2. Skeletal decomposition

Recall that the main idea behind the skeletal decomposition is that under certain conditions we can identify prolific genealogies in the population, and by immigrating non-prolific mass along the trajectories of these prolific genealogies, we can recover the law of the original superprocess. The infinite genealogies are described by a Markov branching process whose initial state is given by a Poisson random measure, while traditionally the immigrants are independent copies of the original process conditioned to become extinct.

In this section we first characterise the two components, then explain how to construct the skeletal decomposition from these building blocks. The results of this section are lifted from [Reference Kyprianou, Pérez and Ren23] and [Reference Eckhoff, Kyprianou and Winkel10].

As we have mentioned, the skeleton is often constructed using the event of extinction, that is, the event $\mathcal{E}_{\textrm{fin}}=\{\langle 1, X_t\rangle=0$ for some $t>0\}$. This guides the skeleton particles into regions where the survival probability is high. If we write $w(x)=-\log\mathbb{P}_{\delta_x}(\mathcal{E_{\textrm{fin}}})$, and assume that $\mu\in\mathcal{M}(E)$ is such that $\langle w,\mu\rangle<\infty$, then it is not hard to see that $\mathbb{P}_\mu(\mathcal{E}_{\textrm{fin}})=\exp \lbrace -\langle w, \mu\rangle \rbrace.$ Furthermore, by conditioning $\mathcal{E}_{\textrm{fin}}$ on $\mathcal{F}_t\,:\!=\sigma(X_s,s\leq t)$, we get that

\begin{equation*}\mathbb{E}_\mu({\textrm{e}}^{-\langle w,X_t\rangle})={\textrm{e}}^{-\langle w,\mu \rangle}.\end{equation*}

Eckhoff, Kyprianou, and Winkel [Reference Eckhoff, Kyprianou and Winkel10] point out that in order to construct a skeletal decomposition along those genealogies that avoid the behaviour specified by w (in this case ‘extinction’), all we need is that the function w gives rise to a multiplicative martingale $(({\textrm{e}}^{-\langle w,X_t\rangle}, t\geq 0),\mathbb{P}_\mu)$. In particular, a skeletal decomposition is given for any choice of a martingale function w which satisfies the following conditions:

1. • for all $x\in E$ we have $w(x)>0$ and $\sup_{x\in E} w(x)<\infty $, and

2. • $\mathbb{E}_\mu({\textrm{e}}^{-\langle w,X_t\rangle})={\textrm{e}}^{-\langle w,\mu \rangle}$ for all $\mu\in\mathcal{M}_c(E)$, $t\geq 0$ (here $\mathcal{M}_c(E)$ denotes the set of finite, compactly supported measures on E).

The condition $w(x)>0$ implicitly hides the notion of supercriticality, as it ensures that survival happens with positive probability. Note, however, that ‘survival’ can be interpreted in many different ways. For example, the choice of $\mathcal{E}_{\textrm{fin}}$ results in skeleton particles that are simply part of some infinite genealogical line of descent, but we could also define surviving genealogies as those that visit a compact domain in E infinitely often.

We will also make the additional assumption that w is in the domain of the generator $\mathcal{L}$. This is predominantly because of the use of partial differential equations in our analysis rather than integral equations.

2.1. Skeleton

First we identify the branching particle system that takes the role of the skeleton in the decomposition of the superprocess. In general, a Markov branching process $Z=(Z_t,t\geq 0)$ takes values in $\mathcal{M}_a(E)$ (the set of finite, atomic measures in E), and it can be characterised by the pair $(\mathcal{P},F)$, where $\mathcal{P}$ is the semigroup of a diffusion and F is the branching generator which takes the form

\begin{equation*}F(x,s)=q(x)\sum_{n\geq 0}p_n(x)(s^n-s),\quad x\in E,\ s\in[0, 1].\end{equation*}

Here q is a bounded, measurable mapping from E to $[0,\infty)$, and $\{ p_n(x),n\geq 0\}$, $x\in E$ are measurable sequences of probability distributions. For $\nu\in\mathcal{M}_a(E)$ we denote the law of the process Z issued from $\nu$ by $\textbf{P}_\nu$. Then we can describe $(Z,\textbf{P}_\nu)$ as follows. We start with initial state $Z_0=\nu$. Particles move according to $\mathcal{P}$, and at a spatially dependent rate $q(x)\,{\textnormal d} t$ a particle is killed and is replaced by n offspring with probability $p_n(x)$. The offspring particles then behave independently and according to the same law as their parent.

In order to specify the parameters of Z we first need to introduce some notation. Let $\xi=(\xi_t,t\geq 0)$ be the diffusion process on $E\cup \{\dagger\}$ (the one-point compactification of E with a cemetery state) corresponding to $\mathcal{P}$, and let us denote its probabilities by $\{\Pi_x,x\in E\}$. (Note that the previously defined martingale function w can be extended to $E\cup \{\dagger\}$ by defining $w(\dagger)=0$). Then, for all $x\in E$,

\begin{equation*}\dfrac{w(\xi_t)}{w(x)}\exp\biggl\lbrace -\int_0^t\dfrac{\psi(\xi_s,w(\xi_s))}{w(\xi_s)}{\textnormal d} s\biggr\rbrace,\quad t\geq 0,\end{equation*}

is a positive local martingale, and hence a supermartingale. (To see why this is true we refer the reader to the discussion in Section 2.1.1 of [Reference Eckhoff, Kyprianou and Winkel10].) Now let $\tau_E=\inf\{ t>0\colon \xi_t\in\{\dagger\}\}$, and consider the following change of measure:

\begin{equation*} \dfrac{{\textnormal d}\Pi_x^w}{{\textnormal d} \Pi_x}\bigg|_{\sigma(\xi_s,s\in[0,t])}=\dfrac{w(\xi_t)}{w(x)}\exp\biggl\lbrace -\int_0^t \dfrac{\psi(\xi_s,w(\xi_s))}{w(\xi_s)}{\textnormal d} s\biggr\rbrace\quad \text{on }\{t<\tau_E\},\ x\in E,\end{equation*}

which uniquely determines a family of (sub)probability measures $\{\Pi_x^w,x\in E\}$ (see e.g. [Reference Ethier and Kurtz14]).

If we let $\mathcal{P}^w$ denote the semigroup of the $E\cup\{\dagger\}$-valued process whose probabilities are $\{\Pi^w_x,x\in E\}$, then it can be shown that the generator corresponding to $\mathcal{P}^w$ is given by

\begin{equation*}\mathcal{L}^w\,:\!=\mathcal{L}_0^w-w^{-1}\mathcal{L}w=\mathcal{L}_0^w-w^{-1}\psi(\cdot,w),\end{equation*}

where $\mathcal{L}_0^w u=w^{-1}\mathcal{L}(wu)$ whenever u is in the domain of $\mathcal{L}$. Note that $\mathcal{L}^w$ is also called an h-transform of the generator $\mathcal{L}$ with $h=w$. The theory of h-transforms for measure-valued diffusions was developed in [Reference Engländer and Pinsky11].

Intuitively, if

(2.1) \begin{equation}w(x)=-\log\mathbb{P}_{\delta_x}(\mathcal{E})\end{equation}

defines a martingale function with the previously introduced conditions for some tail event $\mathcal{E}$, then the motion associated with $\mathcal{L}^w$ forces the particles to avoid the behaviour specified by $\mathcal{E}$. In particular, when $\mathcal{E}=\mathcal{E}_{\textrm{fin}}$ then $\mathcal{P}^w$ encourages $\xi$ to visit domains where the global survival rate is high.

Now we can characterise the skeleton process of $(X,\mathbb{P}_\mu)$ associated with w. In particular, $Z=(Z_t,t\geq 0)$ is a Markov branching process with diffusion semigroup $\mathcal{P}^w$ and branching generator

\begin{equation*}F(x,s)=q(x)\sum_{n\geq 0}p_n(x)(s^n-s),\quad x\in E,\ s\in[0, 1],\end{equation*}

where

(2.2) \begin{equation}q(x)=\psi'(x,w(x))-\dfrac{\psi(x,w(x))}{w(x)},\end{equation}

and $p_0(x)=p_1(x)=0$, and for $n\geq 2$

(2.3) \begin{equation}p_n(x)=\dfrac{1}{w(x)q(x)}\biggl\lbrace \beta(x)w^2(x)\textbf{1}_{\{ n=2\}}+w^n(x)\int_{(0,\infty)}\dfrac{y^n}{n!}\,{\textrm{e}}^{-w(x)y}m(x,{\textnormal d} y)\biggr\rbrace.\end{equation}

Here we used the notation

\begin{equation*}\psi'(x,w(x))\,:\!={\partial_z}\psi(x,z)|_{z=w(x)},\quad x\in E.\end{equation*}

We refer to the process Z as the $(\mathcal{P}^w, F)$ skeleton.

2.2. Immigration

Next we characterise the process that we immigrate along the previously introduced branching particle system. To this end let us define the function

\begin{equation*}\psi^*(x,z)=\psi(x,z+w(x))-\psi(x,w(x)),\quad x\in E,\end{equation*}

which can be written as

(2.4) \begin{equation}\psi^*(x,z)=-\alpha^*(x)z+\beta(x)z^2+\int_{(0,\infty)}({\textrm{e}}^{-zu}-1+zu)m^*(x,{\textnormal d} u),\quad x\in E,\end{equation}

where

\begin{equation*}\alpha^*(x)=\alpha(x)-2\beta(x)w(x)-\int_{(0,\infty)}(1-{\textrm{e}}^{-w(x)u})u\;m(x,{\textnormal d} u)=-\psi'(x,w(x))\end{equation*}

and

\begin{equation*}m^*(x,{\textnormal d} u)={\textrm{e}}^{-w(x)u}m(x,{\textnormal d} u).\end{equation*}

Note that under our assumptions $\psi^*$ is a branching mechanism of the form (1.1). We denote the probabilities of the $(\mathcal{P}, \psi^*)$-superprocess by $(\mathbb{P}^*_\mu)_{\mu\in\mathcal{M}(E)}$.

If $\mathcal{E}$ is the event associated with w (see (2.1)), and $\langle w,\mu\rangle<\infty$, then we have

\begin{equation*}\mathbb{P}^*_{\mu}(\cdot)=\mathbb{P}_\mu(\cdot \mid \mathcal{E}).\end{equation*}

In particular, when $\mathcal{E}=\mathcal{E}_{\textrm{fin}}$, then $\mathbb{P}^*_\mu$ is the law of the superprocess conditioned to become extinct.

