SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\begin{array}{l} \mathbb {E} [ (f (\mathbf {X} _ {t}) - f (\mathbf {X} _ {s})) g (\mathbf {X} _ {t}) ] = \mathbb {E} [ g (t, \mathbf {X} _ {t}) f (\mathbf {X} _ {t}) - f (\mathbf {X} _ {s}) g (s, \mathbf {X} _ {s}) ] \\ = \mathbb {E} [ \int_ {s} ^ {t} \mathcal {A} (f g) (u, \mathbf {X} _ {u}) \mathrm {d} u ] \\ = - \mathbb {E} \left[ \int_ {s} ^ {t} \mathcal {R} (f) (u, \mathbf {X} _ {u}) g (u, \mathbf {X} _ {u}) \mathrm {d} u \right] = - \mathbb {E} \left[ g (\mathbf {X} _ {t}) \int_ {s} ^ {t} \mathcal {R} (f) (u, \mathbf {X} _ {u}) \mathrm {d} u \right], \\ \end{array}
$$
A.2.2 Intuitive Proof of Proposition 1
We recall Proposition 1.
Proposition 1. The time reversal of a forward jump diffusion process given by drift  $\vec{\mathbf{b}}_t$ , diffusion coefficient  $\vec{g}_t$ , rate  $\vec{\lambda}_t(n)$  and transition kernel  $\sum_{i=1}^{n} K^{\mathrm{del}}(i|n) \delta_{\mathrm{del}(\mathbf{X},i)}(\mathbf{Y})$  is given by a jump diffusion process with drift  $\vec{\mathbf{b}}_t^*(\mathbf{X})$ , diffusion coefficient  $\vec{g}_t^*$ , rate  $\vec{\lambda}_t^*(\mathbf{X})$  and transition kernel  $\int_{\mathbf{y}^{\mathrm{add}}} \sum_{i=1}^{n+1} A_t^*(\mathbf{y}^{\mathrm{add}}, i|\mathbf{X}) \delta_{\mathrm{ins}(\mathbf{X},\mathbf{y}^{\mathrm{add}},i)}(\mathbf{Y}) \, \mathrm{d}\mathbf{y}^{\mathrm{add}}$  as defined below
$$
\overleftarrow {\lambda} _ {t} ^ {*} (\mathbf {X}) = \overrightarrow {\lambda} _ {t} (n + 1) \frac {\sum_ {i = 1} ^ {n + 1} K ^ {\mathrm {d e l}} (i | n + 1) \int_ {\mathbf {y} ^ {\mathrm {a d d}}} p _ {t} (\operatorname {i n s} (\mathbf {X} , \mathbf {y} ^ {\mathrm {a d d}} , i)) \mathrm {d} \mathbf {y} ^ {\mathrm {a d d}}}{p _ {t} (\mathbf {X})},
$$
$$
A _ {t} ^ {*} (\mathbf {y} ^ {\mathrm {a d d}}, i | \mathbf {X}) \propto p _ {t} (\operatorname {i n s} (\mathbf {X}, \mathbf {y} ^ {\mathrm {a d d}}, i)) K ^ {\mathrm {d e l}} (i | n + 1).
$$
Diffusion part. Using standard diffusion models arguments such as [18] [45], we get
$$
\overleftarrow {\mathbf {b}} _ {t} ^ {*} (\mathbf {X}) = \overrightarrow {\mathbf {b}} _ {t} (\mathbf {X}) - \overrightarrow {g} _ {t} ^ {2} \nabla_ {\mathbf {x}} \log p _ {t} (\mathbf {X} | n).
$$
Jump part. We use the flux equation from [45] which intuitively relates the probability flow going in the forward direction with the probability flow going the backward direction with equality being achieved at the time reversal.
