June 11, 2026
A generalised inverse first passage problem asks whether, given a probability measure \(p\) on \([0,\infty]\), one can find a boundary \(b:[0,\infty]\to \mathbb{R}\) such that the stopping time: \[\begin{align} \tau:=\inf\left\{t:\Lambda\int_0^t \omega(W_s-b(s))ds \geq U\right\} \end{align}\] has distribution \(p\), where \(U\sim Exp(1)\), \(\Lambda\in(0,\infty)\) and \(\omega\) is a monotonic decreasing function. We construct a branching-selection particle system whose hydrodynamic limit is governed by a free boundary problem and connect this to the generalised inverse first passage problem. In the \(N\)-particle system, particles move as independent Brownian motions, branch at a prescribed rate, and are removed at a rate proportional to their location relative to a position \(b^N(t)\) which is a function of the empirical distribution. We identify the limit of \(b^N\) as the solution of the inverse first passage problem.
The inverse first passage problem (IFPP) is a classical problem in probability which originates from a question posed by Shiryaev. Given a standard Brownian motion \((W_t)_{t\geq 0}\) and a probability distribution \(p\) on \([0,\infty]\), does there exist a boundary \(b(t)\) such that the first passage time \(\tau:=\inf\{t:W_t\leq b(t)\}\) has distribution \(\tau\sim p\). The problem gained interest as it relates to credit default; defaulting being modelled as the point at which the stochastic process crosses the boundary \(b\), and \(p\) being the distribution of default time, which may be observed from the price of credit-default options (see [1], [2], [3], for example).
It was noted by Klump [4] that, in the special case that \(p\) is the Exponential distribution \(Exp(1)\), the IFPP can be related to the hydrodynamic limit of the \(N\)-branching Brownian motion particle system (\(N\)-BBM). In the \(N\)-BBM, \(N\) particles move on \(\mathbb{R}\) as independent Brownian motions, each branching independently at rate \(1\). Simultaneously with branching, the leftmost particle is deleted, so that there remains exactly \(N\) particles at all times. It was shown by De Masi et al. [5] that as \(N\to\infty\), the empirical density of the \(N\)-BBM converges to the solution of the free boundary problem: \[\begin{align} \begin{cases} u_t=\frac{1}{2}u_{xx} + u & x\in(L_t,\infty), t>0, \\ u(x,t)=0 & x\notin (L_t,\infty), t>0, \\ \int_{L_t}^\infty u(y,t)dy=1 & t>0. \\ \end{cases} \end{align}\] By the Feynman-Kac theorem, we can understand the solution \(u\) by \(\int_x^\infty u(y,t)dy = e^t\mathbb{P}(W_t\geq x, \tau>t)\), where \(W\) is a Brownian motion with initial density \(u(x,0)\) and \(\tau:=\inf\{t:W_t\leq L_t\}\). Then taking \(x\to -\infty\), the condition \(\int_{L_t}^\infty u(y,t)dy=1\) gives that \(\mathbb{P}(\tau>t)=e^{-t}\). That is, if \((u,L)\) solves the free boundary problem, then \(L\) solves the inverse first passage problem with \(p=Exp(1)\). Since the location of the leftmost particle of the \(N\)-BBM (\(X_1^N(t)\), say) converges to the free boundary of the PDE, we can think of \(X_1^N(t)\) as approximating the solution of the IFPP. As noted by Klump [4], to obtain a solution to the general IFPP, one must simply consider the \(N\)-BBM with time dependent branching rate \(f(t)\), where \(p([t,\infty))=\exp(-\int_0^t f(s)ds)\). Under this small adaptation, the \(N\)-BBM should have a hydrodynamic limit which solves the PDE \(u_t(x,t)= \frac{1}{2}u_{xx}(x,t)+f(t)u(x,t)\) on \((L_t,\infty)\); correspondingly, if \((u,L)\) solves this PDE, then \(L\) solves the IFPP with \(p((t,\infty)=\exp(-\int_0^t f(s)ds)\).
In [6], [7], a generalisation of the IFPP is studied. They consider the stopping time: \[\begin{align} \tau:=\inf\Big\{t:\Lambda \int_0^t \omega(W_s-b(s))ds > U\Big\}, \end{align}\] where \(\omega\) is a decreasing function with \(\omega(-\infty)=1\) and \(\omega(\infty)=0\) and \(U\sim Exp(1)\). Thinking of \(\tau\) as a default time, this generalisation considers that defaulting does not occur at the immediate instant that \(W_t\leq b(t)\), but as a function of the path of \(W\) up to time \(t\), and is used as a model of defaulting (see [8], for example). If we consider \(\omega(x)=\mathbb{1}_{x\leq 0}\), this is the first time that \(W\) has spent more than \(Exp(1)/\Lambda\) amount of time below \(b\). Then the soft-killed inverse first passage problem (SKIFPP) is the problem of finding \(b\) such that \(\tau\sim p\) given a distribution \(p\) and an initial distribution for \(W_0\). It is already known that, with some additional assumptions on the initial density and on \(p\), the SKIFPP has a unique solution \(b\) when \(\omega(x)=\mathbb{1}_{x\leq 0}\) ([7]) or \(\omega\in C^3(\mathbb{R})\) with \(\{x:\omega(x)\in (0,1)\}\) bounded ([6]). Moreover, the sub-probability density of \(\{W_t,\tau>t\}\) is described by the free boundary problem: \[\begin{align} \begin{cases} u_t(x,t)=\frac{1}{2}u_{xx}(x,t) - \omega(x-b(t))u(x,t) & x\in \mathbb{R}, t>0, \\ u(x,0)=\rho(x) & x\in \mathbb{R}\\ \int_\mathbb{R} u(y,t)dy=p([t,\infty))\\ \end{cases} \end{align}\] The free boundary \(b\) is determined by the integral constraint \(\int_{\mathbb{R}}u(y,t)dy=p([t,\infty))\).
In this paper, we find the branching-selection particle system whose hydrodynamic limit corresponds to the solution of this SKIFPP. In particular, we will construct a branching-selection particle system whose empirical density converges to the solution of the free boundary problem: \[\begin{align} \label{omegaHydroLim} \begin{cases} u_t(x,t)=\frac{1}{2}u_{xx}(x,t) + f(t)u(x,t) - \Lambda\omega(x-b(t))u(x,t) & x\in \mathbb{R}, t>0, \\ u(x,0)=\rho(x) & x\in \mathbb{R}\\ \int_\mathbb{R} u(y,t)dy=1\\ \end{cases} \end{align}\tag{1}\] and show that if \((u,b)\) is a solution, then \(b\) solves the SKIFPP with initial distribution \(\rho\) and \(p([t,\infty))=\exp(-\int_0^t f(s)ds)\). In the particle system, particles branch at rate \(f\) and a particle at location \(x\) is deleted with probability proportional to \(\omega(x-b^N(t))\) where \(b^N(t)\) is a function of the empirical distribution at time \(t\). Accordingly, \(b^N(t)\) can be seen as an approximation of the solution \(b\) of the SKIFPP.
In the specific case that \(\omega(x)=\mathbb{1}_{x\leq 0}\), if we choose \(f(t)\equiv 1\) and \(\Lambda\in (1,\infty)\), then we can combine the PDE and the integral condition to yield the non-linear PDE \[\begin{align} u_t(x,t)=\frac{1}{2} u_{xx}(x,t)+u(x,t)\left(1-\Lambda\mathbb{1}_{\int_{-\infty}^x u(y,t)dy \leq \Lambda^{-1}}\right). \end{align}\] In this case, \(b(t)\) is the point left of which a proportion \(\Lambda^{-1}\) of the mass of \(u\) lies.
Essentially this process will be a branching Brownian motion in which a particle at location \(X(t)\) is deleted at rate \(\Lambda\omega(X(t)-b^N(t))\) where \(b^N:[0,\infty)\to \mathbb{R}\) is a reference point chosen so that the total branching rate at time \(t\) is \(Nf(t)\). Branching and deletion will happen simultaneously so that there are always exaclty \(N\) particles.