2.3. Skeletal path decomposition

Here we give the precise construction of the skeletal decomposition that we introduced in a heuristic way at the beginning of this section. Let $\mathbb{D}([0,\infty)\times \mathcal{M}(E))$ denote the space of measure valued càdlàg function. Suppose that $\mu\in\mathcal{M}(E)$, and let Z be a $(\mathcal{P}^w,F)$-Markov branching process with initial configuration consisting of a Poisson random field of particles in E with intensity $w(x)\mu({\textnormal d} x)$. Next, dress the branches of the spatial tree that describes the trajectory of Z in such a way that a particle at the space-time position $(x,t)\in E\times[0,\infty)$ has a $\mathbb{D}([0,\infty)\times \mathcal{M}(E))$-valued trajectory grafted onto it, say $\omega = (\omega_t, t\geq0)$, with rate

(2.5) \begin{equation}2\beta(x)\,{\textnormal d}\mathbb{Q}_x^*({\textnormal d} \omega)+\int_{(0,\infty)} y \,{\textrm{e}}^{-w(x)y}m(x,{\textnormal d} y)\times {\textnormal d}\mathbb{P}^*_{y\delta_x}({\textnormal d} \omega).\end{equation}

Here $\mathbb{Q}^*_x$ is the excursion measure on the space $\mathbb{D}([0,\infty)\times \mathcal{M}(E))$ which satisfies

\begin{equation*}\mathbb{Q}^*_x(1-{\textrm{e}}^{-\langle{\kern1.3pt} f, X_t\rangle})=u^*_f(x,t)\end{equation*}

for $x\in E$, $t\geq 0$, and $f\in B^+_b(E)$ (the space of non-negative, bounded measurable functions on E), where $u^*_f(x,t)$ is the unique solution to (1.3) with the branching mechanism $\psi$ replaced by $\psi^*$. (For more details of excursion measures see [Reference Dynkin and Kuznetsov9].) Moreover, when a particle in Z dies and gives birth to $n\geq 2$ offspring at spatial position $x\in E$, with probability $\eta_n(x,{\textnormal d} y)\mathbb{P}^*_{y\delta_x}({\textnormal d} \omega)$ an additional $\mathbb{D}([0,\infty)\times \mathcal{M}(E))$-valued trajectory, $\omega$, is grafted onto the space-time branching point, where

(2.6) \begin{equation}\eta_n(x,{\textnormal d} y)=\dfrac{1}{w(x)q(x)p_n(x)}\biggl\lbrace \beta(x)w^2(x)\delta_0({\textnormal d} y)\textbf{1}_{\{ n=2\}}+w^n(x)\dfrac{y^n}{n!}\,{\textrm{e}}^{-w(x)y}m(x,{\textnormal d} y)\biggr\rbrace.\end{equation}

Overall, we have three different types of immigration processes that contribute to the dressing of the skeleton. In particular, the first term of (2.5) is what we call ‘continuous immigration’ along the skeleton, while the second term is referred to as the ‘discontinuous immigration’, and finally (2.6) corresponds to the so-called ‘branch-point immigration’.

Now we define $\Lambda_t$ as the total mass from the dressing present at time t together with the mass present at time t of an independent copy of $(X,\mathbb{P}_\mu^*)$ issued at time 0. We denote the law of $(\Lambda, Z)$ by $\textbf{P}_\mu$. Then Kyprianou et al. [Reference Kyprianou, Pérez and Ren23] showed that $(\Lambda,\textbf{P}_\mu)$ is Markovian and has the same law to $(X,\mathbb{P}_\mu)$. Furthermore, under $\textbf{P}_\mu$, conditionally on $\Lambda_t$, the measure $Z_t$ is a Poisson random measure with intensity $w(x)\Lambda_t({\textnormal d} x)$.

3. SDE representation of the dressed tree

Recall that our main motivation is to reformulate the skeletal decomposition of superprocesses using the language of SDEs. Thus, in this section, after giving an SDE representation of the skeletal process, we derive the coupled SDE for the dressed skeleton, which simultaneously describes the evolution of the skeleton and the total mass in the system.

3.1. SDE of the skeleton

We use the arguments on page 3 of [Reference Xiong34] to derive the SDE for the branching particle diffusion, that will act as the skeleton. Let $(\xi_t,t\geq 0)$ be the diffusion process corresponding to the Feller semigroup $\mathcal{P}$. Since the generator of the motion is given by (1.2), the process $\xi$ satisfies

\begin{equation*}{\textnormal d}\xi_t=b(\xi_t)\,{\textnormal d} t+\sigma(\xi_t)\,{\textnormal d} B_t,\end{equation*}

where $\sigma\colon \mathbb{R}^d\rightarrow\mathbb{R}^d$ is such that $\sigma(x)\sigma^{\texttt{T}}(x)=a(x)$ (where $\texttt{T}$ denotes matrix transpose), and $(B_t,t\geq 0)$ is a d-dimensional Brownian motion (see e.g. Chapter 1 of [Reference Pinsky30]).

It is easy to verify that if $(\tilde{\xi}_t,t\geq 0)$ is the diffusion process under $\mathcal{P}^w$, then it satisfies

\begin{equation*}{\textnormal d}\tilde{\xi}_t=\biggl( b(\tilde{\xi}_t)+\dfrac{\nabla w(\tilde{\xi}_t)}{w(\tilde{\xi}_t)}a(\tilde{\xi}_t) \biggr)\,{\textnormal d} t+\sigma(\tilde{\xi}_t)\,{\textnormal d} B_t,\end{equation*}

where $\nabla w$ is the gradient of w. To simplify computations, define the function $\tilde{b}$ on E given by

\begin{equation*}\tilde{b}(x)\,:\!=b(x)+\dfrac{\nabla w(x)}{w(x)}a(x).\end{equation*}

For $h\in C^2_b(E)$ (the space of bounded, twice differentiable continuous functions on E), using Itô’s formula (see e.g. Section 8.3 of [Reference øksendal29]), we get

\begin{equation*}{\textnormal d} h(\tilde{\xi}_t)=(\nabla h(\tilde{\xi}_t))^{\texttt{T}}\tilde{b}(\tilde{\xi}_t)\,{\textnormal d} t+\dfrac{1}{2}\textrm{Tr}\bigl[ \sigma^{\texttt{T}}(\tilde{\xi}_t) H_{h}(\tilde{\xi}_t)\sigma(\tilde{\xi}_t)\bigr]\,{\textnormal d} t+(\nabla h(\tilde{\xi}_t))^{\texttt{T}} \sigma(\tilde{\xi}_t)\,{\textnormal d} B_t,\end{equation*}

where $x^{\texttt{T}}$ denotes the transpose of x, Tr is the trace operator, and $H_h$ is the Hessian of h with respect to x, that is,

\begin{equation*}H_h(x)_{i,j}=\dfrac{\partial^2}{\partial x_i\partial x_j}h(x).\end{equation*}

Next, summing over all the particles live at time t, the collection of which we denote by $\mathcal{I}_t$, gives

(3.1) \begin{equation}{\textnormal d}\langle h,Z_t\rangle= \langle \nabla h(\cdot)\cdot\tilde{b}(\cdot),Z_t \rangle\,{\textnormal d} t+\biggl\langle \dfrac{1}{2}\textrm{Tr} [\sigma^{\texttt{T}}(\cdot) H_{h}(\cdot)\sigma(\cdot)], Z_t\biggr\rangle\,{\textnormal d} t+\sum_{\alpha\in\mathcal{I}_t} (\nabla h(\xi_t^\alpha))^{\texttt{T}} \sigma(\xi_t^\alpha)\,{\textnormal d} B_t^\alpha,\end{equation}

where for each $\alpha$, $B^\alpha$ is an independent copy of B, and $\xi^\alpha$ is the current position of individual $\alpha\in\mathcal{I}_t$.

If an individual branches at time t, then we have

(3.2) \begin{equation}\langle h,Z_t-Z_{t-}\rangle=\sum_{\alpha\colon \text{death time of }\alpha =t}(k_\alpha-1)h(\xi_t^\alpha).\end{equation}

Here $k_\alpha$ is the number of children of individual $\alpha$, which has distribution $\{p_k,k=0,1,\dots\}$.

Simple algebra shows that

\begin{equation*}\textrm{Tr} [ \sigma^{\texttt{T}}(x)H_h(x)\sigma(x)]=\sum_{ij}a_{ij}(x)\dfrac{\partial^2}{\partial x_i \partial x_j}h(x),\end{equation*}

thus by combining (3.1) and (3.2) we get

(3.3) \begin{equation}\langle h,Z_t\rangle=\langle h,Z_0\rangle+\int_0^t\langle \mathcal{L}^w h,Z_s\rangle\,{\textnormal d} s+V_t^c+\int_0^t\int_{E}\int_{\mathbb{N}}\langle h,(k-1)\delta_x\rangle N^\dagger_s( {\textnormal d} s,{\textnormal d} x, {\textnormal d}\{ k\}),\end{equation}

where $V_t^c$ is a continuous local martingale given by

(3.4) \begin{equation}V_t^c=\int_0^t \sum_{\alpha\in\mathcal{I}_s} (\nabla h(\xi_s^\alpha))^{\texttt{T}} \sigma(\xi_s^\alpha)\,{\textnormal d} B_s^\alpha,\end{equation}

and $N^\dagger_s$ is an optional random measure on $[0,\infty)\times E\times \mathbb{N}$ with predictable compensator of the form $\hat{N}^\dagger({\textnormal d} s, {\textnormal d} x, {\textnormal d} \{k\}) = {\textnormal d} s K^\dagger(Z_{s-}, {\textnormal d} x, {\textnormal d} \{k\})$ such that, for $\mu\in\mathcal{M}(E)$,

\begin{equation*} K^\dagger(\mu,{\textnormal d} x, {\textnormal d} \{k\})= \mu({\textnormal d} x) q(x)p_k(x) \#( {\textnormal d}\{k\}){,}\end{equation*}

where q, $p_k(x)$ are given by (2.2), (2.3) and $\#$ is the counting measure. The reader will note that, for a (random) measure $M\in\mathcal{M}_a(E)$, we regularly interchange the notion of $\sum_{k\in \mathbb{N}} \cdot$ with $\int_{\mathbb{N}}\cdot M({\textnormal d}\{k\})$.

Note that from (3.4) it is easy to see that the quadratic variation of $V_t^c$ is

\begin{equation*}\langle V^c \rangle_t=\int_0^t \sum_{\alpha\in\mathcal{I}_s} (\nabla h(\xi_s^\alpha))^{\texttt{T}} \sigma(\xi_s^\alpha)\sigma(\xi_s^\alpha)^{\texttt{T}}\nabla h(\xi_s^\alpha)\,{\textnormal d} s=\int_0^t \langle( \nabla h)^{\texttt{T}} a \nabla h,Z_s\rangle\,{\textnormal d} s.\end{equation*}

3.2. Thinning of the SDE

Now we will see how to modify the SDE given by (1.4) in order to separate out the different types of immigration processes. We use ideas developed in [Reference Fekete, Fontbona and Kyprianou16].