$$
\begin{array}{l} p _ {t} (\mathbf {X}) \stackrel {\leftarrow} {\lambda} _ {t} ^ {*} (\mathbf {X}) \stackrel {\leftarrow} {K} _ {t} ^ {*} (\mathbf {Y} | \mathbf {X}) = p _ {t} (\mathbf {Y}) \stackrel {\rightarrow} {\lambda} _ {t} (\mathbf {Y}) \stackrel {\rightarrow} {K} _ {t} (\mathbf {X} | \mathbf {Y}) \\ p _ {t} (\mathbf {X}) \stackrel {\leftarrow} {\lambda} _ {t} ^ {*} (\mathbf {X}) \int_ {\mathbf {y} ^ {\text {a d d}}} \sum_ {i = 1} ^ {n + 1} A _ {t} ^ {*} (\mathbf {y} ^ {\text {a d d}}, i | \mathbf {X}) \delta_ {\operatorname {i n s} (\mathbf {X}, \mathbf {y} ^ {\text {a d d}}, i)} (\mathbf {Y}) \mathrm {d} \mathbf {y} ^ {\text {a d d}} \\ = p _ {t} (\mathbf {Y}) \vec {\lambda} _ {t} (\mathbf {Y}) \sum_ {i = 1} ^ {n + 1} K ^ {\mathrm {d e l}} (i | n + 1) \delta_ {\mathrm {d e l} (\mathbf {Y}, i)} (\mathbf {X}). \tag {11} \\ \end{array}
$$
To find  $\overleftarrow{\lambda}_t^* (\mathbf{X})$ , we sum and integrate both sides over  $m$  and  $\mathbf{y}$ , with  $\mathbf{Y} = (m,\mathbf{y})$ ,
$$
\begin{array}{l} \sum_ {m = 1} ^ {N} \int_ {\mathbf {y} \in \mathbb {R} ^ {m d}} p _ {t} (\mathbf {X}) \overleftarrow {\lambda} _ {t} ^ {*} (\mathbf {X}) \int_ {\mathbf {y} ^ {\mathrm {a d d}}} \sum_ {i = 1} ^ {n + 1} A _ {t} ^ {*} (\mathbf {y} ^ {\mathrm {a d d}}, i | \mathbf {X}) \delta_ {\operatorname {i n s} (\mathbf {X}, \mathbf {y} ^ {\mathrm {a d d}}, i)} (\mathbf {Y}) \mathrm {d} \mathbf {y} ^ {\mathrm {a d d}} \mathrm {d} \mathbf {y} \\ = \sum_ {m = 1} ^ {N} \int_ {\mathbf {Y} \in \mathbb {R} ^ {m d}} p _ {t} (\mathbf {Y}) \vec {\lambda} _ {t} (\mathbf {Y}) \sum_ {i = 1} ^ {n + 1} K ^ {\mathrm {d e l}} (i | n + 1) \delta_ {\mathrm {d e l} (\mathbf {Y}, i)} (\mathbf {X}) \mathrm {d} \mathbf {y}. \\ \end{array}
$$
Now we use the fact that  $\delta_{\mathrm{del}(\mathbf{Y},i)}(\mathbf{X})$  is 0 for any  $m\neq n + 1$
$$
\begin{array}{l} p _ {t} (\mathbf {X}) \stackrel {\leftarrow} {\lambda} _ {t} ^ {*} (\mathbf {X}) = \vec {\lambda} _ {t} (n + 1) \int_ {\mathbf {Y} \in \mathbb {R} ^ {(n + 1) d}} p _ {t} (\mathbf {Y}) \sum_ {i = 1} ^ {n + 1} K ^ {\mathrm {d e l}} (i | n + 1) \delta_ {\mathrm {d e l} (\mathbf {Y}, i)} (\mathbf {X}) \mathrm {d} \mathbf {y} \\ = \vec {\lambda} _ {t} (n + 1) \sum_ {i = 1} ^ {n + 1} K ^ {\mathrm {d e l}} (i | n + 1) \int_ {\mathbf {y} \in \mathbb {R} ^ {(n + 1) d}} p _ {t} (\mathbf {Y}) \delta_ {\mathrm {d e l} (\mathbf {Y}, i)} (\mathbf {X}) \mathrm {d} \mathbf {y}. \\ \end{array}
$$
Now letting  $\mathbf{Y} = \mathrm{ins}(\mathbf{X},\mathbf{y}^{\mathrm{add}},i)$
$$
\begin{array}{l} p _ {t} (\mathbf {X}) \stackrel {\leftarrow} {\lambda} _ {t} ^ {*} (\mathbf {X}) = \vec {\lambda} _ {t} (n + 1) \sum_ {i = 1} ^ {n + 1} K ^ {\mathrm {d e l}} (i | n + 1) \int_ {\mathbf {y} ^ {\mathrm {a d d}}} p _ {t} (\operatorname {i n s} (\mathbf {X}, \mathbf {y} ^ {\mathrm {a d d}}, i)) \mathrm {d} \mathbf {y} ^ {\mathrm {a d d}} \\ \overleftarrow {\lambda} _ {t} ^ {*} (\mathbf {X}) x = \overrightarrow {\lambda} _ {t} (n + 1) \frac {\sum_ {i = 1} ^ {n + 1} K ^ {\mathrm {d e l}} (i | n + 1) \int_ {\mathbf {y} ^ {\mathrm {a d d}}} p _ {t} (\operatorname {i n s} (\mathbf {X} , \mathbf {y} ^ {\mathrm {a d d}} , i)) \mathrm {d} \mathbf {y} ^ {\mathrm {a d d}}}{p _ {t} (\mathbf {X})}. \\ \end{array}
$$
To find  $A_{t}^{*}(\mathbf{y}^{\mathrm{add}}, i|\mathbf{X})$ , we start from (11) and set  $\mathbf{Y} = \operatorname{ins}(\mathbf{X}, \mathbf{z}^{\mathrm{add}}, j)$  to get
$$
p _ {t} (\mathbf {X}) \stackrel {\leftarrow} {\lambda} _ {t} ^ {*} (\mathbf {X}) A _ {t} ^ {*} (\mathbf {z} ^ {\mathrm {a d d}}, j | \mathbf {X}) = p _ {t} (\mathbf {Y}) \vec {\lambda} _ {t} (n + 1) K ^ {\mathrm {d e l}} (j | n + 1).
