y\in [0,x]}\left| \frac{1}{\varepsilon } \int _0^{T_y} \mathbf {1}_{\{ 0<H_s \le \varepsilon , m_s =0, \langle \rho _s, 1\rangle \le K\}} \mathrm {d}s\right. \right. \nonumber \\&\qquad -\left. \left. \frac{1}{\varepsilon } \mathbb {E}\left( \int _0^{T_y} \mathbf {1}_{\{ 0<H_s \le \varepsilon , m_s =0, \langle \rho _s, 1\rangle \le K\}} \mathrm {d}s \right) \right| \right) \nonumber \\&\qquad +\, \mathbb {E}\left( \sup _{y\in [0,x]} \bigg | \frac{1}{\varepsilon } \mathbb {E}\left( \int _0^{T_y} \mathbf {1}_{\{ 0<H_s \le \varepsilon , m_s =0, \langle \rho _s, 1\rangle \le K\}} \mathrm {d}s \right) - y\bigg | \right) \end{aligned}$$
(A.5)

The time integral in the above expressions can be written in terms of the excursion point process (6.2) as follows:

$$\begin{aligned} \int _0^{T_y} \mathbf {1}_{\{ 0<H_s \le \varepsilon , m_s =0,\, \langle \rho _s, 1\rangle \le K\}}\mathrm {d}s = \sum _{j\in J_y} \int _0^{\zeta ^j} \mathbf {1}_{\{ 0<H^j_s \le \varepsilon , m^j_s =0,\, \langle \rho _s^j, 1\rangle \le K\}}\mathrm {d}s\, , \end{aligned}$$
(A.6)

where \(J_y:=\{ j\in J: \, \ell ^j \le y\}\), \(H_s^j=H_{\alpha ^j+s}\), \(m_s^j =\mathcal {N}^j_s (\, \cdot \times [0, \Theta (\ell ^j) )) \) and \(\zeta ^j\) is the length of the excursion labelled j. By compensation, the desintegration \(\mathbb {N}(\mathrm {d}\rho \, , \mathrm {d}{{\mathcal {N}}}) = \mathbf {N}(\mathrm {d}\rho ) Q^{H(\rho )}( \mathrm {d}{{\mathcal {N}}} )\) and the very definition of the snake \((\rho ,\mathcal {N})\), we get

$$\begin{aligned} \begin{aligned} \mathbb {E}&\left( \int _0^{T_y} \mathbf {1}_{\{ 0<H_s \le \varepsilon , m_s =0,\, \langle \rho _s, 1\rangle \le K\}} \mathrm {d}s \right) \\ =&\int _0^y \mathrm {d}\ell \, \mathbb {N} \left( \int _0^{\zeta } \mathbf {1}_{\{ 0<H_s(\rho ) \le \varepsilon , \, \mathcal {N}_s (\, \cdot \times [0,\Theta (\ell )))=0\, , \langle \rho _s, 1\rangle \le K\}} \mathrm {d}s \right) \\ =&\int _0^y \mathrm {d}\ell \, \mathbf {N} \left( \int _0^{\zeta } e^{ - \Theta (\ell ) H_s(\rho )} \mathbf {1}_{\{ 0<H_s(\rho ) \le \varepsilon \, , \langle \rho _s, 1\rangle \le K\}} \mathrm {d}s \right) , \end{aligned} \end{aligned}$$

with \(\zeta \) the length of the canonical excursion and \((H_s(\rho ): 0\le s\le \zeta )\) its height process. Thus, the second term in the r.h.s. of (A.5) is bounded by

$$\begin{aligned} \int _0^x \mathrm {d}\ell \, \mathbb {E}\left[ \bigg | \varepsilon ^{-1} \mathbf {N} \left( \int _0^{\zeta } e^{ - \Theta (\ell ) H_s(\rho ) } \mathbf {1}_{\{ 0<H_s(\rho ) \le \varepsilon \, , \langle \rho _s, 1\rangle \le K\}} \mathrm {d}s \right) - 1\bigg |\right] . \end{aligned}$$