Recall that the SDE describing the superprocess $(X,\mathbb{P}_\mu)$ takes the following form:

(3.5) \begin{align}\langle{\kern1.3pt} f,X_t\rangle&=\langle{\kern1.3pt} f,\mu\rangle+\int_0^t\langle \alpha f,X_s\rangle\,{\textnormal d} s +M_t^c({\kern1.3pt}f)\notag\\ &\quad\, +\int_0^t\int_{E}\int_{(0,\infty)}\langle{\kern1.3pt} f, u \delta _x \rangle \tilde{N}( {\textnormal d} s,{\textnormal d} x, {\textnormal d} u)+\int_0^t \langle \mathcal{L}f,X_s \rangle\,{\textnormal d} s,\quad t\geq 0.\end{align}

Here $M_t^c({\kern1.3pt}f)$ is as in (1.4), and $N( {\textnormal d} s, {\textnormal d} x, {\textnormal d} u)$ is an optional random measure on $[0,\infty)\times E\times (0,\infty)$ such that, given $\mu\in\mathcal{M}(E)$, it has predictable compensator given by $\hat{N}( {\textnormal d} s, {\textnormal d} x, {\textnormal d} u)={\textnormal d} s K(X_{s-}, {\textnormal d} x, {\textnormal d} u)$, where

\begin{equation*}K(\mu, {\textnormal d} x, {\textnormal d} u) = \mu({\textnormal d} x) m(x,{\textnormal d} u).\end{equation*}

Moreover, $\tilde{N}( {\textnormal d} s,{\textnormal d} x, {\textnormal d} u)$ is the associated compensated version of $N( {\textnormal d} s, {\textnormal d} x, {\textnormal d} u)$. Let

\begin{equation*} ((s_i,x_i, u_i)\colon i\in\mathbb{N}) \end{equation*}

denote some enumeration of the atoms of $N( {\textnormal d} s,{\textnormal d} x, {\textnormal d} u)$. Next we introduce independent marks to the atoms of N, that is, we define the random measure

\begin{equation*}\mathcal{N}( {\textnormal d} s,{\textnormal d} x, {\textnormal d} u,{\textnormal d}\{k\})=\sum_{i\in\mathbb{N}}\delta_{(s_i,x_i, u_i,k_i)}({\textnormal d} s,{\textnormal d} x, {\textnormal d} u,{\textnormal d}\{k\}),\end{equation*}

whose predictable compensator ${\textnormal d} s\mathcal{K}(X_{s-},{\textnormal d} x, {\textnormal d} u,{\textnormal d}\{k\})$ has the property that, for $\mu\in\mathcal{M}(E)$,

\begin{align*}&\mathcal{K}(\mu,{\textnormal d} x, {\textnormal d} u,{\textnormal d}\{k\})= \mu({\textnormal d} x) m(x,{\textnormal d} u)\dfrac{(w(x)u)^k}{k!}\,{\textrm{e}}^{-w(x)u}\# ({\textnormal d}\{k\}).\end{align*}

Now we can define three random measures by

\begin{align*}N^0({\textnormal d} s,{\textnormal d} x, {\textnormal d} u) & =\mathcal{N}({\textnormal d} s,{\textnormal d} x, {\textnormal d} u,\{k=0\}),\\ N^1({\textnormal d} s,{\textnormal d} x, {\textnormal d} u) & =\mathcal{N}({\textnormal d} s,{\textnormal d} x, {\textnormal d} u,\{k=1\}){,}\intertext{and}N^2({\textnormal d} s,{\textnormal d} x, {\textnormal d} u) & =\mathcal{N}({\textnormal d} s,{\textnormal d} x, {\textnormal d} u,\{k\geq 2\}).\end{align*}

Using Proposition 10.47 of [Reference Jacod21], we see that $N^0$, $N^1$, and $N^2$ are also optional random measures and their compensators ${\textnormal d} s K^0(X_{s-},{\textnormal d} x, {\textnormal d} u)$, ${\textnormal d} s K^1(X_{s-},{\textnormal d} x, {\textnormal d} u)$, and ${\textnormal d} sK^2(X_{s-},{\textnormal d} x, {\textnormal d} u)$ satisfy

\begin{align*}K^0(\mu,{\textnormal d} x, {\textnormal d} u) & = \mu({\textnormal d} x) \,{\textrm{e}}^{-w(x)u}m(x,{\textnormal d} u),\\ K^1(\mu,{\textnormal d} x, {\textnormal d} u) & =\mu({\textnormal d} x) w(x)u\,{\textrm{e}}^{-w(x)u}m(x,{\textnormal d} u),\intertext{and}K^2(\mu,{\textnormal d} x, {\textnormal d} u) & =\mu({\textnormal d} x)\sum_{k=2}^\infty \dfrac{(w(x)u)^k}{k!}\,{\textrm{e}}^{-w(x)u} m(x,{\textnormal d} u)\end{align*}

for $\mu\in\mathcal{M}(E)$. Using these processes we can rewrite (3.5), so we get

(3.6) \begin{align}\langle{\kern1.3pt} f,X_t\rangle &=\langle{\kern1.3pt} f,\mu\rangle +\int_0^t\!\langle \alpha f,X_s\rangle\,{\textnormal d} s +M_t^c({\kern1.3pt}f)+\int_0^t\!\int_E\!\int_{(0,\infty)}\langle{\kern1.3pt} f, u\delta _x \rangle\tilde{N}^0({\textnormal d} s,{\textnormal d} x, {\textnormal d} u)+\int_0^t \!\langle \mathcal{L}f,X_s \rangle\,{\textnormal d} s\notag\\ &\quad\, +\int_0^t\int_E\int_{(0,\infty)}\langle{\kern1.3pt} f,u\delta_x \rangle N^1({\textnormal d} s,{\textnormal d} x, {\textnormal d} u)+\int_0^t\int_E\int_{(0,\infty)}\langle{\kern1.3pt} f, u\delta_x \rangle N^2({\textnormal d} s,{\textnormal d} x, {\textnormal d} u)\notag\\&\quad\, -\int_0^t\biggl\langle \int_{(0,\infty)}uf(\cdot)(1-{\textrm{e}}^{-uw(\cdot)} )m(\cdot,{\textnormal d} u),X_{s-}\biggr\rangle \,{\textnormal d} s\notag\\&=\langle{\kern1.3pt} f,\mu\rangle -\int_0^t\langle \psi'(\cdot,w(\cdot,s)) f(\cdot),X_s\rangle\,{\textnormal d} s +M_t^c({\kern1.3pt}f)+\int_0^t\int_E\int_{(0,\infty)}\langle{\kern1.3pt} f,u\delta_x\rangle\tilde{N}^0({\textnormal d} s,{\textnormal d} x, {\textnormal d} u)\notag\\&\quad\, +\int_0^t \langle \mathcal{L}f,X_s \rangle\,{\textnormal d} s+\int_0^t\int_E\int_{(0,\infty)}\langle{\kern1.3pt} f, u\delta_x \rangle N^1({\textnormal d} s,{\textnormal d} x, {\textnormal d} u)\notag\\ &\quad\, +\int_0^t\int_E\int_{(0,\infty)}\langle{\kern1.3pt} f, u\delta_x \rangle N^2({\textnormal d} s, {\textnormal d} x, {\textnormal d} u)+\int_0^t\langle 2\beta w f,X_{s-}\rangle\,{\textnormal d} s,\end{align}

where we have used the fact that

\begin{equation*}\alpha(x)-\int_{(0,\infty)}(1-{\textrm{e}}^{-w(x)u})u m(x,{\textnormal d} u)=-\psi'(x,w(x))+2\beta(x)w(x).\end{equation*}

Recalling (2.4), we see that the first line of the right-hand side of (3.6) corresponds to the dynamics of a $(\mathcal{P},\psi^*)$-superprocess. Our aim now is to link the remaining three terms to the three types of immigration along the skeleton, and write down a system of SDEs that describe the skeleton and total mass simultaneously. Heuristically speaking, this system of SDEs will consist of (3.3) and a second SDE which looks a little bit like (3.6) (note that the latter has no dependence on the process Z as things stand). To some extent, we can think of the SDE (3.6) as what one might see when ‘integrating out’ (3.3) from the aforesaid second SDE in the coupled system; indeed, this will be one of our main conclusions.

3.3. Coupled SDE

Following the ideas of the previous sections we introduce the following driving sources of randomness that we will use in the construction of our coupled SDE. Our coupled system will describe the evolution of the pair of random measures $(\Lambda, Z) = ((\Lambda_t, Z_t), t\geq 0)$ on $\mathcal{M}(E)\times\mathcal{M}_a(E)$.

1. • Let ${\texttt N}^0({\textnormal d} s,{\textnormal d} x, {\textnormal d} u)$ be an optional random measure on $[0,\infty)\times E\times (0,\infty)$, which depends on $\Lambda$ with predictable compensator $\hat{{\texttt N}}^0({\textnormal d} s,{\textnormal d} x, {\textnormal d} u) = {\textnormal d} s\, \texttt{K}^0(\Lambda_{s-}, {\textnormal d} x, {\textnormal d} u)$, where, for $\mu\in \mathcal{M}(E)$,

   \begin{equation*}\texttt{K}^0(\mu, {\textnormal d} x, {\textnormal d} u)= \mu({\textnormal d} x) \,{\textrm{e}}^{-w(x)u} m(x,{\textnormal d} u),\end{equation*}
   and $\tilde{{\texttt N}}^{0}({\textnormal d} s, {\textnormal d} x, {\textnormal d} u)$ is its compensated version.

2. • Let ${\texttt N}^1({\textnormal d} s,{\textnormal d} x, {\textnormal d} u)$ be an optional random measure on $[0,\infty)\times E\times (0,\infty)$, dependent on Z, with predictable compensator $\hat{{\texttt N}}^1({\textnormal d} s,{\textnormal d} x, {\textnormal d} u) = {\textnormal d} s\, \texttt{K}^1(Z_{s-}, {\textnormal d} x, {\textnormal d} u)$ so that, for $\mu\in\mathcal{M}_a(E)$,

   \begin{equation*}\texttt{K}^1(\mu, {\textnormal d} x, {\textnormal d} u)= \mu({\textnormal d} x) \,{\textrm{e}}^{-w(x)u} m(x,{\textnormal d} u){.}\end{equation*}

3. • Define ${\texttt N}^2({\textnormal d} s,{\textnormal d}\rho, {\textnormal d} x, {\textnormal d} u)$ as an optional random measure on $[0,\infty)\times \mathbb{N}\times E\times (0,\infty)$ also dependent on Z, with predictable compensator

   \begin{equation*}\hat{{\texttt N}}^2({\textnormal d} s,{\textnormal d}\{k\},{\textnormal d} x, {\textnormal d} u)= {\textnormal d} s\, \texttt{K}^2(Z_{s-}, {\textnormal d}\{k\}, {\textnormal d} x, {\textnormal d} u)\end{equation*}
   so that, for $\mu\in\mathcal{M}_a(E)$,
   \begin{equation*}\texttt{K}^2(\mu, {\textnormal d}\{k\}, {\textnormal d} x, {\textnormal d} u)= \mu({\textnormal d} x) q(x) p_k(x)\eta_k(x,{\textnormal d} u) \#({\textnormal d}\{k\}),\end{equation*}
   where q, $p_k({\textnormal d} x)$, and $\eta_k(x,{\textnormal d} u)$ are given by (2.2), (2.3), and (2.6).

Now we can state our main result.