$$
By inspection, we see immediately that
$$
A _ {t} ^ {*} (\mathbf {z} ^ {\mathrm {a d d}}, j | \mathbf {X}) \propto p _ {t} \left(\operatorname {i n s} (\mathbf {X}, \mathbf {z} ^ {\mathrm {a d d}}, j) K ^ {\mathrm {d e l}} (j | n + 1). \right.
$$
With a re-labeling of  $\mathbf{z}^{\mathrm{add}}$  and  $j$  we achieve the desired form
$$
A _ {t} ^ {*} (\mathbf {y} ^ {\text {a d d}}, i | \mathbf {X}) \propto p _ {t} (\operatorname {i n s} (\mathbf {X}, \mathbf {y} ^ {\text {a d d}}, i)) K ^ {\text {d e l}} (i | n + 1).
$$
In this section we prove Proposition 2 using the notation from the main paper by following the framework of [17]. We operate on a state space  $\mathcal{X} = \bigcup_{n=1}^{N}\{n\} \times \mathbb{R}^{nd}$ . On this space the gradient operator  $\nabla : \mathcal{C}(\mathcal{X},\mathbb{R}) \to \mathcal{C}(\mathcal{X},\mathcal{X})$  is defined as  $\nabla f(\mathbf{X}) = \nabla_{\mathbf{x}}^{(nd)}f(\mathbf{X})$  where  $\nabla_{\mathbf{x}}^{(nd)}$  is the standard gradient operator defined as  $\mathcal{C}(\mathbb{R}^{nd},\mathbb{R}) \to \mathcal{C}(\mathbb{R}^{nd},\mathbb{R}^{nd})$  with respect to  $\mathbf{x} \in \mathbb{R}^{nd}$ . We will write integration with respect to a probability measure defined on  $\mathcal{X}$  as an explicit sum over the number of components and integral over  $\mathbb{R}^{nd}$  with respect to a probability density defined on  $\mathbb{R}^{nd}$  i.e.  $\int_{\mathbf{X}} f(\mathbf{X})\mu(\mathrm{d}\mathbf{X}) = \sum_{n=1}^{N}\int_{\mathbf{x} \in \mathbb{R}^{nd}}f(\mathbf{X})p(n)p(\mathbf{x}|n)\mathrm{d}\mathbf{x}$  where, for  $A \subset \mathbb{R}^{nd}$ ,  $\int_{(n,A)}\mu(d\mathbf{X}) = \int_{\mathbf{x} \in A}p(n)p(\mathbf{x}|n)\mathrm{d}\mathbf{x}$ . We will write  $p(\mathbf{X})$  as shorthand for  $p(n)p(\mathbf{x}|n)$ .
Following, [17], we start by augmenting our space with a time variable so that operators become time inhomogeneous on the extended space. We write this as  $\bar{\mathbf{X}} = (\mathbf{X},t)$  where  $\bar{\mathbf{X}}$  lives in the extended space  $S = \mathcal{X}\times \mathbb{R}_{\geq 0}$ . In the proof, we use the infinitesimal generators for the forward and backward processes. An infinitesimal generator is defined as
$$
\mathcal {A} (f) (\bar {\mathbf {X}}) = \lim  _ {t \rightarrow 0} \frac {\mathbb {E} _ {p _ {t | 0} (\bar {\mathbf {Y}} | \bar {\mathbf {X}})} [ f (\bar {\mathbf {Y}}) ] - f (\bar {\mathbf {X}})}{t}
$$
and can be understood as a probabilistic version of a derivative. For our process on the augmented space  $\mathcal{S}$ , our generators decompose as  $\mathcal{A} = \partial_t + \hat{\mathcal{A}}_t$  where  $\hat{\mathcal{A}}_t$  operates only on the spatial components of  $\bar{\mathbf{X}}$  i.e.  $\mathbf{X}$  [17].