Using Proposition 1.2.5 in [12] to compute the integral with respect to \(\mathbf {N}\) for each \(\ell \in [0,x]\), the latter expression is seen to be equal to

$$\begin{aligned} \int _0^x \mathrm {d}\ell \, \left[ 1 -\frac{1}{\varepsilon } \int _0^{\varepsilon } e^{-\alpha b} e^{ - \Theta (\ell ) b } \mathbb {P}( S_b \le K ) \mathrm {d}b \right] \nonumber \\ \le x \left[ 1- \frac{ 1- e^{-(\alpha +{\bar{\theta }} ) \varepsilon }}{(\alpha +{\bar{\theta }} ) \varepsilon } \mathbb {P}( S_\varepsilon \le K ) \right] , \end{aligned}$$
(A.7)

where \((S_b)_{b\ge 0}\) is a subordinator of Laplace exponent \({\widehat{\psi }}(\lambda ):= \frac{\sigma ^2}{2}\lambda + \int _0^\infty (1 - e^{-\lambda } )\Pi ([r, \infty ))\mathrm {d}r\) which does not depend on the drift coefficient \(\alpha \) of the underlying Lévy process X. In particular, the expression on the r.h.s. of (A.7) goes to 0 with \(\varepsilon \).

Let now \(({{\mathcal {Q}}}_\ell )_{\ell \ge 0}\) denote the right continuous completion of the filtration \((\sigma (\mathbf { M}([0,x],d\rho ,d\mathcal {N}): 0 \le x \le \ell ))_{\ell \ge 0}\), with \(\mathbf { M}\) the point process defined in (6.2). Since the first term on the r.h.s. of (A.5) is the expected supremum of the absolute value of a \(({{\mathcal {Q}}}_{\ell })_{\ell \ge 0}\)-martingale, we can bound it using BDG inequality by some universal constant \(C_1\) times

$$\begin{aligned} \sqrt{ Var \left[ \frac{1}{\varepsilon } \int _0^{T_x} \mathbf {1}_{\{ 0<H_s \le \varepsilon , m_s =0, \langle \rho _s, 1\rangle \le K\}} \mathrm {d}s \right] }. \end{aligned}$$

Written in terms of the excursion Poisson point process (6.2), the previous quantity reads

$$\begin{aligned} \sqrt{ Var \left[ \frac{1}{\varepsilon } \sum _{j\in J_x} \int _0^{\zeta _j} \mathbf {1}_{\{ 0<H^j_s \le \varepsilon , m^j_s =0,\, \langle \rho _s^j, 1\rangle \le K\}} \mathrm {d}s \right] } \end{aligned}$$

and can be estimated by the same arguments as in the proof of Lemma 1.3.2 of [12], as follows (see also the proof of Lemma 1.1.3 therein for details on related arguments):

$$\begin{aligned}&Var \left[ \frac{1}{\varepsilon }\int _0^{T_x} \mathbf {1}_{\{ 0<H_s \le \varepsilon , m_s =0, \langle \rho _s, 1\rangle \le K\}}\mathrm {d}s\right] \nonumber \\&\quad =\, \frac{x}{\varepsilon ^2} \mathbb {N}\left( \left( \int _0^{\zeta } \mathbf {1}_{\{ 0<H_s \le \varepsilon , m_s =0, \langle \rho _s, 1\rangle \le K\}}\mathrm {d}s \right) ^2 \right) \nonumber \\&\quad \le \, \frac{x}{\varepsilon ^2} \mathbf {N}\left( \left( \int _0^{\zeta } \mathbf {1}_{\{ 0<H_s \le \varepsilon , \langle \rho _s, 1\rangle \le K\}} \mathrm {d}s \right) ^2 \right) \nonumber \\&\quad \le \, 2 x\mathbb {E}( X_{L^{-1}(\varepsilon )}\wedge K), \end{aligned}$$
(A.8)

where \( \varepsilon \mapsto X_{L^{-1}(\varepsilon )}\) is a subordinator of Laplace exponent \(\exp \left( - t\left( {\widehat{\psi }}(\lambda )+ \alpha \right) \right) \). That is, the same subordinator S as above, but killed at an independent exponential time of parameter \(\alpha \). Thus, we have \(\mathbb {E}( X_{L^{-1}( \varepsilon )}\wedge K)\le \mathbb {E}(S_ \varepsilon \wedge K) + K(1-e^{-\alpha \varepsilon }) \rightarrow 0\) as \( \varepsilon \rightarrow 0\). The statement now follows by bringing together (A.5), (A.8) and (A.7) with the r.h.s. of the latter replaced by its supremum over \(\varepsilon '\in [0,\varepsilon ]\), which is an increasing function of \(\varepsilon \) going to 0 as \(\varepsilon \rightarrow 0\), as required.