Theorem 1. Consider the following system of SDEs for $f,h\in D_0(\mathcal{L})$:

(3.7) \begin{align}\begin{pmatrix}\langle{\kern1.3pt} f,\Lambda_t\rangle\\ \langle h,Z_t\rangle \\ \end{pmatrix}& =\begin{pmatrix}\langle{\kern1.3pt} f,\Lambda_0\rangle \\ \langle h, Z_0\rangle \\ \end{pmatrix} - \int_{0}^{t}\begin{pmatrix}\langle \partial_z\psi^*(\cdot,0)f,\Lambda_{s-} \rangle\\ 0 \\ \end{pmatrix} \,{\textnormal d} s+ \begin{pmatrix}U_t^c({\kern1.3pt}f) \\ V_t^c(h) \\ \end{pmatrix}\notag\\ & \quad\, + \int_{0}^{t}\int_{E}\int_{(0,\infty)}\begin{pmatrix}\langle{\kern1.3pt} f,u\delta_x\rangle \\ 0 \\ \end{pmatrix}\tilde{\!\textit{\texttt N}}^{\,0}({\textnormal d} s, {\textnormal d} x, {\textnormal d} u)+ \int_{0}^{t}\begin{pmatrix}\langle \mathcal{L}f,\Lambda_{s-} \rangle \\ \langle \mathcal{L}^w h,Z_{s-} \rangle \\ \end{pmatrix} \,{\textnormal d} s \notag \\ &\quad\, +\int_{0}^{t}\int_{E}\int_{(0,\infty)} \begin{pmatrix}\langle{\kern1.3pt} f,u\delta_x\rangle \\ 0 \\ \end{pmatrix}{\textit{\texttt N}}^{\,1}({\textnormal d} s,{\textnormal d} x, {\textnormal d} u)\notag\\ &\quad\,+ \int_{0}^{t}\int_{\mathbb{N}}\int_{E}\int_{(0,\infty)}\begin{pmatrix}\langle{\kern1.3pt} f,u\delta_x\rangle \\ \langle h,(k-1)\delta_x \rangle \\ \end{pmatrix} {\textit{\texttt N}}^{\,2}({\textnormal d} s,{\textnormal d} \{k\},{\textnormal d} x, {\textnormal d} u) \notag\\ &\quad\,+ \int_{0}^{t}\begin{pmatrix}\langle 2\beta f,Z_{s-}\rangle \\ 0 \\ \end{pmatrix} \,{\textnormal d} s, \quad t\geq 0,\end{align}

inducing probabilities $\textbf{P}_{(\mu,\nu)}$, $\mu\in\mathcal{M}(E)$, $\nu\in\mathcal{M}_a(E)$, where $(U_t^c({\kern1.3pt}f),t\geq 0)$ is a continuous local martingale with quadratic variation $2 \langle\beta f^2,\Lambda_{t-}\rangle\,{\textnormal d} t$, and $(V_t^c(h),t\geq 0)$ is a continuous local martingale with quadratic variation $\langle (\nabla h)^{\texttt{T}} a\nabla h,Z_{t-}\rangle\,{\textnormal d} t$. (Note that $\partial_z\psi^*(x,0) = \psi'(x,w(x))$ is another way of identifying the drift term in the first integral above.) With an immediate abuse of notation, write $\textbf{P}_\mu = \textbf{P}_{(\mu, \textrm{Po}(w\mu))}$, where $\textrm{Po}(w\mu)$ is an independent Poisson random measure on E with intensity $w\mu$. Then we have the following.

1. (i) There exists a unique weak solution to the SDE (3.7) under each $\textbf{P}_{(\mu,\nu)}$.

2. (ii) Under each $\textbf{P}_\mu$, for $t\geq 0$, conditional on $\mathcal{F}_t^\Lambda=\sigma(\Lambda_s,s\leq t)$, $Z_t$ is a Poisson random measure with intensity $w(x)\Lambda_t({\textnormal d} x)$.

3. (iii) The process $(\Lambda_t,t\geq 0)$, with probabilities $\textbf{P}_{\mu}$, $\mu\in\mathcal{M}(E)$, is Markovian and a weak solution to (3.5).

The rest of the paper is dedicated to the proof of this theorem, which we split over several subsections.

4. Proof of Theorem 1 (i): existence

Consider the pair $(\Lambda, Z)$, where Z is a $(\mathcal{P}^w,F)$ branching Markov process with $Z_0=\nu$ for some $\nu\in\mathcal{M}_a(E)$, and whose jumps are coded by the coordinates of the random measure $\mathtt{N}^2$. Furthermore, we define $\Lambda_t=X^*_t+D_t$, where $X^*$ is an independent copy of the $(\mathcal{P},\psi^*)$-superprocess with initial value $X^*_0=\mu$, $\mu\in\mathcal{M}(E)$, and the process $(D_t,t\geq 0)$ is described by

(4.1) \begin{align}\langle{\kern1.3pt} f,D_t\rangle & = \int_0^t\int_{\mathbb{D}([0,\infty)\times \mathcal{M}(E))}\langle{\kern1.3pt} f,\omega_{t-s}\rangle \mathtt{N}^1({\textnormal d} s,\cdot, \cdot,{\textnormal d}\omega)\notag\\ &\quad\, +\int_0^t\int_{\mathbb{D}([0,\infty)\times \mathcal{M}(E))}\langle{\kern1.3pt} f,\omega_{t-s}\rangle \mathtt{N}^2({\textnormal d} s,\cdot,\cdot, \cdot,{\textnormal d}\omega)\notag\\ &\quad\, +\int_0^t\int_{\mathbb{D}([0,\infty)\times \mathcal{M}(E))}\langle{\kern1.3pt} f,\omega_{t-s}\rangle \mathtt{N}^*({\textnormal d} s,{\textnormal d} \omega),\end{align}

where $f\in D_0(\mathcal{L})$ and, with a slight abuse of the notation that was introduced preceding Theorem 1,

1. • $\mathtt{N}^1$ is an optional random measure on $[0,\infty)\times E \times(0,\infty)\times\mathbb{D}([0,\infty)\times \mathcal{M}(E))$ with predictable compensator

   \begin{equation*}\hat{\mathtt{N}}^1({\textnormal d} s,{\textnormal d} x, {\textnormal d} u,{\textnormal d}\omega)={\textnormal d} s Z_{s-}({\textnormal d} x)u\,{\textrm{e}}^{-w(x)u}m(x,{\textnormal d} u)\mathbb{P}_{u\delta_x}^*({\textnormal d}\omega),\end{equation*}

2. • $\mathtt{N}^2$ is an optional random measure on $[0,\infty)\times \mathbb{N}\times E \times(0,\infty)\times \mathbb{D}([0,\infty)\times \mathcal{M}(E))$ with predictable compensator

   \begin{equation*}\hat{\mathtt{N}}^2({\textnormal d} s,{\textnormal d}\{k\},{\textnormal d} x,{\textnormal d} u, {\textnormal d}\omega)={\textnormal d} s Z_{s-}({\textnormal d} x)q(x)p_k(x)\eta_k(x,{\textnormal d} u)\mathbb{P}_{u\delta_x}^*({\textnormal d}\omega)\#({\textnormal d} \{k\}),\end{equation*}

3. • $\mathtt{N}^*$ is an optional random measure on $[0,\infty)\times \mathbb{D}([0,\infty)\times \mathcal{M}(E))$ with predictable compensator

   (4.2) \begin{equation}\hat{\mathtt{N}}^*({\textnormal d} s,{\textnormal d}\omega)={\textnormal d} s\int_E 2\beta(x) Z_{s-}({\textnormal d} x)\mathbb{Q}^*_{x}({\textnormal d}\omega),\end{equation}
   where
   \begin{equation*}\mathbb{Q}_x^*(1-{\textrm{e}}^{-\langle{\kern1.3pt} f,\omega_t\rangle})=-\log\mathbb{E}_{\delta_x}^*({\textrm{e}}^{-\langle{\kern1.3pt} f, X_t\rangle})=u_f^*(x,t).\end{equation*}

Note that we have used $\cdot$ to denote marginalisation so, for example,

\begin{equation*}\mathtt{N}^1({\textnormal d} s,\cdot, \cdot,{\textnormal d}\omega) = \int_E\int_{(0,\infty)}\mathtt{N}^1({\textnormal d} s,{\textnormal d} x, {\textnormal d} u,{\textnormal d}\omega)\end{equation*}

and, to be consistent with previous notation, we also have, for example,

\begin{equation*}\mathtt{N}^2({\textnormal d} s,{\textnormal d}\{k\},{\textnormal d} x,{\textnormal d} u, \cdot) = \mathtt{N}^2({\textnormal d} s,{\textnormal d}\{k\},{\textnormal d} x,{\textnormal d} u).\end{equation*}

We claim that the pair $(\Lambda, Z)$ as defined above solves the coupled system of SDEs (3.7). To see why, start by noting that Z solves the second coordinate component of (3.7) by definition of it being a spatial branching process; see (3.3). In dealing with the term ${\textnormal d} D_t$, we first note that the random measures $ \mathtt{N}^1$ and $ \mathtt{N}^2$ have finite activity through time, whereas $ \mathtt{N}^*$ has infinite activity. Suppose we write $\mathtt{I}^{(i)}_t$, $i =1,2,3$, for the three integrals on the right-hand side of (4.1), respectively. Taking the case of $\mathtt{I}^{(1)}_t$, if t is a jump time of $ \mathtt{N}^1({\textnormal d} t, {\textnormal d} x, {\textnormal d} u, {\textnormal d} \omega)$, then $\Delta \mathtt{I}^{(1)}_t = \langle{\kern1.3pt} f, \omega_0 \rangle,$ noting in particular that $ \omega_0 = u\delta_x$. A similar argument produces $\Delta \mathtt{I}^{(2)}_t = \langle{\kern1.3pt} f, \omega_0 \rangle = \langle{\kern1.3pt} f, u\delta_x\rangle $, when t is a jump time of $ \mathtt{N}^2({\textnormal d} t ,{\textnormal d} \{k\}, {\textnormal d} x, {\textnormal d} u, {\textnormal d}\omega)$. In contrast, on account of the excursion measures $(\mathbb{Q}^*_{x}, x\in E)$ having the property that $\mathbb{Q}^*_{x}(\omega_0 >0) =0$, we have $\Delta \mathtt{I}^{(3)}_t =0$. Nonetheless, the structure of the compensator (4.2) implies that there is a rate of arrival (of these zero contributions) given by

\begin{align*}\int_{\mathbb{D}([0,\infty)\times \mathcal{M}(E))}\langle{\kern1.3pt} f, \omega_0\rangle \hat{\mathtt{N}}^*({\textnormal d} s, {\textnormal d}\omega) &={\textnormal d} s\int_E 2\beta(x) Z_{s-}({\textnormal d} x)\mathbb{Q}^*_{x}(\langle{\kern1.3pt} f, \omega_0\rangle) \\ &= {\textnormal d} s\int_E 2\beta(x) Z_{s-}({\textnormal d} x)f(x),\end{align*}

where we have used the fact that $\mathbb{E}^*_{\delta_x}[\langle{\kern1.3pt} f, X_t\rangle] = \mathbb{Q}^*_{x}(\langle{\kern1.3pt} f, \omega_t \rangle)$, $t\geq 0$; see e.g. [Reference Dynkin and Kuznetsov9].