We now define the spatial infinitesimal generators for our forward and backward process. We will change our treatment of the time variable compared to the main text. Both our forward and backward processes will run from  $t = 0$  to  $t = T$ , with the true time reversal of  $\mathbf{X}$  following the forward process satisfying  $(\mathbf{Y}_t)_{t\in [0,T]} = (\mathbf{X}_{T - t})_{t\in [0,T]}$ . Further, we will write  $\vec{g}_t$  as  $g_{t}$  and  $\overleftarrow{g}_t = g_{T - t}$  as we do not learn  $g$  and this is the optimal relation from the time reversal. We define
$$
\hat {\mathcal {L}} _ {t} (f) (\mathbf {X}) = \overrightarrow {\mathbf {b}} _ {t} (\mathbf {X}) \cdot \nabla f (\mathbf {X}) + \frac {1}{2} g _ {t} ^ {2} \Delta f (\mathbf {X}) + \overrightarrow {\lambda} _ {t} (\mathbf {X}) \sum_ {m = 1} ^ {N} \int_ {\mathbf {y} \in \mathbb {R} ^ {m d}} f (\mathbf {Y}) (\overrightarrow {K} _ {t} (\mathbf {Y} | \mathbf {X}) - \delta_ {\mathbf {X}} (\mathbf {Y})) \mathrm {d} \mathbf {y},
$$
$$
\hat {\mathcal {K}} _ {t} (f) (\mathbf {X}) = \overleftarrow {\mathbf {b}} _ {t} ^ {\theta} (\mathbf {X}) \cdot \nabla f (\mathbf {X}) + \frac {1}{2} g _ {T - t} ^ {2} \Delta f (\mathbf {X}) + \overleftarrow {\lambda} _ {t} ^ {\theta} (\mathbf {X}) \sum_ {m = 1} ^ {N} \int_ {\mathbf {y} \in \mathbb {R} ^ {m d}} f (\mathbf {Y}) (\overleftarrow {K} _ {t} ^ {\theta} (\mathbf {Y} | \mathbf {X}) - \delta_ {\mathbf {X}} (\mathbf {Y})) \mathrm {d} \mathbf {y}
$$
where  $\Delta = (\nabla \cdot \nabla)$  is the Laplace operator and  $\delta$  is a dirac delta on  $\mathcal{X}$  i.e.  $\sum_{m=1}^{N} \int_{\mathbf{y} \in \mathbb{R}^{md}} \delta_{\mathbf{X}}(\mathbf{Y}) \, \mathrm{d}\mathbf{y} = 1$  and  $\sum_{m=1}^{N} \int_{\mathbf{y} \in \mathbb{R}^{md}} f(\mathbf{Y}) \delta_{\mathbf{X}}(\mathbf{Y}) \, \mathrm{d}\mathbf{y} = f(\mathbf{X})$ .
Verifying Assumption 1. The first step in the proof is to verify Assumption 1 in [17]. Letting  $\nu_{t}(\mathbf{X}) = p_{T - t}(\mathbf{X})$ , we assume we can write  $\partial_t p_t(\mathbf{X}) = \hat{\mathcal{K}}_t^* p_t(\mathbf{X})$  in the form  $\mathcal{M}\nu +c\nu = 0$  for some function  $c:S\to \mathbb{R}$ , where  $\mathcal{M}$  is the generator of another auxiliary process on  $\mathcal{S}$  and  $\hat{\mathcal{K}}_t^*$  is the adjoint operator which satisfies  $\langle \hat{\mathcal{K}}_t^* f,h\rangle = \langle f,\hat{\mathcal{K}}_th\rangle$  i.e.
$$
\sum_ {n = 1} ^ {N} \int_ {\mathbf {x} \in \mathbb {R} ^ {n d}} h (\mathbf {X}) \hat {\mathcal {K}} _ {t} ^ {*} (f) (\mathbf {X}) \mathrm {d} \mathbf {x} = \sum_ {n = 1} ^ {N} \int_ {\mathbf {x} \in \mathbb {R} ^ {n d}} f (\mathbf {X}) \hat {\mathcal {K}} _ {t} (h) (\mathbf {X}) \mathrm {d} \mathbf {x}
$$
We now find  $\hat{\mathcal{K}}_t^*$ . We start by substituting in the form for  $\hat{\mathcal{K}}_t$ ,
$$