We now give a formal construction. Throughout this work, we will assume that \(\omega:\mathbb{R}\to [0,1]\), the selection weighting, is a monotone decreasing function with \(\omega(x)\to 1\) as \(x\to-\infty\) and \(\omega(x)\to 0\) as \(x\to\infty\). We also assume that \(f:[0,\infty)\to [0,\infty)\), the branching rate function,is bounded and that \(\Lambda > \|f\|_\infty\).
Then we will describe the \((\omega,f,\Lambda,N)\)-BBM at time \(t\) by the probability measure \(\mu_t^N:=\frac{1}{N}\sum_{i=1}^N \delta_{X_i^N(t)}\) where \(X^N(t):=(X_1^N(t),\ldots,X_N^N(t))\) denote the locations of the \(N\) particles in the system. Let \(X^N(0)\in \mathbb{R}^N\) be some initial configuration with \(X_1^N(0)\leq X_2^N(0)\leq \cdots\leq X^N_N(0)\) and define \(\rho^N:=\frac{1}{N}\sum_{i=1}^N \delta_{X^N_i(0)}\). Suppose that \(\rho^N\) converges weakly to \(\rho\), with \(\rho\ll\lambda\) (\(\rho\) is absolutely continuous with respect to the Lebesgue measure \(\lambda\)). We will use \(\rho\) to denote both the probability distribution and its density function.
Let \(\mathcal{N}(t)\) be a Poisson process of intensity \(Nf(t)\) and let the discontinuities of \(\mathcal{N}(t)\) be \(\tau_1<\tau_2<\cdots\). These will be the branching times of the process. We construct the process inductively as follows. Initially, the \(N\) particles have configuration \(X^N(0)\). Then on \([0,\tau_1)\), drive the particle at location \(X_i^N(0)\) by Brownian motion \((W_i(t))_{0\leq s<\tau_1}\), so the locations of the \(N\) particles at time \(t\in [0,\tau_1)\) are \(X_i^N(t):=X_i^N(0)+W_i(t)\) for \(i\in [N]\), where \([N]\) denotes the set \(\{1,2,\ldots,N\}\). At each stopping time \(\tau_i\), we will relabel the particles so that they are ordered. That is \(X_1^N(\tau_i)\leq X_2^N(\tau_i)\leq \cdots \leq X_N^N(\tau_i)\) for all \(i\), but this ordering may not hold for \(t\in (\tau_i,\tau_{i+1})\). We define \(\Theta^N:\mathbb{R}^N\to\mathbb{R}^N\) to be the function which orders \(\mathbb{R}^N\)-valued vectors, and say that \(\Theta^N_i(\vec{v})\) is the \(i\)th smallest element of \(\vec{v}\).
Now define \(b^N(t):=\inf\left\{x:\Lambda \langle \mu_t^N,\omega(\cdot-x)\rangle \geq f(t)\right\}\), where the notation \(\langle \mu,g\rangle\) means \(\int_\mathbb{R} g(x)\mu(dx)\). Since \(\omega\) is decreasing, \(\langle \mu_t^N,1\rangle=1\), and \(\Lambda > \|f\|_\infty\), therefore \(b^N(t)\) is well defined. Moreover we can observe that if \(\omega\) is continuous, then \(\frac{\Lambda}{N} \sum_{i=1}^N \omega(X_i^N(t)-b^N(t))\) is exactly \(f(t)\). \(b^N(t)\) defines a moving boundary. Then at time \(\tau_1\), the particle of rank \(I_1\) branches, and simultaneously particle of rank \(J_1\) is deleted, where \(\mathbb{P}(I_1=i)=1/N\) and independently \(\mathbb{P}(J_1=j)=\frac{\omega(\Theta^N_j(X^N(\tau_1-))-b^N(\tau_1-))}{\sum_{\ell=1}^N \omega(\Theta^N_\ell(X^N(\tau_1-))-b^N(\tau_1-))}\)for \(i,j\in [N]\). Repeat this inductively for each interval \([\tau_k,\tau_{k+1}]\).
The main result of this paper is the following hydrodynamic limit, which describes the limiting behaviour of the above described \((\omega,f,\Lambda,N)\)-BBM process by the free boundary problem: \[\begin{align} \label{omegaNBMMHydro} \begin{cases} u_t = \frac{1}{2} u_{xx} + f(t)u - \Lambda\omega(x-b(t))u & x\in \mathbb{R},\;t>0,\\ \int_\mathbb{R} u(y,t)dy = 1 & t>0,\\ u(x,0)=\rho(x) & x\in \mathbb{R}. \end{cases} \end{align}\tag{2}\]
Theorem 1. Let \(\omega:\mathbb{R}\to[0,1]\) be a selection weighting such that either \(\omega\) is continuous or \(\omega(x)=\mathbb{1}_{x\leq 0}\). Let \((\mu_t^N)_{t\geq 0}\) be the measure-valued process describing the \((\omega,f,\Lambda,N)\)-BBM process with initial condition \(\rho^N\) and let \(Q_T^N\) denote its law. Then for any \(T\in (0,\infty)\), the sequence of measures \(Q_T^N\) has a weak limit, \(Q_T^\infty\). Moreover, if \((\mu_t)_{t\in [0,T]}\sim Q_T^\infty\) and \(b(t):=\inf\{x:\Lambda\langle \mu_t,\omega(\cdot-x)\rangle \geq f(t)\}\), then \(\mu_t\) has density \(u(x,t)\) and \((u,b)\) is \(Q_T^\infty\)-a.s. the unique weak solution to the FBP 1 .
Using the results of [6], [7], we also connect this to the SKIFPP.
Theorem 2. Suppose that \(f\) is continuous, and that \(\rho\in C^2(\mathbb{R})\) with \(\rho\), \(\rho'\), and \(\rho''\) bounded. Suppose that either \(\omega(x)=\mathbb{1}_{x\leq 0}\) or \(\omega\in C^3(\mathbb{R})\) is non-increasing and \(\{x:\omega(x)\in (0,1)\}\) is bounded. If \((u,b)\) is the unique weak solution to 2 and \(\tau\) is the stopping time: \[\begin{align} \tau:=\inf\left\{t:\Lambda\int_0^t \omega(W_s-b(s))ds\geq U\right\}, \end{align}\] where \(U\sim Exp(1)\) and \(W_0\sim \rho\), then \(b\) is the unique barrier such that \(\mathbb{P}(\tau>y)=\exp(-\int_0^y f(s)ds)\). That is, \(b\) solves the SKIFPP.
Proof. Let \(\tilde{\omega}(x)=\omega(\Lambda^{-1/2}x)\), and suppose that \((v,c)\) is a weak solution to the PDE: \[\begin{align} \label{pdeFromLit} \begin{cases} v_t(x,t) = \frac{1}{2} v_{xx}(x,t) -\tilde{\omega}(x-c(t))v(x,t) & x\in \mathbb{R},\;t>0,\\ \int_\mathbb{R} v(y,t)dy = G(t):=\exp(-\int_0^{t/\Lambda}f(s)ds) & t>0,\\ \int_{\mathbb{R}} v(y,0)\omega(y-c(0))dy=f(0)\\ v(x,0)=\Lambda^{1/2}\rho(\Lambda^{-1/2}x) & x\in \mathbb{R}.\\ \end{cases} \end{align}\tag{3}\] Then it is easily checked that \[\begin{align} (\tilde{u}(x,t),\tilde{b}(t)):=\left(\Lambda^{1/2}\exp(\int_0^t f(s)ds)v(\Lambda^{1/2}x,\Lambda t),\Lambda^{-1/2}c(\Lambda t)\right) \end{align}\] is a weak solution of 2 , and therefore by assumption coincides with limit \((u,b)\) found in Theorem 1. In the case that \(\omega(x)=\mathbb{1}_{x\leq 0}\), Theorem 1.8 [7] tells us that 3 has a unique weak solution and in the case that \(\omega\in C^3(\mathbb{R})\) and \(\{x:\omega(x)\in (0,1)\}\) is bounded, Corollary 2.7 [6] tells us that 3 has a unique (classical) solution. Theorem 1.8 [7] and Corollary 2.7 [6] also tell that \(c\) is the unique continuous barrier such that \[\begin{align} \mathbb{P}\left(\inf\left\{t:\int_0^t \omega(W_s-c(s))ds \geq U\right\}>y\right)=G(y)=\exp(-\int_0^{y/\Lambda}f(s)ds). \end{align}\] Then using the fact that \(\Lambda^{-1/2}W_{\Lambda z}\overset{d}{=}W_z\) and \(c(s)=\Lambda^{1/2}b(s/\Lambda)\), this implies that \[\begin{align} \mathbb{P}\left(\inf\left\{t:\Lambda \int_0^t \omega(W_s - b(s))ds\geq U\right\}>y\right)\exp(-\int_0^y f(s)ds), \end{align}\] which is to say that \(b\) solves the SKIFPP. ◻
The method in this section follows similarly to Demircigil and Tomasevic [9] and the author’s paper [10]. As in [10], define the set of test functions \(\mathcal{BC}^{2,1}(\mathbb{R}\times [0,\infty),\mathbb{R})\) to be the set of bounded and continuous functions \(f(x,t):\mathbb{R}\times [0,\infty)\to\mathbb{R}\) with bounded and continuous derivatives \(f_x\), \(f_{xx}\), and \(f_t\). We also use \(C_b(\mathbb{R})\) to denote the continuous bounded functions \(\mathbb{R}\to\mathbb{R}\). For a (local) martingale \((M_t)_{t\geq 0}\), we denote the quadratic variation process by \(([M]_t)_{t\geq 0}\).