Now suppose that t is not a jump time of $\mathtt{I}^{(i)}_t$, $i =1,2,3$. In that case, we note that $\langle{\kern1.3pt} f, \Lambda_{t}\rangle = \langle{\kern1.3pt} f, X^*_{t} \rangle + \langle{\kern1.3pt} f, D_{t} \rangle $ is simply the aggregation of mass that has historically immigrated and evolved under $\mathbb{P}^*$. As such (comparing with (1.4), for example),

(4.3) \begin{align}{\textnormal d}\langle{\kern1.3pt} f, \Lambda_t\rangle & = -\langle\partial_z\psi^*(\cdot,0)f, \Lambda_{t-}\rangle \,{\textnormal d} t + {\textnormal d} U_t^c({\kern1.3pt}f) + {\textnormal d} U_t^d({\kern1.3pt}f)+ \langle\mathcal{L}f,\Lambda_{t-} \rangle\,{\textnormal d} t \notag\\ &\quad\,+\int_E\int_{(0,\infty)}\langle{\kern1.3pt} f, u\delta_x \rangle \mathtt{N}^1({\textnormal d} t,{\textnormal d} x,{\textnormal d} u)\notag\\&\quad\, +\int_{\mathbb{N}}\int_E\int_{(0,\infty)}\langle{\kern1.3pt} f, u\delta_x\rangle \mathtt{N}^2({\textnormal d} t,{\textnormal d}\{k\}, {\textnormal d} x, {\textnormal d} u)\notag\\ &\quad\, +\langle 2\beta f,Z_{t-} \rangle\,{\textnormal d} t, \quad t\geq 0,\end{align}

where

\begin{equation*}U_t^d({\kern1.3pt}f) = \int_0^t\int_{E}\int_{(0,\infty)}\langle{\kern1.3pt} f, u\delta_x\rangle\tilde{N}^0_s( {\textnormal d} s,{\textnormal d} x, {\textnormal d} u), \quad t\geq 0,\end{equation*}

and $U^c({\kern1.3pt}f)$ was defined immediately above Theorem 1. As such, we see from (4.3) that the pair $(\Lambda,Z)$ defined in this section provides a solution to (3.7).

5. Some integral and differential equations

The key part of our reasoning in proving parts (ii) and (iii) of Theorem 1 will be to show that

(5.1) \begin{equation}\mathbb{E}_\mu [{\textrm{e}}^{-\langle{\kern1.3pt} f,\Lambda_t\rangle-\langle h,Z_t\rangle}]=\mathbb{E}_\mu [ {\textrm{e}}^{-\langle{\kern1.3pt} f+w (1-{\textrm{e}}^{-h}),X_t\rangle}],\end{equation}

where X satisfies (3.5). Moreover, the key idea behind the proof of (5.1) is to fix $T>0$ and $f,h\in D_0(\mathcal{L})$, and choose time-dependent test functions $f^T$ and $h^T$ in such a way that the processes

(5.2) \begin{equation}F_t^T={\textrm{e}}^{-\langle{\kern1.3pt} f^T(\cdot,T-t),\Lambda_t\rangle-\langle h^T(\cdot,T-t),Z_t\rangle},\quad t\in[0,T],\end{equation}

and

(5.3) \begin{equation}G_t^T= {\textrm{e}}^{-\langle{\kern1.3pt} f^T(\cdot,T-t)\ +\ w (1-{\textrm{e}}^{-h^T(\cdot,T-t)}),X_t\rangle},\quad t\in[0,T],\end{equation}

have constant expectations on [0, T]. The test functions are defined as solutions to some partial differential equations with final value conditions $f^T(x,T)=f(x)$ and $h^T(x,T)=h(x)$. This, together with the fact that $\Lambda_0=X_0=\mu$, and that $Z_0$ is a Poisson random measure with intensity $w(x)\Lambda_0({\textnormal d} x)$, will then give us (5.1).

Thus, to prove (5.1) and hence parts (ii) and (iii) of Theorem 1, we need the existence of solutions of two differential equations. Recall from Section 2 that in the skeletal decomposition of superprocesses the total mass present at time t has two main components. The first one corresponds to an initial burst of subcritical mass, which is an independent copy of $(X,\mathbb{P}^*_\mu)$, and the second one is the accumulated mass from the dressing of the skeleton. As we will see in the next two results below, one can associate the first differential equation, i.e. the equation defining $f^T$, to $(X,\mathbb{P}^*_\mu)$, while the equation defining $h^T$ has an intimate relation to the dressed tree defined in the previous section.

Lemma 1. Fix $T>0$, and let $f\in D_0(\mathcal{L})$. Then the following differential equation has a unique non-negative solution:

(5.4) \begin{align}\dfrac{\partial}{\partial t}f^T(x,t)&=-\mathcal{L}f^T(x,t)+\psi^*(x,f^T(x,t)), \quad 0\leq t\leq T,\end{align}

\begin{align}\hspace*{-58.5pt} f^T(x,T)&=f(x),\notag\hspace*{58.5pt}\end{align}

where $\psi^*$ is given by (2.4).

Proof. Recall that $(X,\mathbb{P}^*_\mu)$ is a $(\mathcal{P}, \psi^*)$-superprocess, and as such its law can be characterised through an integral equation. More precisely, for all $\mu\in \mathcal{M}( E)$ and $f\in B^+( E)$, we have

\begin{equation*}\mathbb{E}^*_\mu [{\textrm{e}}^{-\langle{\kern1.3pt} f, X_t\rangle}]=\exp\lbrace -\langle u^*_f(\cdot,t)\mu\rangle\rbrace,\quad t\geq 0,\end{equation*}

where $u^*_f(x,t)$ is the unique non-negative solution to the integral equation

(5.5) \begin{equation}u^*_f(x,t)=\mathcal{P}_t[f](x)-\int_0^t \,{\textnormal d} s\cdot \mathcal{P}_s[\psi^*(\cdot,u^*_f(\cdot,t-s))](x),\quad x\in E,\ t\geq 0.\end{equation}

Li [Reference Li27, Theorem 7.11] showed that this integral equation is equivalent to the following differential equation:

(5.6) \begin{align}\dfrac{\partial}{\partial t}u^*_f(x,t)&=\mathcal{L}u^*_f(x,t)-\psi^*(x,u^*_f(x,t)),\\[5pt] u^*_f(x,0)&=f(x).\nonumber \end{align}

Thus (5.6) also has a unique non-negative solution. If for each fixed $T>0$ we define $f^T(x,t)=u_f^*(x,T-t)$, then it is not hard to see that the lemma holds.

Theorem 2. Fix $T>0$, and take $f,h\in D_0(\mathcal{L})\cap B^+_b(E)$. If $f^T$ is the unique solution to (5.4), then the following differential equation has a unique non-negative solution:

(5.7) \begin{align}{\textrm{e}}^{-h^T(x,t)}w(x)\dfrac{\partial}{\partial t}h^T(x,t)&=\mathcal{L}\big( w(x)\,{\textrm{e}}^{-h^T(x,t)}\big)\notag \\[1pt] &\quad\, +\big(\psi^*\big(x,-w(x)\,{\textrm{e}}^{-h^T(x,t)}+f^T(x,t)\big)-\psi^*\big(x,f^T(x,t)\big)\big),\\[1pt] h^T(x,T)&=h(x),\nonumber \end{align}

where $\psi^*$ is given by (2.4), and w is a martingale function that satisfies the conditions in Section 2.

Proof. Recall the process (D, Z) constructed in Section 4. For every $\mu\in\mathcal{M}(E)$, $\nu\in\mathcal{M}_a(E)$, and $f,h\in B^+_b(E)$, we have

\begin{equation*}\mathbb{E}_{(\mu,\nu)} [ {\textrm{e}}^{-\langle{\kern1.3pt} f, D_t\rangle-\langle h, Z_t\rangle }]={\textrm{e}}^{ -\langle v_{f,h}(\cdot,t),\nu\rangle},\end{equation*}

where $\textrm{exp}\{-v_{f,h}(x,t)\}$ is the unique [0, 1]-valued solution to the integral equation

(5.8) \begin{align}&w(x)\,{\textrm{e}}^{-v_{f,h}(x,t)}=\mathcal{P}_t [ w(\cdot)\,{\textrm{e}}^{-h(\cdot)}](x) \notag \\ &\quad +\int_0^t \textrm{d}s\cdot \mathcal{P}_{s} [\psi^*(\cdot,-w(\cdot)\,{\textrm{e}}^{-v_{f,h}(\cdot,t-s)}+u^*_f(\cdot,t-s))-\psi^*(\cdot,u^*_f(\cdot,t-s))](x),\end{align}

and $u^*_f$ is the unique non-negative solution to (5.5). Indeed, this claim is a straightforward adaptation of the proof of Theorem 2 in [Reference Kyprianou, Pérez and Ren23], the details of which we leave to the reader. Note also that a similar statement has appeared in [Reference Berestycki, Kyprianou and Murillo-Salas2] in the non-spatial setting.

Next suppose that $f,h\in D_0(\mathcal{L})\cap B^+_b(E)$. We want to show that solutions to the integral equation (5.8) are equivalent to solutions of the following differential equation:

(5.9) \begin{align}{\textrm{e}}^{-v_{f,h}(x,t)}w(x)\dfrac{\partial}{\partial t}v_{f,h}(x,t)&=-\mathcal{L}[ w(\cdot)\,{\textrm{e}}^{-v_{f,h}(\cdot,t)}](x)\notag \\ &\quad -(\psi^*(x,-w(x)\,{\textrm{e}}^{-v_{f,h}(x,t)}+u^*_f(x,t))-\psi^*(x,u^*_f(x,t))),\\ v_{f,h}(x,0)&=h(x).\nonumber \end{align}

The reader will note that the statement and proof of this claim are classical. However, we include them here for the sake of completeness. One may find similar computations in the Appendix of [Reference Dynkin6], for example.

We first prove the claim that solutions to the integral equation (5.8) are solutions to the differential equation (5.9). To this end consider (5.8). Note that since $\mathcal{P}$ is a Feller semigroup the right-hand side is differentiable in t, and thus $v_{f,h}(x,t)$ is also differentiable in t. To find the differential version of the equation, we can use the standard technique of propagating the derivative at zero using the semigroup property of $v_{f,h}$ and $u^*_f$. Indeed, on one hand the semigroup property can easily be verified using

\begin{align*}\mathbb{E}_{(\mu,\nu)}\big[{\textrm{e}}^{-\langle{\kern1.3pt} f,\Lambda_{t+s}\rangle -\langle h,Z_{t+s}\rangle}\big]&=\mathbb{E}_{(\mu,\nu)}\big[\mathbb{E}\big[{\textrm{e}}^{-\langle{\kern1.3pt} f,\Lambda_{t+s}\rangle -\langle h,Z_{t+s}\rangle}\mid \mathcal{F}_t\big]\big]\\ &=\mathbb{E}_{(\mu,\nu)} \big[\mathbb{E}_{(\Lambda_t,Z_t)} \big[{\textrm{e}}^{-\langle{\kern1.3pt} f,\Lambda_s\rangle -\langle h,Z_s\rangle}\big]\big]\\ &=\mathbb{E}_{(\mu,\nu)}\big[{\textrm{e}}^{-\langle u_f^*(\cdot,t),\Lambda_s \rangle-\langle v_{f,h}(\cdot,t),Z_s\rangle} \big]\\ &={\textrm{e}}^{-\langle u^*_{u^*_f(\cdot,t)}(\cdot,s),\mu \rangle-\langle v_{u^*_f(\cdot,t),v_{f,h}(\cdot,t)}(\cdot,s),\nu\rangle},\end{align*}

that is, we have $v_{u^*_f(\cdot,t),v_{f,h}(\cdot,t)}(\cdot,s)=v_{f,h}(\cdot,t+s)$, and $u^*_{u_f^*(\cdot,t)}(\cdot,s)=u_f^*(\cdot,t+s)$. This implies