\begin{array}{l} \sum_ {n = 1} ^ {N} \int_ {\mathbf {x} \in \mathbb {R} ^ {n d}} f (\mathbf {X}) \hat {\mathcal {K}} _ {t} (h) (\mathbf {X}) \mathrm {d} \mathbf {x} = \sum_ {n = 1} ^ {N} \int_ {\mathbf {x} \in \mathbb {R} ^ {n d}} f (\mathbf {X}) \{(\stackrel {\leftarrow} {\mathbf {b}} _ {t} ^ {\theta} (\mathbf {x}) \cdot \nabla h) (\mathbf {X}) + \frac {1}{2} g _ {T - t} ^ {2} \Delta h (\mathbf {X}) + \\ \stackrel {\leftarrow} {\lambda} _ {t} ^ {\theta} (\mathbf {X}) \sum_ {m = 1} ^ {N} \int_ {\mathbf {y} \in \mathbb {R} ^ {m d}} h (\mathbf {Y}) \left(\stackrel {\leftarrow} {K} _ {t} ^ {\theta} (\mathbf {Y} | \mathbf {X}) - \delta_ {\mathbf {X}} (\mathbf {Y})) \mathrm {d} \mathbf {y} \right\} \mathrm {d} \mathbf {x} \\ \end{array}
$$
We first focus on the RHS terms corresponding to the diffusion part of the process
$$
\begin{array}{l} \sum_ {n = 1} ^ {N} \int_ {\mathbf {x} \in \mathbb {R} ^ {n d}} f (\mathbf {X}) \left\{\left(\overleftarrow {\mathbf {b}} _ {t} ^ {\theta} \cdot \nabla h\right) (\mathbf {X}) + \frac {1}{2} g _ {T - t} ^ {2} \Delta h (\mathbf {X}) \right\} \mathrm {d} \mathbf {x} \\ = \sum_ {n = 1} ^ {N} \int_ {\mathbf {x} \in \mathbb {R} ^ {n d}} f (\mathbf {X}) \left(\overleftarrow {\mathbf {b}} _ {t} ^ {\theta} \cdot \nabla h\right) (\mathbf {X}) + \frac {1}{2} g _ {T - t} ^ {2} f (\mathbf {X}) \nabla \cdot \nabla h (\mathbf {X}) \mathrm {d} \mathbf {x} \\ = \sum_ {n = 1} ^ {N} \int_ {\mathbf {x} \in \mathbb {R} ^ {n d}} f (\mathbf {X}) \left(\overleftarrow {\mathbf {b}} _ {t} ^ {\theta} \cdot \nabla h\right) (\mathbf {X}) + \frac {1}{2} g _ {T - t} ^ {2} h (\mathbf {X}) \nabla \cdot \nabla f (\mathbf {X}) \mathrm {d} \mathbf {x} \\ = \sum_ {n = 1} ^ {N} \int_ {\mathbf {X} \in \mathbb {R} ^ {n d}} - h (\mathbf {X}) \nabla \cdot (f \overleftarrow {\mathbf {b}} _ {t} ^ {\theta}) (\mathbf {X}) + \frac {1}{2} g _ {T - t} ^ {2} h (\mathbf {X}) \nabla \cdot \nabla f (\mathbf {X}) \mathrm {d} \mathbf {x} \\ = \sum_ {n = 1} ^ {N} \int_ {\mathbf {X} \in \mathbb {R} ^ {n d}} h (\mathbf {X}) \left\{- \nabla \cdot \left(f \overleftarrow {\mathbf {b}} _ {t} ^ {\theta}\right) (\mathbf {X}) + \frac {1}{2} g _ {T - t} ^ {2} \nabla \cdot \nabla f (\mathbf {X}) \right\} \mathrm {d} \mathbf {x} \\ = \sum_ {n = 1} ^ {N} \int_ {\mathbf {X} \in \mathbb {R} ^ {n d}} h (\mathbf {X}) \{- f (\mathbf {X}) \nabla \cdot \stackrel {\leftarrow} {\mathbf {b}} _ {t} ^ {\theta} (\mathbf {X}) - \nabla f (\mathbf {X}) \cdot \stackrel {\leftarrow} {\mathbf {b}} _ {t} ^ {\theta} (\mathbf {X}) + \frac {1}{2} g _ {T - t} ^ {2} \nabla \cdot \nabla f (\mathbf {X}) \} \mathrm {d} \mathbf {x}. \\ \end{array}
$$
where we apply integration by parts twice to arrive at the third line and once to arrive at the fourth line. We now focus on the RHS term corresponding to the jump part of the process
$$