(b) Observe first that by continuity of \(s\mapsto L_s^0\)

$$\begin{aligned}&\forall t < T_x, \forall n\ge \dfrac{1}{x}, \exists y_n,z_n\in [\dfrac{1}{n},x] \text { such that } T_{y_n}\le t\le T_{z_n}\nonumber \\&\quad \text { and } z_n- \dfrac{1}{n} \le L_t^0 \le y_n+ \dfrac{1}{n}. \end{aligned}$$

We deduce that

$$\begin{aligned} |L_t^0- L_t^{\varepsilon }(m) | \le {\left\{ \begin{array}{ll} y_n- L^{\varepsilon }_{T_{y_n}}(m) + \dfrac{1}{n} &{} \text { if } \, L_t^{\varepsilon }(m)< L_t^0 \\ L^{\varepsilon }_{T_{z_n}}(m) -z_n + \dfrac{1}{n} &{} \text { if } \, L_t^{\varepsilon }(m)> L_t^0 \end{array}\right. } . \end{aligned}$$

Therefore, we have \(\sup \limits _{t\in [0,T_x]}| L_t^0- L_t^{\varepsilon }(m) |\le \sup \limits _{y\in [0,x]} \left| L^{\varepsilon }_{T_y}(m)- y \right| \) and it is enough to establish the required upper bound for the quantity

$$\begin{aligned} \mathbb {E}\left[ \sup _{y\in [0,x]} \left| L^{\varepsilon }_{T_y}(m) - y \right| \mathbf {1}_{\{\tau ^K> T_x\}} \right] . \end{aligned}$$
(A.9)

We have

$$\begin{aligned} \begin{aligned} L^\varepsilon _{T_y}(m)-y&= \left[ L^\varepsilon _{T_y}(m) -\frac{1}{\varepsilon } \int _0^{T_y} \mathbf {1}_{\{ \varepsilon<H_s \le 2\varepsilon , m_s((0,\varepsilon ))=0\}} \mathrm {d}s \right] \\&\quad + 2\left[ \frac{1}{2\varepsilon } \int _0^{T_y} \mathbf {1}_{\{ 0<H_s \le 2\varepsilon ,m_s((0,\varepsilon ))=0\}} \mathrm {d}s - y \right] \\&\quad + \left[ y- \frac{1}{\varepsilon } \int _0^{T_y} \mathbf {1}_{\{ 0<H_s \le \varepsilon ,m_s((0,\varepsilon ))=0\}} \mathrm {d}s \right] , \end{aligned} \end{aligned}$$

and the absolute value of the second term on the right hand side is bounded by

$$\begin{aligned} 2 \left| \frac{1}{2\varepsilon } \int _0^{T_y} \mathbf {1}_{\{ 0<H_s \le 2\varepsilon ,m_s((0,2\varepsilon ))=0\}} \mathrm {d}s - y \right| + 2 \left| \frac{1}{2\varepsilon } \int _0^{T_y} \mathbf {1}_{\{ 0<H_s \le 2\varepsilon \}} \mathrm {d}s - y \right| , \end{aligned}$$

thanks to the inequalities \(\mathbf {1}_{\{ 0<H_s \le 2\varepsilon \}}\ge \mathbf {1}_{\{ 0<H_s \le 2 \varepsilon ,m_s ((0,\varepsilon ))=0\}} \ge \mathbf {1}_{ \{0<H_s \le 2\varepsilon ,m_s((0,2\varepsilon ))=0\}}\). It follows from part a) that expression (A.9) is upper bounded by

$$\begin{aligned}&\left( 2 \hat{{\mathcal {C}}}({{\bar{\theta }}} ,K,2\varepsilon ) + 2\hat{{\mathcal {C}}}(0 ,K,\varepsilon ) + \hat{{\mathcal {C}}}({{\bar{\theta }}} ,K,\varepsilon ) \right) ( x+ \sqrt{x}) \nonumber \\&\quad +\, \mathbb {E}\left[ \sup _{y\in [0,x]} \left| L^\varepsilon _{T_y}(m) -\frac{1}{\varepsilon } \int _0^{T_y} \mathbf {1}_{\{ \varepsilon < H_s \le 2\varepsilon , m_s((0,\varepsilon ))=0\}} \mathrm {d}s \right| \mathbf {1}_{\{\tau ^K> T_x\}} \right] \quad \quad \end{aligned}$$
(A.10)