Proposition 3. Let \((\mu_t^N)_{t\geq 0}\) be the measure-valued process describing the \((\omega,f,\Lambda,N)\)-BBM process, and define \(b^N(t):=\inf\{x:\Lambda\langle \mu_t^N,\omega(\cdot-x)\rangle\geq f(t)\}\). Then for any bounded \(\omega\) and any test function \(\phi(x,t)\in \mathcal{BC}^{2,1}(\mathbb{R}\times [0,\infty),\mathbb{R})\) we have \[\begin{align} \label{probRepIto} \langle \mu_t^N,\phi(\cdot,t)\rangle = \langle \rho^N, \phi(\cdot,0)\rangle &+ \int_0^t \langle \mu_s^N, \frac{1}{2} \phi_{xx}(\cdot,s) + \phi_t(\cdot,s)+f(s)\phi(\cdot,s)\rangle \\ &- \frac{f(s)\langle \mu_s^N,\phi(\cdot,s)\omega(\cdot-b^N(s))}{\langle \mu_s^N,\omega(\cdot-b^N(s))\rangle} ds \nonumber + M_t^{N,W} + M_t^{N,b}, \end{align}\qquad{(1)}\] where \(M^{N,W}_t\) is a continuous local martingale and \(M^{N,b}_t\) is a local martingale, with \(\mathbb{E}[[M^{N,W}]_t],\mathbb{E}[[M^{N,b}]_t]\to 0\) as \(N\to\infty\) for fixed \(t\geq 0\).
Proof. Let the branching times of the particle system be \(\tau_1<\tau_2<\cdots\). Then by Ito’s formula, for \(t\in (\tau_{k-1},\tau_k)\), we have: \[\phi(X_i^N(t),t)=\phi(X_i^N(\tau_{k-1}),\tau_{k-1})+\int_{\tau_{k-1}}^t \phi_t(X_i^N(s),s)+\frac{1}{2} \phi_{xx}(X_i^N(s),s)\rangle ds + \int_{\tau_{k-1}}^t \phi_x(X_i^N(s),s)dW_i(s),\] so we have that: \[\langle \mu_t^N,\phi(\cdot,t)\rangle = \langle \mu_{\tau_{k-1}},\phi(\cdot,\tau_{k-1})\rangle + \int_{\tau_{k-1}}^t \langle \mu_s^N, \phi_t(\cdot,s)+\frac{1}{2} \phi_{xx}(\cdot,s)ds + \sum_{i=1}^N\int_{\tau_{k-1}}^t \phi_x(X_i^N(s),s) dW_i(s).\] Then, at time \(\tau_k\), the quantity \(\langle \mu_t^N, \phi(\cdot,t)\rangle\) increases by \(\frac{1}{N}(\phi(\Theta^N_{I_i}(X^N(\tau_k-)),\tau_k)-\phi(\Theta^N_{J_i}(X^N(\tau_k-)),\tau_k))\) due to branching and deletion. Therefore integrating over \([0,t]\), combining the terms due to Brownian motion and the terms due to branching and selection, we yield: \[\begin{align} \langle \mu_t^N,\phi(\cdot,t)\rangle = \langle \rho^N,\phi(\cdot,0)\rangle &+ \int_0^t \langle \mu_s^N,\frac{1}{2} \phi_{xx}(\cdot,s)+\phi_t(\cdot,s)\rangle + M^{N,W}_t \\ &+ \frac{1}{N}\int_0^t \phi(\Theta^N_{I_{\mathcal{N}(s)}}(X^N(s-)),s)-\phi(\Theta^N_{J_{\mathcal{N}(s)}}(X^N(s-)),s)\mathcal{N}(ds) \end{align}\] where \(M^{N,W}_t:=\frac{1}{N} \sum_{i=1}^N \int_0^t \phi_x(X_i^N(s),s)dW_i(s)\) is a continuous local martingale. Next we prove that \[\begin{align} M^{N,b}_t:=\frac{1}{N} \int_0^t \phi(\Theta^N_{I_{\mathcal{N}(s)}}&(X^N(s-)),s)-\phi(\Theta^N_{J_{\mathcal{N}(s)}}(X^N(s-)),s)\mathcal{N}(ds) \\ &- \int_0^t f(s)\langle \mu_s^N,\phi(\cdot,s)\rangle - \frac{f(s)\langle \mu_s^N,\phi(\cdot,s)\omega(\cdot-b^N(s))\rangle}{\langle \mu_s^N,\omega(\cdot-b^N(s))\rangle}ds \end{align}\] is a local martingale. Certainly if \(\mathcal{F}_t:=\sigma(W_i(s),\mathcal{N}(s),I_j,J_j:s\leq t, i\in [N],j\leq \mathcal{N}(t))\) is the natural filtration of the process, then \(\mathbb{E}[M^{N,b}_t-M^{N,b}_s|\mathcal{F}_s]=0\). This follows by the independence of \(I\) and \(J\) from \(\mathcal{N}\) so that: \[\begin{align} \mathbb{E}\Bigg[&\frac{1}{N}\int_s^t \phi(\Theta^N_{I_{\mathcal{N}(u)}}(X^N(u-)),u)-\phi(\Theta^N_{J_{\mathcal{N}(u)}}(X^N(u-)),u)\mathcal{N}(du)|\mathcal{F}_s\Big]\\ &=\mathbb{E}\Bigg[\frac{1}{N}\int_s^t \phi(\Theta^N_{I_{\mathcal{N}(u)}}(X^N(u-)),u)-\phi(\Theta^N_{J_{\mathcal{N}(u)}}(X^N(u-)),u)\mathcal{N}(du)\Bigg]\\ &=\mathbb{E}\Bigg[\frac{1}{N}\sum_{i,j\in[N]}\int_s^t \big(\phi(\Theta^N_i(X^N(u-)),u)-\phi(\Theta^N_j(X^N(u-)),u)\big)\mathbb{P}(I_{\mathcal{N}(u)}=i,J_{\mathcal{N}(u)}=j)\mathcal{N}(du)]\\ &=\mathbb{E}\Bigg[\frac{1}{N^2}\sum_{i,j\in[N]}\int_s^t \big( \phi(\Theta^N_i(X^N(u-)),u)-\phi(\Theta^N_j(X^N(u-)),u)\big)\frac{\omega(\Theta^N_j(X^N(u-))-b^N(u-))}{\sum_{k=1}^N\omega(\Theta^N_k(X^N(u-))-b^N(u-))}\mathcal{N}(du)\Bigg]\\ &=\mathbb{E}\Bigg[\int_s^t f(u)\langle \mu_u^N,\phi(\cdot,u)\rangle - \frac{f(u)\langle \mu_u^N,\phi(\cdot,u)\omega(\cdot-b^N(u))\rangle}{\langle \mu_u^N,\omega(\cdot-b^N(u))\rangle}du\Bigg], \end{align}\] where the final equality follows from the fact that \(\mathcal{N}(t)-N\int_0^t f(s)ds\) is a martingale (the ‘compensated Poisson process’). Therefore \[\begin{align} \langle \mu_t^N,\phi(\cdot,t)\rangle = \langle \rho^N,&\phi(\cdot,0)\rangle + \int_0^t \langle \mu_s^N,\frac{1}{2} \phi_{xx}(\cdot,s)+\phi_t(\cdot,s)\rangle ds + M^{N,W}_t + M^{N,b}_t \\ &+\int_0^t f(s)\langle \mu_s^N,\phi(\cdot,s)\rangle - \frac{f(s)\langle \mu_s^N,\phi(\cdot,s)\omega(\cdot-b^N(s))\rangle}{\langle \mu_s^N,\omega(\cdot-b^N(s))\rangle} ds \end{align}\] It just remains to prove that \(\mathbb{E}[[M^{N,W}]_t]\) and \(\mathbb{E}[[M^{N,b}]_t]\) converge to \(0\) as \(N\to\infty\). For fixed \(t\geq 0\), Itô’s isometry gives: \[\mathbb{E}\left[ \frac{1}{N} \sum_{i=1}^N \int_0^t \phi(X_i^N(s),s)dW_i(s)\right]_t = \frac{1}{N^2}\sum_{i=1}^N \int_0^t \phi_x(X_i^N(s),s)^2 d[W_i]_s = \frac{1}{N} \int_0^t \langle \mu_s^N, \phi_x(\cdot,s)^2\rangle ds.