\begin{array}{l} \sum_ {n = 1} ^ {N} \int_ {\mathbf {x} \in \mathbb {R} ^ {n d}} f (\mathbf {X}) \left\{\stackrel {\leftarrow} {\lambda} _ {t} ^ {\theta} (\mathbf {X}) \sum_ {m = 1} ^ {N} \int_ {\mathbf {y} \in \mathbb {R} ^ {m d}} h (\mathbf {Y}) \left(\stackrel {\leftarrow} {K} _ {t} ^ {\theta} (\mathbf {Y} | \mathbf {X}) - \delta_ {\mathbf {X}} (\mathbf {Y})\right) \mathrm {d} \mathbf {y} \right\} \mathrm {d} \mathbf {x} \\ = \sum_ {m = 1} ^ {N} \int_ {\mathbf {y} \in \mathbb {R} ^ {m d}} h (\mathbf {Y}) \left\{\sum_ {n = 1} ^ {N} \int_ {\mathbf {x} \in \mathbb {R} ^ {n d}} f (\mathbf {X}) \overleftarrow {\lambda} _ {t} ^ {\theta} (\mathbf {X}) \left(\overleftarrow {K} _ {t} ^ {\theta} (\mathbf {Y} | \mathbf {X}) - \delta_ {\mathbf {X}} (\mathbf {Y})\right) \mathrm {d} \mathbf {x} \right\} \mathrm {d} \mathbf {y} \\ = \sum_ {n = 1} ^ {N} \int_ {\mathbf {x} \in \mathbb {R} ^ {n d}} h (\mathbf {X}) \left\{\sum_ {m = 1} ^ {N} \int_ {\mathbf {y} \in \mathbb {R} ^ {m d}} f (\mathbf {Y}) \stackrel {{\leftarrow}} {\lambda} _ {t} ^ {\theta} (\mathbf {Y}) \left(\stackrel {{\leftarrow}} {K} _ {t} ^ {\theta} (\mathbf {X} | \mathbf {Y}) - \delta_ {\mathbf {Y}} (\mathbf {X})\right) \mathrm {d} \mathbf {y} \right\} \mathrm {d} \mathbf {x}, \\ \end{array}
$$
where on the last line we have relabelled  $\mathbf{X}$  to  $\mathbf{Y}$  and  $\mathbf{Y}$  to  $\mathbf{X}$ . Putting both re-arranged forms for the RHS together, we obtain
$$
\begin{array}{l} \sum_ {n = 1} ^ {N} \int_ {\mathbf {x} \in \mathbb {R} ^ {n d}} h (\mathbf {X}) \hat {\mathcal {K}} _ {t} ^ {*} (f) (\mathbf {X}) \mathrm {d} \mathbf {x} = \\ \sum_ {n = 1} ^ {N} \int_ {\mathbf {X} \in \mathbb {R} ^ {n d}} h (\mathbf {X}) \{- f (\mathbf {X}) \nabla \cdot \overleftarrow {\mathbf {b}} _ {t} ^ {\theta} (\mathbf {X}) - \nabla f (\mathbf {X}) \cdot \overleftarrow {\mathbf {b}} _ {t} ^ {\theta} (\mathbf {X}) + \frac {1}{2} g _ {T - t} ^ {2} \nabla \cdot \nabla f (\mathbf {X}) + \\ \sum_ {m = 1} ^ {N} \int_ {\mathbf {y} \in \mathbb {R} ^ {m d}} f (\mathbf {Y}) \stackrel {{\leftarrow}} {\lambda} _ {t} ^ {\theta} (\mathbf {Y}) \left(\stackrel {{\leftarrow}} {K} _ {t} ^ {\theta} (\mathbf {X} | \mathbf {Y}) - \delta_ {\mathbf {Y}} (\mathbf {X})) \mathrm {d} \mathbf {y} \right\} \mathrm {d} \mathbf {x}. \\ \end{array}
$$
We therefore have
$$
\begin{array}{l} \hat {\mathcal {K}} _ {t} ^ {*} (f) (\mathbf {X}) = - f (\mathbf {X}) \nabla \cdot \overleftarrow {\mathbf {b}} _ {t} ^ {\theta} (\mathbf {X}) - \nabla f (\mathbf {X}) \cdot \overleftarrow {\mathbf {b}} _ {t} ^ {\theta} (\mathbf {X}) + \frac {1}{2} g _ {T - t} ^ {2} \Delta f (\mathbf {X}) + \\ \sum_ {m = 1} ^ {N} \int_ {\mathbf {y} \in \mathbb {R} ^ {m d}} f (\mathbf {Y}) \stackrel {{\leftarrow}} {\lambda} _ {t} ^ {\theta} (\mathbf {Y}) \bigl (\stackrel {{\leftarrow}} {K} _ {t} ^ {\theta} (\mathbf {X} | \mathbf {Y}) - \delta_ {\mathbf {X}} (\mathbf {Y}) \bigr) \mathrm {d} \mathbf {y}. \\ \end{array}
$$
Now we re-write  $\partial_t p_t(\mathbf{X}) = \hat{\mathcal{K}}_t^* p_t(\mathbf{x})$  in the form  $\mathcal{M}\nu +c\nu = 0$ . We start by re-arranging
$$
\partial_ {t} p _ {t} (\mathbf {X}) = \hat {\mathcal {K}} _ {t} ^ {*} p _ {t} (\mathbf {X}) \Longrightarrow 0 = \partial_ {t} \nu_ {t} (\mathbf {X}) + \hat {\mathcal {K}} _ {T - t} ^ {*} \nu_ {t} (\mathbf {X}).