(5.10) \begin{equation}\dfrac{\partial}{\partial t}u_f^*(x,t+)=\dfrac{\partial}{\partial s}u_{u_f^*(\cdot,t)}(x,s)\bigg|_{s\downarrow 0}=\dfrac{\partial}{\partial s}u_{u_f^*(\cdot,t)}(x,0+)\end{equation}

and

(5.11) \begin{equation}\dfrac{\partial}{\partial t}v_{f,h}(\cdot,t+)=\dfrac{\partial}{\partial s}v_{u^*_f(\cdot,t),v_{f,h}(\cdot,t)}(x,s)\bigg|_{s\downarrow 0},\end{equation}

providing that the two derivatives at zero exist from the right. One may similarly use the semigroup property, splitting at time s and $t-s$ to give the left derivatives at time $t>0$. On the other hand, differentiating (5.8) in t and taking $t\downarrow 0$ gives

\begin{align*} -w(x)\,{\textrm{e}}^{-v_{f,h}(x,0+)}\dfrac{\partial}{\partial t}v_{f,h}(x,t)\bigg|_{t=0+}&=\mathcal{L}[ w(\cdot)\,{\textrm{e}}^{-h(\cdot)}](x)\notag \\ &\quad\, +\psi^*( x,-w(x)\,{\textrm{e}}^{-v_{f,h}(x,0+)}+u_f^*(x,0+) )\notag\\ &\quad\, -\psi^*(x,u_f^*(x,0+)) ,\end{align*}

which, recalling $v_{f,h}(x,0)=h(x)$ and $u_f^*(x,0)=f(x)$, can be rewritten as

\begin{align*}\dfrac{\partial}{\partial t}v_{f,h}(x,0+)&=-\dfrac{1}{w(x)}\,{\textrm{e}}^{h(x)}\mathcal{L}[ w(\cdot)\,{\textrm{e}}^{-h(\cdot)}](x)\\ &\quad\, -\dfrac{1}{w(x)}\,{\textrm{e}}^{h(x)}\psi^*( x,-w(x)\,{\textrm{e}}^{-h(x)}+f(x))+\dfrac{1}{w(x)}\,{\textrm{e}}^{h(x)}\psi^*(x,f(x)).\end{align*}

Hence, combining the previous observations in (5.10) and (5.11), we get

\begin{align*}\dfrac{\partial}{\partial t}v_{f,h}(x,t)=&-\dfrac{1}{w(x)}\,{\textrm{e}}^{v_{f,h}(x,t)}\mathcal{L} [ w(\cdot)\,{\textrm{e}}^{-v_{f,h}(x,t)}](x)\\ &-\dfrac{1}{w(x)}\,{\textrm{e}}^{v_{f,h}(x,t)}\psi^*( x,-w(x)\,{\textrm{e}}^{-v_{f,h}(x,t)}+u_f^*(x,t))\\ &+\dfrac{1}{w(x)}\,{\textrm{e}}^{v_{f,h}(x,t)}\psi^*(x,u_f^*(x,t)).\end{align*}

To see why the differential equation (5.9) implies the integral equation (5.8), define

\begin{equation*}g(x,s)=\mathcal{P}_{t-s}(w(x)\,{\textrm{e}}^{-v_{f,h}(x,s)}), \quad 0\leq s\leq t.\end{equation*}

Then differentiating with respect to the time parameter gives

\begin{align*}\dfrac{\partial}{\partial s}g(x,s)&=-\mathcal{P}_{t-s}w(x)\,{\textrm{e}}^{v_{f,h}(x,s)}\dfrac{\partial}{\partial s}v_{f,h}(x,s)-\mathcal{P}_{t-s}\mathcal{L}(w(x)\,{\textrm{e}}^{-v_{f,h}(x,s)})\\ &=\mathcal{P}_{t-s}[\psi^*(x,-w(x)\,{\textrm{e}}^{-v_{f,h}(x,s)}+u^*_f(x,s))-\psi^*(x,u^*_f(x,s))],\end{align*}

which we can then integrate over [0, t] to get (5.8).

To complete the proof we fix $T>0$ and define $h^T(x,t)\,:\!= v_{f,h}(x,T-t)$, and the result follows.

6. Proof of Theorem 1 (ii) and (iii)

The techniques we use here are similar in spirit to those in the proof of Theorem 2.1 in [Reference Fekete, Fontbona and Kyprianou16], in the sense that we use stochastic calculus to show the equality (5.1). However, what is new in the current setting is the use of the processes (5.2) and (5.3).

Fix $T>0$, and let $f^T$ be the unique non-negative solution to (5.4), and let $h^T$ be the unique non-negative solution to (5.7). Define $F_t^T\,:\!= {\textrm{e}} ^{-\langle{\kern1.3pt} f^T(\cdot,t),\Lambda_t\rangle-\langle h^T(\cdot,t),Z_t\rangle}$, $t\leq T$. Using stochastic calculus, we first verify that our choice of $f^T$ and $h^T$ results in the process $F_t^T$, $t\leq T$, having constant expectation on [0, T]. In the definition of $F^T$ both $\langle{\kern1.3pt} f^T(\cdot,t),\Lambda_t\rangle$ and $\langle h^T(\cdot,t),Z_t\rangle$ are semi-martingales, thus we can use Itô’s formula (see e.g. Theorem 32 in [Reference Protter31]) to get

\begin{align*}{\textnormal d} F_t^T&=- F_{t-}^T\,{\textnormal d}\Lambda_t^{f^T}- F_{t-}^T\,{\textnormal d} Z_t^{h^T}+\dfrac{1}{2} F_{t-}^T\,{\textnormal d}\Big[\Lambda^{f^T},\Lambda^{f^T}\Big]_t^c+\dfrac{1}{2} F_{t-}^T\,{\textnormal d}\Big[Z^{h^T},Z^{h^T}\Big]_t^c\\ &\quad\, + F_{t-}^T\,{\textnormal d}\Big[\Lambda^{f^T},Z^{h^T}\Big]_t^c+\Delta F_t^T+ F_{t-}^T\Delta\Lambda_t^{f^T}+ F_{t-}^T\Delta Z_t^{h^T},\quad 0\leq t\leq T,\end{align*}

where $\Delta \Lambda_t^{f^T}=\langle{\kern1.3pt} f^T(\cdot,t),\Lambda_t-\Lambda_{t-}\rangle$, and to avoid heavy notation we have written $\Lambda_t^{f^T}$ instead of $\langle{\kern1.3pt} f^T(\cdot,t),\Lambda_t\rangle$ and $Z_t^{h^T}$ instead of $\langle h^T(\cdot,t),Z_t\rangle$. Note that without the movement Z is a pure jump process, and since the interaction between $\Lambda$ and Z is limited to the time of the immigration events, we have $[\Lambda^{f^T},Z^{h^T}]_t^c=0$. Taking advantage of

\begin{equation*}F_t^T=F_{t-}^T\,{\textrm{e}}^{-\Delta\Lambda_t^{f^T}-\Delta Z_t^{h^T}},\end{equation*}

we may thus write in integral form

\begin{align*}F_t^T&=F_0^T-\int_0^t F_{s-}^T\,{\textnormal d}\Lambda^{f^T}_s-\int_0^t F_{s-}^T\,{\textnormal d} Z^{h^T}_s+\int_0^t F_{s-}^T\langle \beta(\cdot) ({\kern1.3pt}f^T(\cdot,s))^2,\Lambda_{s-}\rangle \,{\textnormal d} s\\ &\quad\, +\int_0^t F_{s-}^T\langle(\nabla h^T(\cdot,s))^{\texttt{T}} a\nabla h^T(\cdot,s),Z_{s-}\rangle \,{\textnormal d} s+\sum_{s\leq t}\Big\{ \Delta F_s^T+F_{s-}^T\Delta\Lambda_s^{f^T}+ F_{s-}^T\Delta Z_s^{h^T}\Big\}.\end{align*}

To simplify the notation we used the fact that both $f^T(x,t)$ and $h^T(x,t)$ are continuous in t, thus $f^T(x,t)=f^T(x,t-)$ and $h^T(x,t)=h^T(x,t-)$.

We can split up the last term, that is, the sum of discontinuities according to the optional random measure in (3.7) responsible for this discontinuity. Thus, writing $\Delta^{(i)}$, $i=0,1,2$, to mean an increment coming from each of the three random measures,

\begin{align*}F_t^T&=F_0^T-\int_0^t F_{s-}^T\,{\textnormal d}\Lambda^{f^T}_s-\int_0^t F_{s-}^T\,{\textnormal d} Z^{h^T}_s+\int_0^t F_{s-}^T\langle \beta(\cdot) ({\kern1.3pt}f^T(\cdot,s))^2,\Lambda_{s-}\rangle \,{\textnormal d} s\\ &\quad\, +\int_0^t F_{s-}^T\langle(\nabla h^T(\cdot,s))^{\texttt{T}} a\nabla h^T(\cdot,s),Z_{s-}\rangle \,{\textnormal d} s+\sum_{s\leq t}F_{s-}^T\Big\{ {\textrm{e}}^{-\Delta^{(0)}\Lambda_s^{f^T}}-1+\Delta^{(0)}\Lambda_s^{f^T}\Big\}\\ &\quad\, +\sum_{s\leq t}F_{s-}^T\Big\{{\textrm{e}}^{-\Delta^{(1)}\Lambda_s^{f^T}}-1+\Delta^{(1)}\Lambda_s^{f^T}\Big\}\\ &\quad\, +\sum_{s\leq t}F_{s-}^T\Big\{ {\textrm{e}}^{-\Delta^{(2)}\Lambda_s^{f^T}-\Delta Z_s^{h^T}}-1+\Delta^{(2)}\Lambda_s^{f^T}+ \Delta Z_s^{h^T}\Big\}.\end{align*}

Next, plugging in ${\textnormal d}\Lambda_s^{f^T}$ and ${\textnormal d} Z_s^{h^T}$ gives

(6.1) \begin{align}F_t^T&=F_0^T+\int_0^t F_{s-}^T\langle \psi'(\cdot,w(\cdot))f^T(\cdot,s),\Lambda_{s-}\rangle \,{\textnormal d} s+\int_0^t F_{s-}^T\langle \beta(\cdot)({\kern1.3pt}f^T(\cdot,s))^2,\Lambda_{s-}\rangle\,{\textnormal d} s \notag \\ &\quad\, -\eta\int_0^t F_{s-}^T\langle \mathcal{L}f^T(\cdot,s),\Lambda_{s-}\rangle\,{\textnormal d} s-\int_0^t F_{s-}^T\biggl\langle \dfrac{\partial}{\partial s}f^T(\cdot,s),\Lambda_{s-}\biggr\rangle\,{\textnormal d} s \notag \\&\quad\, -\int_0^t F_{s-}^T\biggl\langle \dfrac{\partial}{\partial s}h^T(\cdot,s),Z_{s-}\biggr\rangle\,{\textnormal d} s-\int_0^t F_{s-}^T\langle \mathcal{L}^w h^T(\cdot,s),Z_{s-}\rangle\,{\textnormal d} s \notag \\&\quad\, -\int_0^t F_{s-}^T \langle 2\beta(\cdot) f^T(\cdot,s),Z_{s-}\rangle \,{\textnormal d} s+\sum_{s\leq t}F_{s-}^T \Big\{ {\textrm{e}}^{-\Delta^{(0)}\Lambda_s^f}-1+\Delta^{(0)}\Lambda_s^{f^T} \Big\} \notag \\ &\quad\, +\sum_{s\leq t}F_{s-}^T \Big\{ {\textrm{e}}^{-\Delta^{(1)}\Lambda_s^{f^T}}-1 \Big\}+\sum_{s\leq t}F_{s-}^T \Big\{ {\textrm{e}}^{-\Delta^{(2)}\Lambda_s^{f^T}-\Delta Z_s^{h^T}}-1 \Big\} \notag \\ &\quad\, +\int_0^t F_{s-}^T\langle (\nabla h^T(\cdot,s))^{\texttt{T}} a\nabla h^T(\cdot,s),Z_{s-}\rangle \,{\textnormal d} s+M_t^{\textrm{loc}},\end{align}

where $M_t^{\textrm{loc}}$ is a local martingale corresponding to the terms $U_t^c({\kern1.3pt}f^T)$, $V_t^c(h^T)$ and the integral with respect to the random measure $\tilde{{\texttt N}}^{0}$ in (3.7). Note that the two terms with the time-derivative are due to the extra time dependence of the test functions in the integrals $\langle{\kern1.3pt} f^T(\cdot,s),\Lambda_s\rangle$ and $\langle h^T(\cdot,s),Z_s\rangle$. In particular, a change in $\langle{\kern1.3pt} f^T(s,\cdot),\Lambda_s\rangle$ corresponds to either a change in $\Lambda_s$ or a change in $f^T(\cdot,s)$.