\] Thus since \(\phi_x\) is bounded, this quantity converges to \(0\) as \(N\to\infty\). We also calculate that: \[\begin{align} \mathbb{E}&[[M^{N,b}]_t] \\ &= \mathbb{E}\left[\int_0^t \frac{1}{N^2}\sum_{i,j\in[N]}\frac{\omega(X_j^N(s-)-b^N(s-))(\phi(X_j^N(s-),s)-\phi(X_i^N(s-),s))}{\sum_{k=1}^N \omega(\Theta^N_k(X^N(s-))-b^N(s-))}(\mathcal{N}(ds)-Nf(s)ds)\right]_t \\ &= \int_0^t \left(\frac{1}{N^2}\sum_{i,j\in[N]}\frac{\omega(X_j^N(s-)-b^N(s-))(\phi(X_j^N(s-),s)-\phi(X_i^N(s-),s))}{\sum_{k=1}^N\omega(\Theta^N_k(X^N(s-))-b^N(s-))}\right)^2 d[\mathcal{N}(\cdot)-Nf(\cdot)]_s \end{align}\] Now \(d[\mathcal{N}(\cdot)-Nf(\cdot)]_s=Nf(s)ds\), and since \(\phi\) is bounded, say by \(\Phi\), we have \[\begin{align} \Bigg(\frac{1}{N^2}\sum_{i,j\in[N]}&\frac{\omega(X_j^N(s-)-b^N(s-))(\phi(X_j^N(s-),s)-\phi(X_i^N(s-),s))}{\sum_{k=1}^N\omega(\Theta^N_k(X^N(s-))-b^N(s-))}\Bigg)^2 \\ &\leq \left( \sum_{i \in [N]} \frac{2\Phi}{N^2}\frac{\sum_{j=1}^N\omega(X_j^N(s-)-b^N(s-))}{\sum_{k=1}^N \omega(X_k^N(s-)-b^N(s-))}\right)^2 = \left(\sum_{i=1}^N \frac{2\Phi}{N^2}\right)^2 = \frac{4\Phi^2}{N^2}, \end{align}\] therefore, since \(f\) is bounded by \(\Lambda\), \[\mathbb{E}[M^{N,b}]_t \leq \int_0^t \frac{4\Phi^2}{N}f(s)ds \leq \frac{4\Lambda\Phi^2 t}{N} \xrightarrow[N\to\infty]{} 0,\] thus concluding the proof. ◻
Since we seek the limit of the sequence \(((\mu_t^N)_{t\geq 0})_{N=1,2,\ldots}\), we then need to show that the sequence of laws is tight, which is the substance of the next theorem. This proposition follows the structure of Proposition 5, [10]. In particular, we show tightness in the space \(\mathcal{P}(\mathcal{D}([0,T],\mathcal{M}_F^w(\bar{\mathbb{R}})))\), where \(\mathcal{M}_F^w(\bar{\mathbb{R}})\) is the space of finite measures on the extended real line under the weak topology. We do this as it will help to prove a compact containment condition.
Proposition 4. Fix \(T>0\), and let \(Q^N_T\) denote the law of \((\mu_t^N)_{t\in [0,T]}\). Then the sequence \((Q^N_T)_{N=1,2,\ldots}\) is tight in \(\mathcal{P}(\mathcal{D}([0,T],\mathcal{M}^w_F(\bar{\mathbb{R}})))\) with respect to the Skorokhod topology.
Proof. By Theorem 1.18, [11], in order to show tightness, it is sufficient to show the following conditions:
Compact containment: For all \(\epsilon > 0\) there exists a compact subset \(K_\epsilon\subseteq \mathcal{M}_F^w(\bar{\mathbb{R}})\) such that \[\inf_{N\in \mathbb{N}} \mathbb{P}(\mu_t^N \in K_\epsilon\; \forall \, t\in [0,T])>1-\epsilon.\]
Tightness of real-valued processes: The sequence of laws of \((\langle \mu_t^N,f\rangle)_{0\leq t\leq T}\) is tight in \(\mathcal{P}(\mathcal{D}([0,T],\mathbb{R}))\) for every function \(f\) in a dense subset of \(C_b(\bar{\mathbb{R}})\).
Since the set \(\{\mu:\mu(\bar{\mathbb{R}})=1\}\) is a compact subset of \(\mathcal{M}_F^w(\bar{\mathbb{R}})\) and \(\mu_t^N(\bar{\mathbb{R}})=1\) for all \(t\geq 0\) and \(N\in \mathbb{N}\), thus condition (i) holds immediately. Next we show condition (ii). Since each \(\mu_t^N\) is supported on \(\mathbb{R}\), condition (ii) is equivalent to showing tightness of the sequence of laws of \((\langle \mu_t^N,f\rangle )_{t\in [0,T]}\) for all \(f\) in a dense subset of \(C_b(\mathbb{R})\); we will take \(\mathcal{BC}^{2,1}(\mathbb{R}\times [0,\infty),\mathbb{R})\cap C_b(\mathbb{R})\) for our dense subset. By Aldous’ tightness criterion (see, for example, Theorem 16.10, [12]), the sequence of laws of \(((\langle \mu_t^N,\phi\rangle)_{t\in [0,T]})_{N=1,2,\ldots}\) is tight if:
For every \(m>0\), \(\lim_{a\to\infty}\limsup_{N\to\infty}\mathbb{P}(\sup_{0\leq t\leq m}|\langle \mu_t^N,\phi\rangle|\geq a)=0.\)
For each \(\epsilon, \eta, m>0\), there exists \(\delta_0,N_0\) such that if \(\delta\leq \delta_0\) and \(N\geq N_0\) and \(\tau\) is a stopping time \(\tau\leq m\) then \(\mathbb{P}(|\langle \mu_{\tau+\delta}^N,\phi\rangle - \langle \mu_\tau^N,\phi\rangle |\geq \epsilon)\leq \eta\).