$$
Substituting in our form for  $\hat{\mathcal{K}}_t^*$  we obtain
$$
\begin{array}{l} 0 = \partial_ {t} \nu_ {t} (\mathbf {X}) - \nu_ {t} (\mathbf {X}) \nabla \cdot \overleftarrow {\mathbf {b}} _ {T - t} ^ {\theta} (\mathbf {X}) - \overleftarrow {\mathbf {b}} _ {T - t} ^ {\theta} (\mathbf {X}) \cdot \nabla \nu_ {t} (\mathbf {X}) + \frac {1}{2} g _ {t} ^ {2} \Delta \nu_ {t} (\mathbf {X}) \\ + \sum_ {m = 1} ^ {N} \int_ {\mathbf {y} \in \mathbb {R} ^ {m d}} \nu_ {t} (\mathbf {Y}) \overleftarrow {\lambda} _ {T - t} ^ {\theta} (\mathbf {Y}) \left(\overleftarrow {K} _ {T - t} ^ {\theta} (\mathbf {X} | \mathbf {Y}) - \delta_ {\mathbf {X}} (\mathbf {Y})\right) \mathrm {d} \mathbf {y} \tag {12} \\ \end{array}
$$
We define our auxiliary process to have generator  $\mathcal{M} = \partial_t + \hat{\mathcal{M}}_t$  with
$$
\hat {\mathcal {M}} _ {t} (f) (\mathbf {X}) = \mathbf {b} _ {t} ^ {M} (\mathbf {X}) \cdot \nabla f (\mathbf {X}) + \frac {1}{2} g _ {t} ^ {2} \Delta f (\mathbf {X}) + \lambda_ {t} ^ {M} (\mathbf {X}) \sum_ {m = 1} ^ {N} \int_ {\mathbf {Y} \in \mathbb {R} ^ {m d}} f (\mathbf {Y}) (K _ {t} ^ {M} (\mathbf {Y} | \mathbf {X}) - \delta_ {\mathbf {X}} (\mathbf {Y})) \mathrm {d} \mathbf {y}
$$
which is a jump diffusion process with drift  $\mathbf{b}_t^M$ , diffusion coefficient  $g_{t}$ , rate  $\lambda_t^M$  and transition kernel  $K_{t}^{M}$ . Then if we have  $\mathbf{b}_t^M = -\stackrel{\leftarrow}{\mathbf{b}}_T^\theta_{T-t}$ ,
$$
\lambda_ {t} ^ {M} (\mathbf {X}) = \sum_ {m = 1} ^ {N} \int_ {\mathbf {y} \in \mathbb {R} ^ {m d}} \stackrel {\leftarrow} {\lambda} _ {T - t} ^ {\theta} (\mathbf {Y}) \stackrel {\leftarrow} {K} _ {T - t} ^ {\theta} (\mathbf {X} | \mathbf {Y}) \mathrm {d} \mathbf {y}, \tag {13}
$$
$$
K _ {t} ^ {M} (\mathbf {Y} | \mathbf {X}) \propto \overleftarrow {\lambda} _ {T - t} ^ {\theta} (\mathbf {Y}) \overleftarrow {K} _ {T - t} ^ {\theta} (\mathbf {X} | \mathbf {Y}). \tag {14}
$$
Then we have (12) can be rewritten as
$$
\begin{array}{l} 0 = \partial_ {t} \nu_ {t} (\mathbf {X}) - \nu_ {t} (\mathbf {X}) \nabla \cdot \overleftarrow {\mathbf {b}} _ {T - t} ^ {\theta} (\mathbf {X}) + \mathbf {b} _ {t} ^ {M} (\mathbf {X}) \cdot \nabla \nu_ {t} (\mathbf {X}) + \frac {1}{2} g _ {t} ^ {2} \Delta \nu_ {t} (\mathbf {X}) \\ - \nu_ {t} (\mathbf {X}) \stackrel {\leftarrow} {\lambda} _ {T - t} ^ {\theta} (\mathbf {X}) + \nu_ {t} (\mathbf {X}) \sum_ {m = 1} ^ {N} \int_ {\mathbf {Y} \in \mathbb {R} ^ {m d}} \stackrel {\leftarrow} {\lambda} _ {T - t} ^ {\theta} (\mathbf {Y}) \stackrel {\leftarrow} {K} _ {T - t} ^ {\theta} (\mathbf {X} | \mathbf {Y}) \mathrm {d} \mathbf {y} + \\ \lambda_ {t} ^ {M} (\mathbf {x}) \sum_ {m = 1} ^ {N} \int_ {\mathbf {y} \in \mathbb {R} ^ {m d}} \nu_ {t} (\mathbf {Y}) \left(K _ {t} ^ {M} (\mathbf {Y} | \mathbf {X}) - \delta_ {\mathbf {X}} (\mathbf {Y})\right) \mathrm {d} \mathbf {y} \\ \end{array}
$$
which is in the form  $\mathcal{M}(\nu)(\bar{\mathbf{X}}) + c(\bar{\mathbf{X}})\nu (\bar{\mathbf{X}}) = 0$  if we let
$$
c (\bar {\mathbf {X}}) = - \nabla \cdot \stackrel {\leftarrow} {\mathbf {b}} _ {T - t} ^ {\theta} (\mathbf {X}) - \stackrel {\leftarrow} {\lambda} _ {T - t} ^ {\theta} (\mathbf {X}) + \sum_ {m = 1} ^ {N} \int_ {\mathbf {y} \in \mathbb {R} ^ {m d}} \stackrel {\leftarrow} {\lambda} _ {T - t} ^ {\theta} (\mathbf {Y}) \stackrel {\leftarrow} {K} _ {T - t} ^ {\theta} (\mathbf {X} | \mathbf {Y}) \mathrm {d} \mathbf {y}.