and it only remains us to bound this last expectation. Notice to that end that the inner supremum can be written in terms of the Poisson excursions point process pruned below the level \(a=\varepsilon \) as considered in (6.4), namely

$$\begin{aligned} \sum _{i\in I^{\vartheta ^\varepsilon }} \delta _{(L_{\alpha ^{(i)}}^\varepsilon (m) , \rho ^{(i)},\, {{\mathcal {N}}}^{(i)})} \, , \end{aligned}$$
(A.11)

where \(I^{\vartheta ^{\varepsilon }}= \{ i\in I \, : \, m^{\vartheta ^{\varepsilon }}_{\alpha ^{(i)}}([0,\varepsilon ))= 0\}\). As before, let \({\widehat{L}}^0_t\), \({\widehat{T}}_x\), \({\widehat{H}}_t\) and \({\widehat{\tau }}^K\) denote the corresponding local time at 0, inverse local time at 0, height process, and the stopping time \({\widehat{\tau }}^K:=\inf \{s>0 \, : \langle {\widehat{\rho }}_s, 1\rangle \ge K\}\), all of them now associated with \(({\widehat{\rho }}, {{\widehat{\mathcal {N}}}})\), the snake process associated with (A.11).

Then, writing in a similar way as in (A.6) the time integral in (A.10), that is, as a sum of integrals over (now) non marked excursion intervals above level \(\varepsilon \), we get

$$\begin{aligned} L^\varepsilon _{T_y}(m) -\frac{1}{\varepsilon } \int _0^{T_y} \mathbf {1}_{\{ \varepsilon<H_s \le 2\varepsilon , m_s((0,\varepsilon ))=0\}} \mathrm {d}s= {\widehat{L}}^0_{{\widehat{T}}_{L^\varepsilon _{T_y}(m)}} -\frac{1}{\varepsilon } \int _0^{{\widehat{T}}_{L^\varepsilon _{T_y}(m)}} \mathbf {1}_{\{ 0<{\widehat{H}}_s \le \varepsilon \}} \mathrm {d}s. \end{aligned}$$

Since \(T_x\ge {\widehat{T}}_{L^\varepsilon _{T_x}(m)}\) and \(\sup \limits _{t\le T_x} \langle \rho _t, 1\rangle \ge \sup \limits _{t\le {\widehat{T}}_{L^a_{T_x}} (m)} \langle {\widehat{\rho }}_t, 1\rangle \), the expectation in (A.10) is seen to be less than

$$\begin{aligned}&\mathbb {E}\left[ \sup _{z\in [0, L^\varepsilon _{T_x}(m) ]} \left| z -\frac{1}{\varepsilon } \int _0^{{\widehat{T}}_z} \mathbf {1}_{\{ 0<{\widehat{H}}_s \le \varepsilon \}} \mathrm {d}s \right| \mathbf {1}_{\{{{\widehat{\tau }}}^K> {\widehat{T}}_{L^\varepsilon _{T_x}(m)} \}} \right] \\&\quad \le \hat{{\mathcal {C}}}(0 ,K,\varepsilon ) \mathbb {E}\left[ L^\varepsilon _{T_x}(m)+\sqrt{L^\varepsilon _{T_x}(m)} \right] , \end{aligned}$$

by applying, thanks to Lemma 6.2, the previous part a) (with \(m=0\) or equivalently \({{\bar{\theta }}} =0\)) conditionally on \({{\mathcal {E}}}_{\varepsilon }\). With the obvious bounds \(L^\varepsilon _{T_x}(m)\le L^\varepsilon _{T_x}\) a.s., \(\mathbb {E}\left[ \sqrt{L^\varepsilon _{T_x}}\right] \le \sqrt{\mathbb {E}\left[ L^\varepsilon _{T_x}\right] }\) and the equalities \(\mathbb {E}( L^{\varepsilon }_{T_x} )= x \mathbf {N}\left( L^{\varepsilon }_{\zeta } \right) = x e^{- \alpha {\varepsilon }} \) following from Corollary 1.3.4 in [12], the desired result is seen to hold with \({{\mathcal {C}}}({{\bar{\theta }}} ,K,\varepsilon )= \left( 2\hat{{\mathcal {C}}}({{\bar{\theta }}} ,K,2\varepsilon ) + 3\hat{{\mathcal {C}}}(0 ,K,\varepsilon ) + \hat{{\mathcal {C}}}({{\bar{\theta }}} ,K,\varepsilon ) \right) \).