Next we show that the local martingale term is in fact a real martingale, which will then disappear when we take expectations. First note that due to the boundedness of the drift and diffusion coefficients of the branching mechanism, and the conditions we had on its Lévy measure, the branching of the superprocess can be stochastically dominated by a finite-mean CSBP. This means that the CSBP associated with the Esscher-transformed branching mechanism $\psi^*$ is almost surely finite on any finite time interval [0, T], and thus the function $f^T$ is bounded on [0, T]. Using the boundedness of $f^T$ and the drift coefficient $\beta$, the quadratic variation of the integral

(6.2) \begin{equation}\int_0^tF_{s-}^T \,{\textnormal d} U_s^c({\kern1.3pt}f^T)\end{equation}

can be bounded from above as follows:

\begin{align*}\int_0^t 2 F_{s-}^T \langle \beta(\cdot) ({\kern1.3pt}f^T(\cdot,s))^2,\Lambda_{s-}\rangle \,{\textnormal d} s &\leq \int_0^t \,{\textrm{e}}^{-\langle{\kern1.3pt} f^T(\cdot,s),\Lambda_{s-}\rangle} \langle C ({\kern1.3pt}f^T(\cdot,s))^2,\Lambda_{s-}\rangle \,{\textnormal d} s\\ &\leq \int_0^t \,{\textrm{e}}^{-\widetilde{C}\|\Lambda_{s-}\|}\widehat{C}\|\Lambda_{s-}\|\,{\textnormal d} s,\end{align*}

where C, $\widehat{C}$, and $\widetilde{C}$ are finite constants. Since the function $x\mapsto {\textrm{e}}^{-\widetilde{C}x}x$ is bounded on $[0,\infty)$, the previous quadratic variation is finite, and so the process (6.2) is a martingale on [0, T].

To show the martingale nature of the stochastic integral

(6.3) \begin{equation}\int_0^t F_{s-}\,{\textnormal d} V_s^c(h^T){,}\end{equation}

we note that due to construction, $h^T\in \mathcal{D}_0(\mathcal{L})$, and is bounded on [0, T]. Thus, $V_t^c(h^T)$ is in fact a martingale on [0, T], and since $F_{s-}\leq 1$, $s\in[0, T]$, the quadratic variation of (6.3) is also finite, which gives the martingale nature of (6.3) on [0, T].

Finally, we consider the integral

(6.4) \begin{equation}\int_0^t\int_{E}\int_{(0,\infty)} F^T_{s-} \langle{\kern1.3pt} f^T(\cdot,s),u\delta_x\rangle \tilde{N}^0({\textnormal d} s, {\textnormal d} x, {\textnormal d} u).\end{equation}

Note that for compactly supported $\mu\in\mathcal{M}(E)^\circ$,

\begin{align*}Q_t&\,:\!= \mathbb{E}_\mu\biggl[\int_0^t\int_{E}\int_{(0,\infty)} \big(F^T_{s-}\langle{\kern1.3pt} f^T(\cdot,s),u\delta \rangle\big)^2\hat{N}^0({\textnormal d} s,{\textnormal d} x, {\textnormal d} u)\biggr]\\ &= \mathbb{E}_\mu\biggl[\int_0^t \int_E\int_{(0,\infty)}\big(F^T_{s-} u f^T(x,s)\big)^2 \,{\textrm{e}}^{-w(x)u}m(x,{\textnormal d} u) \Lambda_{s-}({\textnormal d} x){\textnormal d} s\biggr]\\&\leq \mathbb{E}_\mu\biggl[\int_0^t \,{\textrm{e}}^{-2C \|\Lambda_{s-}\|}C\biggl\langle \int_{(0,\infty)}u^2\,{\textrm{e}}^{-w(x)u}m(x,{\textnormal d} u),\Lambda_{s-}\biggr\rangle \,{\textnormal d} s\biggr]\\&\leq \mathbb{E}_\mu\biggl[\int_0^t \,{\textrm{e}}^{-\widetilde{C}\|\Lambda_{s-}\|} \widehat{C}\|\Lambda_{s-}\|\,{\textnormal d} s\biggr]\\ &\leq C' t{,}\end{align*}

where C, $\widetilde{C}$, $\widehat{C}$, and C ^′ are finite constants. Thus $Q_t<\infty$ on [0, T], and we can refer to page 63 of [Reference Ikeda and Watanabe20] to conclude that the process (6.4) is indeed a martingale on [0, T].

Thus, after taking expectations and gathering terms in (6.1), we get

(6.5) \begin{align}\mathbb{E}_\mu [F_t^T] &=\mathbb{E}_\mu [F_0^T] +\int_0^t \mathbb{E}_\mu [F_{s-}^T\langle A(\cdot,f^T(\cdot,s)),\Lambda_{s-}\rangle ]\,{\textnormal d} s\notag \\ &\quad\, -\int_0^t \mathbb{E}_\mu\biggl[F_{s-}^T\biggl\langle \dfrac{\partial}{\partial s}f^T(\cdot,s),\Lambda_{s-}\biggr\rangle\biggr]\,{\textnormal d} s \notag \\&\quad\, +\int_0^t \mathbb{E}_\mu[F_{s-}^T\langle B(\cdot,h^T(\cdot,s),f^T(\cdot,s)),Z_{s-}\rangle]\,{\textnormal d} s \notag \\ &\quad\, -\int_0^t \mathbb{E}_\mu\biggl[F_{s-}^t\biggl\langle \dfrac{\partial}{\partial s}h^T(\cdot,s),Z_{s-}\biggr\rangle\biggr]\,{\textnormal d} s,\quad 0\leq t\leq T,\end{align}

where

(6.6) \begin{align}A(x,f)&=\psi'(x,w(x))f+\beta(x)f^2-\mathcal{L}f+\int_{(0,\infty)}({\textrm{e}}^{-u f}-1+u f)\,{\textrm{e}}^{-w(x)u}m(x,{\textnormal d} u)\\[4pt] &=-\mathcal{L}f+\psi^*(x,f)\notag , \end{align}

and

(6.7) \begin{align}B(x,h,f)&=(\nabla h)^{\texttt{T}}a\nabla h-\mathcal{L}^w h-2\beta(x) f+\int_{(0,\infty)}({\textrm{e}}^{- u f}-1) u\,{\textrm{e}}^{-w(x) u}m(x,{\textnormal d} u)\notag \\ &\quad\, +\sum_{k=2}^\infty \int_{(0,\infty)}({\textrm{e}}^{- u f-(k-1)h}-1)\dfrac{1}{w(x)}\notag \\ &\quad\, \times\biggl\{ \beta(x) w^2(x)\delta_0({\textnormal d} u)\textbf{1}_{\{ k=2\}} +w^k(x)\dfrac{u^k}{k!}\,{\textrm{e}}^{-w(x)u}m(x,{\textnormal d} u)\biggr\}.\end{align}

We can see immediately that $A(x,f^T(x,t))$ is exactly what we have on the right-hand side of (5.4). Furthermore, using the fact that

(6.8) \begin{equation}(\nabla h)^{\texttt{T}}a\nabla h-\mathcal{L}^w h={\textrm{e}}^h\dfrac{1}{w}\mathcal{L}( w \,{\textrm{e}}^{-h})-\dfrac{1}{w}\psi(\cdot,w),\end{equation}

we can also verify that

\begin{equation*}B(x,h,f)={\textrm{e}}^{h}\dfrac{1}{w}\mathcal{L}( w\,{\textrm{e}}^{-h})+{\textrm{e}}^{h}\dfrac{1}{w}(\psi^*( x,-w(x)\,{\textrm{e}}^{-h}+f) -\psi^*(x,f)),\end{equation*}

that is, $B(x,h^T(x,t),f^T(x,t))$ is equal to the right-hand side of (5.7). Hence, recalling the defining equations of $f^T$ (5.4) and $h^T$ (5.7), we get that the last four terms of (6.5) cancel, and thus $\mathbb{E}_\mu[F_t^T]=\mathbb{E}_\mu[F_0^T]$ for $t\in[0, T]$, as required. In particular, using the boundary conditions for $f^T$ and $h^T$, we get

(6.9) \begin{align}\mathbb{E}_\mu \big[F_T^T\big]=\mathbb{E}_\mu \big[ {\textrm{e}}^{-\langle{\kern1.3pt} f,\Lambda_T\rangle-\langle h,Z_T\rangle}\big]=\mathbb{E}_\mu \Big[ {\textrm{e}}^{-\langle{\kern1.3pt} f^T(\cdot,0),\Lambda_0\rangle -\langle h^T(\cdot,0),Z_0\rangle}\Big]=\mathbb{E}_\mu \big[F_0^T\big].\end{align}

Note that by construction we can relate the right-hand side of this previous expression to the superprocess. In particular, using the Poissonian nature of $Z_0$, and the fact that $X_0=\Lambda_0=\mu$ is deterministic, we have

(6.10) \begin{equation}\mathbb{E}_\mu \Big[{\textrm{e}}^{-\langle{\kern1.3pt} f^T(\cdot,0),\Lambda_0\rangle-\langle h^T(\cdot,0),Z_0\rangle}\Big]=\mathbb{E}_\mu \Big[{\textrm{e}}^{-\langle{\kern1.3pt} f^T(\cdot,0)+w(\cdot) (1-{\textrm{e}}^{-h^T(\cdot,0)}),X_0\rangle}\Big],\end{equation}

where $X_t$ is a solution to (3.6). Thus, by choosing the right test functions, we could equate the value of $F_t^T$ at T to its initial value, which in turn gave a connection with the superprocess. The next step is to show that the process

\begin{equation*}{\textrm{e}}^{-\langle{\kern1.3pt} f^T(\cdot,t)+w(\cdot)( 1-{\textrm{e}}^{-h^T(\cdot,t)}),X_t\rangle}, \quad t\in[0, T],\end{equation*}

has constant expectation on [0, T], which would then allow us to deduce

\begin{equation*}\mathbb{E}_\mu [ {\textrm{e}}^{-\langle{\kern1.3pt} f,\Lambda_T\rangle-\langle h ,Z_T\rangle}]=\mathbb{E}_\mu [ {\textrm{e}}^{-\langle{\kern1.3pt} f +w (1-{\textrm{e}}^{-h }),X_T\rangle} ].\end{equation*}

To simplify the notation let $\kappa^T(x,t)\,:\!=f^T(x,t)+w(x)( 1-{\textrm{e}}^{-h^T(x,t)})$, and define $G_t^T\,:\!={\textrm{e}}^{-\langle \kappa^T(\cdot,t),X_t\rangle}$. As the argument here is the exact copy of the previous analysis, we only give the main steps of the calculus, and leave it to the reader to fill in the gaps.