Since \(|\langle \mu_t^N,\phi\rangle|\leq \|\phi\|_\infty\), condition A follows immediately. For condition B, note that by ?? and the boundedness of \(f\), \(\phi\), \(\phi_t\), \(\phi_{xx}\), \(\omega\), and \(\Lambda\), there exists constant \(\kappa\) such that \[\begin{align} |\langle \mu_{\tau+\delta}^N,\phi\rangle - \langle \mu_\tau^N,\phi\rangle |\leq \kappa\delta + |M^{N,W}_{\tau+\delta}-M^{N,W}_{\tau}|+|M^{N,b}_{\tau+\delta}-M^{N,b}_{\tau}|, \end{align}\] therefore by the Markov inequality and the Burkholder-Davis-Gundy inequality, there exists a constant \(\hat{\kappa}\) \[\begin{align} \mathbb{P}(|\langle \mu_{\tau+\delta}^N,\phi,\rangle-\langle \mu_\tau^N,\phi\rangle|\geq \epsilon)&\leq \mathbb{1}_{\kappa\delta\geq \epsilon/3}+\mathbb{P}(|M_{\tau+\delta}^{N,W}-M^{N,W}_\tau|^2\geq \epsilon^2/9)+\mathbb{P}(|M_{\tau+\delta}^{N,b}-M_{\tau}^{N,b}|^2\geq \epsilon^2/9)\\ &\leq \mathbb{1}_{\kappa\delta\geq \epsilon/3}+\frac{9}{\epsilon^2}\mathbb{E}[|M_{\tau+\delta}^{N,W}-M^{N,W}_\tau|^2]+\frac{9}{\epsilon^2}\mathbb{E}[|M_{\tau+\delta}^{N,b}-M^{N,b}_\tau|^2]\\ &\leq \mathbb{1}_{\kappa\delta \geq \epsilon/3}+\frac{9\hat{\kappa}}{\epsilon^2}(\mathbb{E}[[M^{N,W}]_\delta]+\mathbb{E}[[M^{N,b}]_\delta])\xrightarrow[\delta\to 0,N\to\infty]{}0. \end{align}\] This shows condition B, and therefore \(((\langle \mu_t^N,\phi\rangle)_{t\in [0,T]})_{N=1,2,\ldots}\) is tight in \(\mathcal{D}([0,T],\mathbb{R})\). Hence condition (ii) holds, and thus the sequence \((Q^N_T)_{N=1,2,\ldots}\) is tight in \(\mathcal{D}([0,T],\mathcal{M}^w_F(\bar{\mathbb{R}}))\). ◻
We need not worry about the generalisation from \(\mathcal{M}_F^w(\mathbb{R})\) to \(\mathcal{M}_F^w(\bar{\mathbb{R}})\) since the next theorem shows that if \(Q^\infty_T\) is a subsequential limit of \((Q^N_T)_{N\geq 1}\) then \(Q^\infty_T\)-almost all measure-valued processes have no mass on \(\{\pm\infty\}\) for any \(t\in [0,T]\). That is, if \((\mu_t)_{t\in [0,T]}\sim Q^\infty_T\), then \(\mu:[0,T]\mapsto \mathcal{M}_F^w(\mathbb{R})\) almost surely. It also shows that \(\mu_t\) has a density for all \(t\in [0,T]\) \(Q^\infty_T\)-almost surely.
Proposition 5. Let \(Q^\infty_T\) be a subsequential limit of \((Q^N_T)_{N\geq 1}\) and let \((\mu_t)_{t\in [0,T]}\sim Q^\infty_T\). Then \(\mu_t\ll\lambda\) and \(\mu_t(\{-\infty,\infty\})=0\) for all \(t\in [0,T]\) \(Q_T^\infty\)-a.s.
Proof. Consider a branching Brownian motion (BBM) \((X_u(t):u\in\mathcal{U}(t))_{t\geq 0}\) starting from initial configuration \(\rho^N\) and branching at the time-dependent rate \(f\). Let \(\mathcal{U}(t)\) denote the set of particles at time \(t\) and \(\mathcal{U}^i(t)\) denote the subset of \(\mathcal{U}(t)\) consisting of the descendants of the particle which starts at the location of the \(i\)th leftmost particle of \(\rho^N\). We will colour the particles of the BBM blue and red, and at each branching event, a particle will branch into two particles of like colour. Initially colour all particles blue. Then subsequently, at each branching time \(\tau_k\) of a blue particle, colour exactly one blue particle red, choosing the \(j\)th leftmost blue particle with probability \(\frac{\omega(\Theta^N_j(X^N(\tau_k-))-b^N(\tau_k-))}{\sum_{k=1}^N\omega(\Theta^N_k(X^N(\tau_k-))-b^N(\tau_k-))}\). By construction, the subset of blue particles describes a \((\omega,f,\Lambda,N)\)-BBM process. Let \(\hat{\mu}^N(t)=\frac{1}{N}\sum_{u\in\mathcal{U}(t)}\delta_{X_u(t)}\) be the rescaled empirical measure of this coloured BBM at time \(t\). By the above coupling, it is clear that \(\mu_t^N(A)\leq \hat{\mu}_t^N(A)\) for any measurable set \(A\). It is well known (see Appendix of [10] for a proof of the exact statement) that if \(\hat{Q}^N_T\) is the law of \((\hat{\mu}^N_t)_{t\in [0,T]}\), then \(\hat{Q}^N_T\) converges weakly to \(\hat{Q}^\infty_T\), and if \((\hat{\mu}_t)_{t\in [0,T]}\sim \hat{Q}^\infty_T\), then \(\hat{\mu}_t(dx)=u(x,t)dx\) \(\hat{Q}^\infty_T\)-a.s., where \(u(x,t)=\exp(\int_0^t f(s)ds)\frac{\partial}{\partial x}\mathbb{P}_\rho(B(t)\leq x)\) for Brownian motion \(B\) with initial distribution \(\rho\). In particular, \(u\) is the unique classical solution to the PDE \(u_t = \frac{1}{2} u_{xx}+f(t)u\). Then since \(\hat{\mu}_t\ll \lambda\) and \(\hat{\mu}_t(\{-\infty,\infty\})=0\), and \(\hat{\mu}^N_t\) dominates \(\mu_t^N\) for all \(N\), thus \(\mu_t\ll \lambda\) and \(\mu_t(\{-\infty,\infty\})=0\) for all \(t\in [0,T]\). ◻
Proposition 6. Let \(Q^\infty_T\) be a subsequential limit of \((Q^N_T)_{N=1,2,\ldots}\) and let \((\mu_t)_{t\in [0,T]}\sim Q^\infty_T\). Then the map \(t\mapsto \mu_t\) is continuous on \([0,T]\) with respect to the weak topology \(Q^\infty_T\)-a.s.