1.7 Lamperti-type representation

Proposition A.1

Let \(X_t\) be a Lévy process with Laplace exponent satisfying (1.1) and (1.2), and assume that the locally bounded competition mechanisms g is such that \(\lim \limits _{z\rightarrow 0} \frac{g(z)}{z}\) exists. For each \(x>0\) there is a unique strong solution \(U_t\) to the SDE (2.4). Moreover, if we set

$$\begin{aligned} V_{t}:= \left\{ \begin{array}{ll} U_{C_{t}} &{}\quad \text {if}\, 0\le t<\eta _{\infty }, \\ 0 &{}\quad \text {if}\,\, \eta _{\infty }<\infty \wedge t\ge \eta _{\infty }, \end{array} \right. \end{aligned}$$
(A.12)

with \(C_t\) the right inverse of \(\eta _{t}:=\inf \left\{ s>0:\int _{0}^{s}\frac{\mathrm {d}r}{U_r }>t \right\} \), there exists in some enlarged probability space a Brownian motion \(B^V\) and an independent Poisson point process \(N^V\) in \([0,\infty )^3\) with intensity measure \(\mathrm {d}t\otimes \mathrm {d}\nu \otimes \Pi (\mathrm {d}r)\), such that

$$\begin{aligned} V_{t}= & {} x - \alpha \int _{0}^{t}V_{s}\mathrm {d}s + \sigma \int _{0}^{t}\sqrt{V_{s}}\mathrm {d}B_{s}^{V} + \int _{0}^{t}\int _{0}^{V_{s-}}\int _{0}^{\infty }r{\tilde{N}}^{V}(\mathrm {d}s, \mathrm {d}\nu , \mathrm {d}r)\nonumber \\&- \int _{0}^{t}G(V_{s})\mathrm {d}s , \end{aligned}$$
(A.13)

for all \(t\ge 0\). Last, pathwise uniqueness (and then also in law) holds for (A.13).

Since \((Z_t(x): t\ge 0)\) in (1.5) satisfies (A.13) with the Brownian motion

$$\begin{aligned} B_t:=\int _0^t \int _{0}^{Z_{s-}(x)} \left( Z_{s-}(x)\right) ^{-{\frac{1}{2}}} W(\mathrm {d}s,\mathrm {d}u)\, , \end{aligned}$$

we conclude that \((Z_t(x): t\ge 0)\) and \((V_t :t\ge 0)\) are equal in law.

Proof

Let \(B^X\) and \(N^X\) respectively denote a standard Brownian motion and a Poisson random measure on \([0,\infty )^2\) with intensity \(\mathrm {d}s\otimes \Pi (\mathrm {d}r)\) such that \(\mathrm {d}X_t= -\alpha \mathrm {d}t + \sigma \mathrm {d}B^X_t + \int _0^{\infty } r{\widetilde{N}}^X (\mathrm {d}t,\mathrm {d}r)\). Standard localization arguments using a sequence \(G^R\) of globally Lipschitz functions equal to G on [0, R] (the local Lipschitz character of G following from the assumptions) show the existence of a unique strong solution U until some random explosion time. For each \(R,K\ge 0\), set \(\tau ^R=\inf \{s\ge 0 \, : U^2_s\ge R^2\}\) and \(\theta _K=\inf \{s\ge 0 \, : [U,U]_s \ge K\}\). Applying Itô’ s formula to the solution of equation (2.4) with \(G_R\) instead of G while keeping in mind the sign of G, we get using Gronwall’s lemma that \(\mathbb {E}(U_{t\wedge \theta _K\wedge \tau ^{R}}^2) \le c+ c'K\) for some finite constants \(c,c'>0\) depending on tx and the characteristics of X but not on G. Fatou’s Lemma then yields \(t\wedge \theta _K\le \tau ^{\infty } =\sup _{R\ge 0}\tau ^{R}\) a.s., from where \(\tau ^{\infty }=\infty \) a.s. and

$$\begin{aligned} \mathrm {d}U_t =- \alpha \mathrm {d}t + \sigma \mathrm {d}B^X_t + \int _0^{\infty } r {\widetilde{N}}^X (\mathrm {d}t,\mathrm {d}r) - \frac{G(|U_t| )}{U_t} \mathrm {d}t . \end{aligned}$$
(A.14)