$$
Verifying Assumption 2. Now that we have verified Assumption 1, the second step in the proof is Assumption 2 from [17]. We assume there is a bounded measurable function  $\alpha : S \to (0, \infty)$  such
that  $\alpha \mathcal{M}f = \mathcal{L}(f\alpha) - f\mathcal{L}\alpha$  for all functions  $f:\mathcal{X}\to \mathbb{R}$  such that  $f\in \mathcal{D}(\mathcal{M})$  and  $f\alpha \in \mathcal{D}(\mathcal{L})$ . Substituting in  $\mathcal{M}$  and  $\mathcal{L}$  we get
$$
\begin{array}{l} \alpha_ {t} (\mathbf {X}) [ \partial_ {t} f (\mathbf {X}) - \overleftarrow {\mathbf {b}} _ {T - t} ^ {\theta} (\mathbf {X}) \cdot \nabla f (\mathbf {X}) \\ + \frac {1}{2} g _ {t} ^ {2} \Delta f (\mathbf {X}) + \lambda_ {t} ^ {M} (\mathbf {X}) \sum_ {m = 1} ^ {N} \int_ {\mathbf {y} \in \mathbb {R} ^ {m d}} f (\mathbf {Y}) (K _ {t} ^ {M} (\mathbf {Y} | \mathbf {X}) - \delta_ {\mathbf {X}} (\mathbf {Y})) \mathrm {d} \mathbf {y} ] \\ = \partial_ {t} (f \alpha_ {t}) (\mathbf {X}) + \overrightarrow {\mathbf {b}} _ {t} (\mathbf {X}) \cdot \nabla (f \alpha_ {t}) (\mathbf {X}) + \frac {1}{2} g _ {t} ^ {2} \Delta (f \alpha_ {t}) (\mathbf {X}) \\ + \vec {\lambda} _ {t} (\mathbf {X}) \sum_ {m = 1} ^ {N} \int_ {\mathbf {y} \in \mathbb {R} ^ {m d}} f (\mathbf {Y}) \alpha_ {t} (\mathbf {Y}) \left(\vec {K} _ {t} (\mathbf {Y} | \mathbf {X}) - \delta_ {\mathbf {X}} (\mathbf {Y})\right) \mathrm {d} \mathbf {y} \\ - f (\mathbf {X}) [ \partial_ {t} \alpha_ {t} (\mathbf {X}) + \overrightarrow {\mathbf {b}} _ {t} (\mathbf {X}) \cdot \nabla \alpha_ {t} (\mathbf {X}) + \frac {1}{2} g _ {t} ^ {2} \Delta \alpha_ {t} (\mathbf {X}) \\ + \vec {\lambda} _ {t} (\mathbf {X}) \sum_ {m = 1} ^ {N} \int_ {\mathbf {y} \in \mathbb {R} ^ {m d}} \alpha_ {t} (\mathbf {Y}) \left(\vec {K} _ {t} (\mathbf {Y} | \mathbf {X}) - \delta_ {\mathbf {X}} (\mathbf {Y})\right) \mathrm {d} \mathbf {y} ] \tag {15} \\ \end{array}
$$
Since  $f$  does not depend on time,  $\partial_t f(\mathbf{X}) = 0$  and  $\partial_t (f\alpha_t)(\mathbf{X}) = f(\mathbf{X})\partial_t\alpha_t(\mathbf{X})$  thus the  $\partial_t$  terms on the RHS also cancel out. Comparing terms on the LHS and RHS relating to the diffusion part of the process we obtain
$$
- \alpha_ {t} (\mathbf {X}) \left(\overleftarrow {\mathbf {b}} _ {T - t} ^ {\theta} (\mathbf {X}) \cdot \nabla f (\mathbf {X})\right) + \frac {1}{2} \alpha_ {t} (\mathbf {X}) g _ {t} ^ {2} \Delta f (\mathbf {X}) =
$$
$$
\overrightarrow {\mathbf {b}} _ {t} (\mathbf {X}) \cdot \nabla (f \alpha_ {t}) (\mathbf {X}) + \frac {1}{2} g _ {t} ^ {2} \Delta (f \alpha_ {t}) (\mathbf {X}) - f (\mathbf {X}) \overrightarrow {\mathbf {b}} _ {t} (\mathbf {X}) \cdot \nabla \alpha_ {t} (\mathbf {X}) - \frac {1}{2} f (\mathbf {X}) g _ {t} ^ {2} \Delta \alpha_ {t} (\mathbf {X}).
$$