Proof. Recall the Kolmogorov continuity theorem, which states that if \(X:[0,\infty)\times \Omega\to S\) is a stochastic process and \((S,d)\) a metric space, and there exist positive \(\alpha,\beta,K\) such that \(\mathbb{E}[d(X_s,X_t)^\alpha]\leq K|t-s|^{1+\beta}\) for all \(s,t\), then there exists a modification of \(X\) which is continuous. So fix \(s,t\) and \(\phi\in \mathcal{BC}^{2,1}(\mathbb{R}\times [0,\infty),\mathbb{R})\). Since \(Q^N_T\) converges to \(Q_T^\infty\) in the Skorokhod topology, there exists a sequence \(\ell_N:[0,T]\to [0,T]\) such that \(\sup_{t\in [0,T]}|\ell_N(t)-t|\) and \(\sup_{t\in [0,T]}|\langle \phi,\lambda^N_{\ell_N(t)}\rangle-\langle \phi,\mu_t\rangle|\) converge to \(0\) as \(N\to\infty\). Then by Fatou’s lemma: \[\mathbb{E}[|\langle \phi,\mu_s\rangle - \langle \phi,\mu_t\rangle|^2]=\mathbb{E}[\lim_{N\to\infty}|\langle \phi,\mu^N_{\ell_N(s)}\rangle - \langle \phi,\mu^N_{\ell_N(t)}\rangle|^2]\leq \liminf_{N\to\infty}\mathbb{E}[|\langle \phi,\mu_{\ell_N(s)}^N\rangle - \langle \phi,\mu^N_{\ell_N(t)}\rangle |^2].\] Then as \(\omega\), \(\Lambda\), \(f\), \(\phi\), \(\phi_t\), and \(\phi_{xx}\) are bounded, and \(\langle\mu_s^N,1\rangle\equiv 1\), thus ?? yields that there exists a constant \(\kappa\) such that \[\begin{align} \mathbb{E}[|\langle \phi,\mu_{\ell_N(s)}^N\rangle - \langle \phi, \mu_{\ell_N(t)}^N\rangle|^2]\leq 9\left(\kappa|\ell_N(s)-\ell_N(t)|^2+\mathbb{E}[|M^{N,W}_{\ell_N(s)}-M^{N,W}_{\ell_N(t)}|^2]+\mathbb{E}[|M^{N,b}_{\ell_N(s)}-M^{N,b}_{\ell_N(t)}|^2]\right). \end{align}\] By the Bukrholder-Davis-Gundy inequality and Proposition 3, \(\mathbb{E}[|M^{N,W}_{\ell_N(s)}-M^{N,W}_{\ell_N(t)}|^2]\) and \(\mathbb{E}[|M^{N,b}_{\ell_N(s)}-M^{N,b}_{\ell_N(t)}|^2]\) converge to \(0\) as \(N\to\infty\). Moreover, as \(\sup_{t\in [0,T]}|\ell_N(t)-t|\xrightarrow[N\to\infty]{}0\), thus \[\begin{align} \liminf_{N\to\infty}\mathbb{E}[|\langle \phi,\mu_{\ell_N(s)}^N\rangle - \langle \phi, \mu_{\ell_N(t)}^N\rangle|^2]\leq 9\kappa|s-t|^2. \end{align}\] Therefore by the Kolmogorov continuity theorem, \(t\mapsto\langle \phi,\mu_s\rangle\) has a continuous modification on \([0,T]\), and thus, since it is a cadlag process, is almost surely continuous on \([0,T]\) (see Theorem 1, [13]). Therefore \(t\mapsto\langle \phi,\mu_t\rangle\) is continuous for all \(\phi\) in a countable dense subset of \(C_b(\mathbb{R})\) and therefore \((\mu_s)_{t\in [0,T]}\) is almost surely continuous with respect to the weak topology. ◻
Lemma 1. Suppose that the measure \(\mu^N_s\) converges weakly to the measure \(\mu_s\ll\lambda\) almost surely and \(b(s):=\inf\{x:\Lambda\langle \mu_s,\omega(\cdot-x)\rangle \geq f(s)\}\). If either \(\omega\) is continuous or \(\omega = \mathbb{1}_{\{x\leq 0\}}\), then for any \(\phi\in C_b(\mathbb{R})\), \(\langle \mu^N_s,\phi(\cdot)\omega(\cdot-b^N(s))\rangle\to\langle \mu_s, \phi(\cdot)\omega(\cdot-b(s))\rangle\) almost surely.
Proof. Fix an element \(\sigma\in \Omega\) in our state space at which \(\mu_s^N(\sigma)\) converges weakly to \(\mu_s(\sigma)\). We will omit dependence on \(\sigma\) here onwards for readability.
Consider the first case that \(\omega\) is continuous. Define the function \(B(x,\nu):=\langle \omega(\cdot-x),\nu\rangle\). Then for any probability measure on \(\mathbb{R}\), \(B\) is a continuous and non-decreasing function of \(x\) with \(\lim_{x\to\infty}B(x,\nu)=1\), \(\lim_{x\to-\infty}B(x,\nu)=0\), and \(B(b^N(s),\mu^N_s)=f(s)/\Lambda=B(b(s),\mu_s)\) for any \(s\). Since \(\omega\) is continuous and bounded, thus by weak convergence, \(B(x,\mu^N_s)\to B(x,\mu_s)\). Moreover, since \(B(x,\mu^N_s)\) is monotonic increasing for each \(N\) and \(s\) and converges pointwise to \(B(x,\mu_s)\), thus convergence is uniform in \(x\) (see [14], §8.2, Lemma 3, for example). Now let \(b^\star(s)=\lim_{k\to\infty}b^{N_k}(s)\) be a subsequential limit. Therefore \(B(b^{N_i}(s),\mu^N_s)\) converges to \(B(b^{N_i}(s),\mu_s)\) uniformly for all \(i\). Finally, as \(B(x,\nu)\) is continuous in \(x\), we can conclude that \[\begin{align} B(b(s),\mu)= f(s)/\Lambda = \lim_{k\to\infty}B(b^{N_k}(s),\mu^{N_k})=\lim_{k\to\infty}B(b^{N_k}(s),\mu)=B(\lim_{k\to\infty}b^{N_k}(s),\mu)=B(b^\star(s),\mu) \end{align}\] Therefore \(b^\star(s)\geq b(s)\). Then as \(x\mapsto \omega(y-x)\) is non-decreasing, \(\omega(y-b^\star(s))-\omega(y-b(s))\) is non-negative for all \(y\). Then \(0=f(s)/\Lambda- f(s)/\Lambda=B(b^\star(s),\mu)-B(b(s),\mu)=\int_{\mathbb{R}}\omega(y-b^\star(s))-\omega(y-b(s))\mu(dy)\), which is to say that \(\omega(y-b^\star(s))=\omega(y-b(s))\) for \(\mu\)-almost all \(y\). By the triangle inequality: \[\begin{align} |\langle &\mu_s^{N_k},\phi(\cdot)\omega(\cdot-b^{N_k}(s))\rangle - \langle \mu_s,\phi(\cdot)\omega(\cdot-b^\star(s))\rangle| \\ &\leq |\langle \mu_s^{N_k},\phi(\cdot)\omega(\cdot-b^{N_k}(s))\rangle - \langle \mu_s,\phi(\cdot)\omega(\cdot-b^{N_k}(s))\rangle| + |\langle \mu_s,\phi(\cdot)\omega(\cdot-b^{N_k}(s))\rangle - \langle \mu_s,\phi(\cdot)\omega(\cdot-b^\star(s))\rangle|. \end{align}\] The first term on the right hand side of the inequality converges to \(0\) by weak convergence, since \(x\mapsto \phi(x)\omega(x-b^N(s))\) is bounded and continuous. Moreover, since \(\omega\) is bounded, continuous, and monotonic, thus \(\omega\) is uniform continuous, and therefore \(\omega(x-b^{N_k})\) converges to \(\omega(x-b^\star)\) uniformly for \(x\in \mathbb{R}\). Therefore the second term is converges to \(0\) as \(k\to\infty\). Finally, we note that, as \(\omega(y-b^\star(s))=\omega(y-b(s))\) \(\mu\)-a.s., thus \(\langle \mu_s,\phi(s)\omega(\cdot-b^\star(s))\rangle = \langle \mu_s,\phi(s)\omega(\cdot-b(s))\rangle\).
Next consider the case \(\omega=\mathbb{1}_{x\leq 0}\), so \(b(s)=\inf\{x:\Lambda\mu_s((-\infty,x])\geq f(s)\}\) and \(\omega(x-b(s))=\mathbb{1}_{\{\Lambda\mu_s((-\infty,x])<f(s)\}}\). Similarly \(b^N(s)=\inf\{x:\Lambda\mu_s^N((-\infty,x])\geq f(s)\}\) and \(\omega(x-b^N(s))=\mathbb{1}_{\{\Lambda\mu_s^N((-\infty,x])<f(s)\}}\).
Since, \(\mu_s\ll\lambda\), so \(\mu_s((-\infty,x])\) is continuous, thus again, as \(\mu_s^N((-\infty,x])\) is monotonic, \(\mu^N_s((-\infty,x])\) converges to \(\mu_s((-\infty,x])\) uniformly in \(x\) (see [14], §8.2, Lemma 3, for example). Moreover, for any continuous and bounded function \(\phi\), \(\int_{-\infty}^z \phi(x)\mu_s^{N_k}(dx)\) is the distribution function of a rescaled probability measure (which we will call \(\phi\mu_s^{N_k}\)) and as an immediate consequence of the weak convergence \(\mu_s^{N_k}\Rightarrow \mu_s\), we have \(\phi\mu_s^{N_k}\Rightarrow \phi\mu_s\). Therefore we also have that \(\int_{-\infty}^z \phi(x)\mu_s^{N_k}(dx)\) converges to \(\int_{-\infty}^z \phi(x)\mu_s(dx)\) uniformly in \(z\).