Let us now set \(T_0:= \inf \{t>0:U_t=0\}\) and \(T:= \inf \{t>0:V_t=0\}= \inf \{t>0: U_{C_t}=0\}\wedge \eta _{\infty }.\) As \(C_t\) is right-continuous, we have that \(U_{C_T}=0,\) so that \(C_T = T_0\) since \(\eta _{r}= \int _0^{r \wedge T_0} \dfrac{\mathrm {d}s}{U_s}= \eta _{\infty }\) for all \(r \ge T_0\). In order to show that the time-changed process \(V = (U_{C_{t}}: t \ge 0)\) is solution of (A.13), we follow the arguments of Caballero et al. [10, Proposition 4] providing the existence of a standard Brownian motion B such that

$$\begin{aligned} \int _0^t \sqrt{V_s} \mathrm {d}B_s = B^{X}(C_t\wedge T_0) \end{aligned}$$
(A.15)

and of a Poisson random measure N with intensity \(\mathrm {d}s \otimes \mathrm {d}\nu \otimes \Pi (\mathrm {d}r)\) such that

$$\begin{aligned} \sum \limits _{\{n:t^X_n<C_t\}} r^X_n = \sum \limits _{\{n:t_n<t\}}\Delta _n = \int _0^t \int _0^{V_{s-}}\int _0^{\infty } r\mathbf {1}_{\{r \ge 1\}}N(\mathrm {d}s,\mathrm {d}\nu ,\mathrm {d}r), \end{aligned}$$

where \(((\Delta _n, t_n) : n \in \mathbb {N})\) is a fixed but arbitrary labeling of the jump times and sizes of V and \(((r^X_n, t^X_n) : n \in \mathbb {N})\) are the atoms of \(N^X\). We have in an \(L^2\) sense that

$$\begin{aligned}&\lim \limits _{\varepsilon \searrow 0}\left[ \sum \limits _{\{n:t_n<t\}} \Delta _n \mathbf {1}_{\{\Delta _n> \varepsilon \}} - \int _0^t V_s \mathrm {d}s \int _{\varepsilon }^{\infty } r\Pi (\mathrm {d}r)\right] \nonumber \\&\quad = \lim \limits _{\varepsilon \searrow 0}\left[ \sum \limits _{\{n:t^X_n<C_t\}}r^X_n \mathbf {1}_{\{r^X_n> \varepsilon \}} - \int _0^{C_t} \mathrm {d}s \int _{\varepsilon }^{\infty } r\Pi (\mathrm {d}r)\right] , \end{aligned}$$
(A.16)

so that the compensated measures satisfy

$$\begin{aligned} \int _0^t \int _0^{V_{s^{-}}}\int _0^\infty r {\widetilde{N}}(\mathrm {d}s,\mathrm {d}\nu ,\mathrm {d}r) = \int _0^{C_t}\int _0^{\infty } r{\widetilde{N}}^X(\mathrm {d}s,\mathrm {d}r). \end{aligned}$$
(A.17)

Inserting the identities (A.15), (A.16) and (A.17) into equation (A.14), we deduce that

$$\begin{aligned} \mathrm {d}U_{C_t} = -\alpha \mathrm {d}C_t + \sigma \sqrt{V_t} \mathrm {d}B_t + \int _0^{V_{t-}}\int _0^{\infty } r {\widetilde{N}} (\mathrm {d}t,\mathrm {d}\nu ,\mathrm {d}r) - cU_{C_t} \mathrm {d}C_t. \end{aligned}$$

By (A.12), \(\mathrm {d}C_t = V_t \mathrm {d}t \) and \( \dfrac{G(|U_{C_t}| )}{U_{C_t}} \mathrm {d}C_t = G(|V_t|)\mathrm {d}t\), thus \(V=(V_t: t\ge 0)\) is a solution of (A.13). Uniqueness for (A.13) follows from general results in [14]. \(\square \)

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Berestycki, J., Fittipaldi, M.C. & Fontbona, J. Ray–Knight representation of flows of branching processes with competition by pruning of Lévy trees. Probab. Theory Relat. Fields 172, 725–788 (2018). https://doi.org/10.1007/s00440-017-0819-4

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  • DOI: https://doi.org/10.1007/s00440-017-0819-4

Keywords

  • Continuous state branching processes
  • Competition
  • Stochastic flows
  • Lévy-tree exploration and local times
  • Poisson-snake
  • Pruning
  • Ray–Knight representation

Mathematics Subject Classification

  • 60J25
  • 60G57
  • 60J80