Since $\langle \kappa^T(\cdot,t),X_t\rangle$, $t\leq T$, is a semi-martingale, we can use Itô’s formula to get

(6.11) \begin{align}G_t^T &=G_0^T+\int_0^t G_{s-}^T\langle \psi'(\cdot,w(\cdot))\kappa^T(s,\cdot),X_{s-}\rangle \,{\textnormal d} s+\int_0^t G_{s-}^T\langle \beta(\cdot)(\kappa^T(\cdot,s))^2,X_{s-}\rangle\,{\textnormal d} s \notag \\ &\quad\, -\int_0^t G_{s-}^T\langle 2\beta(\cdot) w(\cdot) \kappa^T(\cdot,s),X_{s-}\rangle \,{\textnormal d} s-\int_0^t G_{s-}^T\langle \mathcal{L}\kappa^T(\cdot,s),X_{s-}\rangle \,{\textnormal d} s \notag \\&\quad\, +\int_0^t G_{s-}^T\biggl\langle \int_0^\infty ( {\textrm{e}}^{-u\kappa^T(\cdot,s)}-1+u\kappa^T(\cdot,s))\,{\textrm{e}}^{-w(\cdot)u}m(\cdot,{\textnormal d} u),X_{s-}\biggr\rangle\,{\textnormal d} s \notag \\&\quad\, +\int_0^t G_{s-}^t\biggl\langle \int_0^\infty ( {\textrm{e}}^{-u\kappa^T(\cdot,s)}-1)w(\cdot)u\,{\textrm{e}}^{-w(\cdot)u}m(\cdot,{\textnormal d} u),X_{s-}\biggr\rangle\,{\textnormal d} s \notag \\&\quad\, +\int_0^t G_{s-}^T\biggl\langle \int_0^\infty ( {\textrm{e}}^{-u \kappa^T(\cdot,s)}-1)\sum_{k=2}^\infty \dfrac{(w(\cdot)u)^k}{k!}\,{\textrm{e}}^{-w(\cdot) u}m(\cdot,{\textnormal d} u),X_{s-}\biggr\rangle\,{\textnormal d} s \notag \\ &\quad\, -\int_0^t G_{s-}^T\biggl\langle \dfrac{\partial}{\partial s}\kappa^T(\cdot,s),X_{s-}\biggr\rangle \,{\textnormal d} s+M_t^{\textrm{loc}}.\end{align}

where $M_t^{\textrm{loc}}$ is a local martingale corresponding to the term $M_t^c({\kern1.3pt}f)$, and the integral with respect to the random measure $\tilde{N}^{0}$ in (3.6). Note that the reasoning that led to the martingale nature of the local martingale term of (6.1) can also be applied here, which gives that $M_t^{\textrm{loc}}$ in (6.11) is in fact a true martingale on [0, T], which we denote by $M_t$.

Next we plug in $\kappa^T$, and after some laborious amount of algebra get

\begin{align*}G_t^T&= G_0^T+\int_0^t G_{s-}^T\langle \psi'(\cdot,w(\cdot))f^T(\cdot,s),X_{s-}\rangle\,{\textnormal d} s+\int_0^t G_{s-}^T\langle \beta(\cdot)({\kern1.3pt}f^T(\cdot,s))^2,X_{s-}\rangle\,{\textnormal d} s\\ &\quad\, -\int_0^t G_{s-}^T\langle \mathcal{L}f^T(\cdot,s),X_{s-}\rangle\,{\textnormal d} s\\ &\quad\,+\int_0^t G_{s-}^T\biggl\langle \int_{(0,\infty)}({\textrm{e}}^{-u f^T(\cdot,s)}-1+u f^T(\cdot,s))\,{\textrm{e}}^{-w(\cdot)u}m(\cdot,{\textnormal d} u),X_{s-} \biggr\rangle\,{\textnormal d} s\\ &\quad\,-\int_0^t G_{s-}^T\langle 2\beta(\cdot) f^T(\cdot,s)\,{\textrm{e}}^{-h^T(\cdot,s)}w(\cdot),X_{s-} \rangle\,{\textnormal d} s\\ &\quad\,+\int_0^t G_{s-}^T\biggl\langle\int_{(0,\infty)}({\textrm{e}}^{- u f^T(\cdot,s)}-1) u\,{\textrm{e}}^{-w(\cdot) u}m(\cdot,{\textnormal d} u)\,{\textrm{e}}^{-h^T(\cdot,s)}w(\cdot),X_{s-}\biggr\rangle \,{\textnormal d} s\displaybreak\\ &\quad\,+\int_0^t G_{s-}^T\biggl\langle \sum_{k=2}^\infty \int_{(0,\infty)}({\textrm{e}}^{- u f^T(\cdot,s)-(k-1)h^T(\cdot,s)}-1)\dfrac{1}{w(\cdot)}\\ &\quad\,\times \biggl\lbrace \beta(\cdot) w^2(\cdot)\delta_0({\textnormal d} u)\textbf{1}_{\{ k=2\}}+w^k(\cdot)\dfrac{u^k}{k!}\,{\textrm{e}}^{-w(\cdot)u}m(\cdot,{\textnormal d} u)\biggr\rbrace \,{\textrm{e}}^{-h^T(\cdot)}w(\cdot), X_{s-}\biggr\rangle\,{\textnormal d} s\\ &\quad\,+\int_0^t G_{s-}^T\biggl\langle (1-{\textrm{e}}^{-h^T(\cdot,s)})\psi(\cdot,w(\cdot))-\mathcal{L}w(\cdot)(1-{\textrm{e}}^{-h^T(\cdot,s)}), X_{s-}\biggr\rangle \,{\textnormal d} s\\ &\quad\,-\int_0^t G_{s-}^T\biggl\langle \dfrac{\partial}{\partial s}\kappa^T(\cdot,s),X_{s-}\biggr\rangle \,{\textnormal d} s+M_t.\end{align*}

Once again using the identity (6.8), and taking expectations, gives

(6.12) \begin{align}\mathbb{E}_{\mu}[G_t^T]&=\mathbb{E}_{\mu}[G_0^T]+\int_0^t \mathbb{E}_{\mu}[G_{s-}^T\langle A(\cdot,f^T(\cdot,s)),X_{s-}\rangle]\,{\textnormal d} s\end{align}

\begin{align}&\quad\, +\int_0^t \mathbb{E}_{\mu}[G_{s-}^T\langle {\textrm{e}}^{-h^T(\cdot,s)}w(\cdot) B(\cdot,h^T(\cdot,s),f^T(\cdot,s), X_{s-}\rangle]\,{\textnormal d} s \notag \\*&\quad\, -\int_0^t \mathbb{E}_{\mu}\biggl[G_{s-}^T\biggl\langle \dfrac{\partial}{\partial s}\kappa^t(s,\cdot),X_{s-}\biggr\rangle\biggr] \,{\textnormal d} s,\notag\end{align}

where A and B are given by (6.6) and (6.7). Finally, noting

\begin{align*}\dfrac{\partial}{\partial s}\kappa^T(x,s)=\dfrac{\partial}{\partial s}f^T(x,s)+w(x)\,{\textrm{e}}^{-h^T(x,s)}\dfrac{\partial}{\partial s}h^T(x,s)\end{align*}

gives

\begin{equation*}\dfrac{\partial}{\partial s}\kappa^T(s,x)=-A(x,f^T(x,s))-w(x)\,{\textrm{e}}^{-h^T(x,s)}B(x,h^T(x,s),f^T(x,s)),\end{equation*}

which results in the cancellation of the last three terms in (6.12), and hence verifies the constant expectation of $G_t^T$ on [0, T]. In particular, we have proved that

(6.13) \begin{align}\mathbb{E}_\mu[G_T^T]&=\mathbb{E}_\mu [ {\textrm{e}}^{-\langle{\kern1.3pt} f+w(1-{\textrm{e}}^{-h}),X_T \rangle}] \notag \\*&=\mathbb{E}_\mu [ {\textrm{e}}^{- \langle{\kern1.3pt} f^T(\cdot,0)+w(1-{\textrm{e}}^{-h^T(\cdot,0)}),X_0\rangle}] \notag \\*&=\mathbb{E}_\mu[G_0^T].\end{align}

In conclusion, combining the previous observations (6.9) and (6.10) with (6.13) gives

\begin{equation*}\mathbb{E}_\mu [ {\textrm{e}}^{-\langle{\kern1.3pt} f,\Lambda_T\rangle-\langle h,Z_T\rangle}]=\mathbb{E}_\mu [ {\textrm{e}}^{-\langle{\kern1.3pt} f+w (1-{\textrm{e}}^{-h}),X_T\rangle}].\end{equation*}

Since $T>0$ was arbitrary, the above equality holds for any time $T>0$.

Then we have the following implications. First, by setting $h=0$ we find that

\begin{equation*}\mathbb{E}_\mu [ {\textrm{e}}^{-\langle{\kern1.3pt} f,\Lambda_T\rangle}]=\mathbb{E}_\mu[ {\textrm{e}}^{-\langle{\kern1.3pt} f,X_T\rangle}],\end{equation*}

which not only shows that under our conditions $(\Lambda_t,t\geq 0)$ is Markovian, but also that its semigroup is equal to the semigroup of $(X,\mathbb{P}_\mu)$, and hence proves that $(\Lambda_t,t\geq 0)$ is indeed a weak solution to (3.5).

Furthermore, choosing h and f not identical to zero, we get that the pair $(\Lambda_t,Z_t)$ under $\textbf{P}_{\mu}$ has the same law as $(X_t,\textrm{Po}(w(x)X_t({\textnormal d} x)))$ under $\mathbb{P}_\mu$, where $\textrm{Po}(w(x)X_t({\textnormal d} x))$ is an autonomously independent Poisson random measure with intensity $w(x)X_t({\textnormal d} x)$. Thus $Z_t$ given $\Lambda_t$ is indeed a Poisson random measure with intensity $w(x)\Lambda_t({\textnormal d} x)$.

7. Proof of Theorem 1 (i): uniqueness

If we review the calculations that led to (6.9), we observe that any solution $(\Lambda, Z)$ to the coupled SDE (3.7) has the property that, for $\mu\in\mathcal{M}(E)$ and $\nu\in\mathcal{M}_a(E)$,

\begin{equation*}\mathbb{E}_{(\mu, \nu)} [F_T^T]=\mathbb{E}_{(\mu, \nu)} [ {\textrm{e}}^{-\langle{\kern1.3pt} f,\Lambda_T\rangle-\langle h,Z_T\rangle}]={\textrm{e}}^{-\langle{\kern1.3pt} f^T(\cdot,0),\mu\rangle -\langle h^T(\cdot,0),\nu\rangle}.\end{equation*}

Hence, since any two solutions to (3.7) are Markovian, the second equality above identifies their transitions as equal thanks to the uniqueness of the PDEs in Lemma 1 and Theorem 2. In other words, there is a unique weak solution to (3.7).