Now suppose that \(b^\star(s)=\lim_{k\to\infty}b^{N_k}(s)\) is a subsequential limit. Therefore \(\mu^{N_k}_s((-\infty,b^{N_k}(s)])\to\mu_s((-\infty,b^\star(s)])\). Moreover, for any \(N\) and \(\epsilon>0\), \(\Lambda\mu^N_s((-\infty,b^N(s)])\geq f(s)\) and \(\Lambda\mu^N_s((-\infty,b^N(s)-\epsilon])<f(s)\). Therefore \(\Lambda\mu_s((-\infty,b^\star(s)])\geq f(s)\) and \(\Lambda\mu_s((-\infty,b^\star(s)-\epsilon])\leq f(s)\) for any \(\epsilon\). Thus \(\Lambda\mu_s((-\infty,b^\star(s)])=f(s)\). Therefore if \(b^\star(s)>b(s)\), then \(\mu_s((b(s),b^\star(s)])=0\). Then \[\begin{align} \nonumber |\langle \mu_s^{N_k},\phi(\cdot)\omega(\cdot-b^{N_k}(s))\rangle - \langle \mu_s,\phi(\cdot)\omega(\cdot-b(s))\rangle| = \Big|\int_{-\infty}^{b^{N_k}(s)}\phi(x)\mu^{N_k}_s(dx)-\int_{-\infty}^{b(s)}\phi(x)\mu_s(dx)\Big|\\ \nonumber\leq \Big|\int_{-\infty}^{b^{N_k}(s)}\phi(x)\mu^{N_k}_s(dx)-\int_{-\infty}^{b^{N_k}(s)}\phi(x)\mu_s(dx)\Big| + \Big|\int_{-\infty}^{b^{N_k}(s)}\phi(x)\mu_s(dx)-\int_{-\infty}^{b^\star(s)}\phi(x)\mu_s(dx)\Big|\\ +\Big|\int_{-\infty}^{b^\star(s)}\phi(x)\mu_s(dx)-\int_{-\infty}^{b(s)}\phi(x)\mu_s(dx)\Big|\\ \label{triangleBConv}\leq \sup_{z\in \mathbb{R}}\Big|\int_{-\infty}^z \phi(x)\mu_s^{N_k}(dx)-\int_{-\infty}^z \phi(x)\mu_s(dx)\Big| + \|\phi\|_\infty (\mu_s((b^{N_k}(s),b^\star(s)])+\mu_s((b(s),b^\star(s)])). \end{align}\tag{4}\] By uniform convergence in \(z\), the first term converges to \(0\); by the facts that \(b^{N_k}(s)\to b^\star(s)\) and \(\mu_s\ll\lambda\), the second term converges to \(0\); and the third term is \(0\). This proves the theorem. ◻
Proposition 7. Let \(Q^\infty_T\) be a subsequential limit of \((Q^N_T)_{N=1,2,\ldots}\) and let \((\mu_t)_{t\in[0,T]}\sim Q^\infty_T\). Suppose that \(w\) is either continuous or \(w(x)=\mathbb{1}_{\{x\leq 0\}}\). Then for any \(\phi(x,t)\in \mathcal{BC}^{2,1}(\mathbb{R}\times [0,\infty),\mathbb{R})\) we have: \[\begin{align} \label{convOfWeakRep}\langle \mu_t,\phi(\cdot,t)\rangle = \langle \rho, \phi(\cdot,0)\rangle + \int_0^t \langle \mu_s, \frac{1}{2} \phi_{xx}(\cdot,s)+\phi_t(\cdot,s)+f(s)\phi(\cdot,s)-\Lambda\phi(\cdot,s)\omega(\cdot-b^\mu(s))\rangle ds, \end{align}\qquad{(2)}\] \(Q^\infty_T\)-almost surely.
Proof. Without loss of generality, we will consider the convergent subsequence to be \((Q^N_T)_{N=1,2,\ldots}\) to avoid using sub-subscript. For measure-valued process \(\mu=(\mu_t)_{t\geq 0}\), define \(b^\mu(t):=\inf\{x:\Lambda\langle\mu_s,\omega(\cdot-x)\rangle\geq f(t)\}\) and note that \(b^N=b^{\mu^N}\). Then define the function: \[\begin{align} G(\mu,\phi,t):=\langle \mu_t,\phi(\cdot,t)\rangle - \langle \rho, \phi(\cdot,0)\rangle - \int_0^t \langle \mu_s, \frac{1}{2} \phi_{xx}(\cdot,s)+\phi_t(\cdot,s)+f(s)\phi(\cdot,s)\rangle-\\\frac{f(s)\langle\mu_s,\phi(\cdot,s)\omega(\cdot-b^\mu(s))\rangle}{\langle \mu_s,\omega(\cdot-b^\mu(s))\rangle} ds. \end{align}\] By Proposition ?? , \(G(\mu^N,\phi,t)=M^{N,W}_t+M^{N,b}_t\), and so by the Burkholder-Davis-Gundy inequality, since \(\mathbb{E}[[M^{N,W}]_t]\) and \(\mathbb{E}[[M^{N,b}]_t]\) both converge to \(0\) as \(N\to\infty\), there exists a constant \(\kappa\) such that \[\mathbb{E}[G(\mu^N,\phi,t)^2]=\mathbb{E}[(M_t^{N,W}+M_t^{N,b})^2]\leq 4\mathbb{E}[(M_t^{N,W})^2+(M_t^{N,b})^2]\leq 4\kappa(\mathbb{E}[[M^{N,W}]_t]+\mathbb{E}[[M^{N,b}]_t])\xrightarrow[N\to\infty]{}0.\] Now by the Skorokhod representation theorem, there exists a sequence of random measure-valued processes \((\nu_t^N)_{t\in [0,T]}\) and \((\nu_t)_{t\in [0,T]}\) defined on the same probability space \((\Omega,\mathcal{F},\mathbb{P})\) such that \((\nu_t^N)_{t\in [0,T]}\) converges in the Skorokhod topology to \((\nu_t)_{t\in [0,T]}\) \(\mathbb{P}\)-almost surely, with \((\nu^N_t)_{t\in [0,T]}\sim Q_T^N\) and \((\nu_t)_{t\in [0,T]}\sim Q_T^\infty\). By the definition of convergence in the Skorokhod topology on \(\mathcal{D}([0,T],\mathcal{M}^w_F(\mathbb{R}))\), there exists a sequence \(\ell_N:[0,T]\to[0,T]\) such that \(\sup_{t\in [0,T]}|\ell_N(t)-t|\xrightarrow[N\to\infty]{}0\) and \(\sup_{t\in [0,T]}|\langle \nu_s,\phi(\cdot)\rangle - \langle \nu_{\ell_N(s)}^N,\phi(\cdot)\rangle|\xrightarrow[N\to\infty]{}0\) for all \(\phi\in \mathcal{BC}^{2,1}\) \(\mathbb{P}\)-a.s. By Proposition 6 the map \(t\mapsto \nu_t\) is continuous on \([0,T]\) \(Q^\infty_T\)-a.s., so we have that for \(\phi\in \mathcal{BC}^{2,1}\): \[|\langle \phi,\nu_s^N\rangle-\langle \phi,\nu_s\rangle|\leq |\langle \phi,\nu_s^N\rangle -\langle \phi, \nu_{\ell_N^{-1}(s)}\rangle|+|\langle \phi,\nu_{\ell_N^{-1}(s)}\rangle - \langle \phi,\nu_s\rangle|\xrightarrow[N\to\infty]{\mathbb{P}-a.s.}0,\] which is to say that \(\nu_s^N\) converges to \(\nu_s\) for all \(s\in [0,T]\) in the weak topology. Therefore for \(\phi\in \mathcal{BC}^{2,1}\), \(\langle \nu^N_t,\phi(\cdot,t) \rangle \xrightarrow[N\to\infty]{\mathbb{P}-a.s.} \langle \nu_t,\phi(\cdot,t)\rangle\) and, by Lemma 1 and the dominated convergence theorem, \[\begin{align} \int_0^t \langle \nu_s^N, \frac{1}{2} \phi_{xx}(\cdot,s)+\phi_t(\cdot,s)+f(s)\phi(\cdot,s)\rangle ds &\xrightarrow[N\to\infty]{\mathbb{P}-a.s.}\int_0^t \langle \nu_s, \frac{1}{2} \phi_{xx}(\cdot,s)+\phi_t(\cdot,s)+f(s)\phi(\cdot,s)\rangle ds,\\ \int_0^t \frac{f(s)\langle \nu_s^N, \phi(\cdot,)\omega(\cdot-b^N(s))\rangle}{\langle \nu_s,\omega(\cdot-b^N(s))\rangle}ds &\xrightarrow[N\to\infty]{\mathbb{P}-a.s.}\int_0^t \Lambda \langle \nu_s,\phi(\cdot,s)\omega(\cdot-b^\nu(s))\rangle ds. \end{align}\] Finally, observe that \(G(\nu^N,\phi,t)\) is bounded uniformly in \(N\) by \(\|\phi\|_\infty + t(\frac{1}{2}\|\phi_{xx}\|_\infty+\|\phi_t\|_\infty+(\Lambda+\|f\|_\infty)\|\phi\|_\infty)\). Therefore \((|G(\nu^N,\phi,t)|)_{N\geq 0}\) is uniformly integrable and converges to \(0\) in expectation, and hence converges to \(0\) almost surely (see Appendixes, Proposition 2.3 in [15]). Therefore \(\lim_{N\to\infty}G(\mu^N,\phi,t)=0\) \(\mathbb{P}\)-a.s. Since \(\nu_s\ll\lambda\), thus \(\langle \nu_s,\omega(\cdot-b^\nu(s))\rangle = f(s)/\Lambda\), and so: \[\begin{align} \langle \mu_t,\phi(\cdot,t)-\langle \rho,\phi(\cdot,0)\rangle - \int_0^t \langle \mu_s,\frac{1}{2}\phi_{xx}(\cdot,s)+\phi_t(\cdot,s)+f(s)\phi(\cdot,s)-\Lambda \phi(\cdot,s)\omega(\cdot-b(s))\rangle ds=0 \end{align}\] \(\mathbb{P}\)-a.s. Under the measure \(\mathbb{P}\), \((\nu_s)_{s\in [0,T]}\sim Q_T^\infty\), therefore ?? holds \(Q_T^\infty\)-almost surely, as required. ◻
Finally we show that the equation ?? has a unique solution when \(\omega\) is continuous. In the case that \(\omega(x)=\mathbb{1}_{x\leq 0}\), we already know by Theorem 1.8 of [6] that the weak solution is unique.
Proposition 8. Suppose that \(\omega\) is continuous. Then there exists at most one measure-valued process \((\mu_t)_{t\in [0,T]}\) satsifying ?? for all test functions \(\phi\in \mathcal{BC}^{2,1}(\mathbb{R}\times [0,\infty),\mathbb{R})\).
Proof. Let \((\mu_t)_{t\in [0,T]}\) and \((\nu_t)_{t\in [0,T]}\) both satisfy ?? for all \(\phi\in \mathcal{BC}^{2,1}\), with \(\mu_0=\rho=\nu_0\). Let \(\psi\in C_b(\mathbb{R})\) be a continuous test function with \(\|\psi\|_\infty\leq 1\), and let \(\phi(x,t)\) be the solution of the backwards heat equation \(\phi_t + \frac{1}{2} \phi_{xx}=0\) with terminal condition \(\phi(x,t)=\psi(x)\). So \(|\phi(x,s)|\leq 1\) for all \(x\in \mathbb{R}\) and \(s\in [0,t]\). Then ?? yields that: \[\begin{align} \langle \mu_t,\psi\rangle - \langle \nu_t,\psi\rangle = \int_0^t \langle \mu_s - \nu_s, f(s)\phi(\cdot,s)\rangle - \Lambda\langle \mu_s,\phi(\cdot,s)\omega(\cdot-b^\mu(s))\rangle + \Lambda \langle \nu_s,\phi(\cdot,s)\omega(\cdot-b^\nu(s))\rangle ds. \end{align}\] Then by the triangle inequality \[\begin{align} |\langle \nu_s,\phi(\cdot,s)\omega(\cdot-b^\nu(s))\rangle &- \langle \mu_s,\phi(\cdot,s)\omega(\cdot-b^\mu(s))\rangle| \\ &\leq |\langle \nu_s - \mu_s, \phi(\cdot,s)\omega(\cdot-b^\mu(s))\rangle|+|\langle \nu_s,\phi(\cdot,s)(\omega(\cdot-b^\nu(s))-\omega(\cdot-b^\mu(s)))\rangle|. \end{align}\] Now define the distance \(d(\mu,\nu):=\sup_{\{g\in C(\mathbb{R}):\|g\|_\infty \leq 1\}}|\langle \mu,g\rangle - \langle \nu,g\rangle|\) between two measures. So then as the function \(\phi(\cdot,s)\omega(\cdot-b^\mu(s))\) is continuous in \(x\) and bounded by \(1\), thus \(\|\langle \nu_s-\mu_s,\phi(\cdot,s)\omega(\cdot-b^\mu(s))\rangle|\leq d(\mu_s,\nu_s)\). Now suppose without loss of generality that \(b^\mu(s)\leq b^\nu(s)\). Since \(\omega\) is monotone decreasing, thus \(\omega(x-b^\nu(s))-\omega(x-b^\mu(s))\geq 0\) for all \(x\), so that: \[\begin{align} |\langle \nu_s, \phi(\cdot,s)\big(\omega(\cdot-b^\nu(s))-\omega(\cdot-b^\mu(s)))\rangle| \leq \|\phi(\cdot,s)\|_\infty\langle \nu_s, \omega(\cdot-b^\nu(s))-\omega(\cdot-b^\mu(s))\rangle \end{align}\] Next, we observe that by the continuity of \(\omega\), \(\langle \nu_s,\omega(\cdot-b^\nu(s))\rangle = f(s)/\Lambda = \langle \mu_s,\omega(\cdot-b^\mu(s))\rangle\), so that the right hand side above is \[\begin{align} \|\phi(\cdot,s)\|_\infty |\langle \mu_s,\omega(\cdot-b^\mu(s))\rangle - \langle \nu_s,\omega(\cdot-b^\mu(s))\rangle|\leq d(\mu_s,\nu_s). \end{align}\] Putting this together, using the fact that \(f\) is bounded by \(\Lambda\), yields: \[\begin{align} |\langle \mu_t,\psi\rangle-\langle \nu_t,\psi\rangle|\leq 3\Lambda\int_0^t d(\mu_s,\nu_s)ds \end{align}\] for any continuous \(\psi\) bounded by \(1\). Therefore \(d(\mu_t,\nu_t)\leq 3\Lambda \int_0^t d(\mu_s,\nu_s)\) and hence by Grönwall’s inequality, \(d(\mu_s,\nu_s)\equiv 0\), so that we can conclude \(\mu_s\equiv \nu_s\), and thus the solution is unique. ◻
Proof. (Of Theorem 1) By Proposition 4, the sequence \((Q^N_T)_{N=1,2,\ldots}\) has subsequential limits, and by Proposition 7, if \((\mu_t)_{t\in [0,T]}\sim Q^\infty_T\), then \(\mu_t\) is a weak solution to the FBP 2 . By Theorem 1.8 of [6] (in the case that \(\omega(x)=\mathbb{1}_{x\leq 0}\)) and Proposition 8 (in the case that \(\omega\) is continuous), this solution is unique. Therefore as all subsequential limits are the same, \(\mu_t\) is the unique weak solution to 2 . ◻
jacob.mercer@maths.ox.ac.uk, Department of Statistics, University of Oxford↩︎