Stochastic integration by parts for random point processes

Question

I'm trying to understand this proof of the following specifing integration by parts.

Introduction

Let $\Omega=Point_{\mathbb{R}}$ the set of point distributions in $\mathbb{R}^3$ (i.e an element $w \in \Omega$ is a locally finite subset of $\mathbb{R}^3$). We equip this space with a canonic tribe $\mathcal{A}$ and a probability $\mathbb{P}$.

Let $\phi : \Omega \rightarrow \mathbb{R}^3$ a measurable function. For $w \in \Omega$, we define the realization of $\phi$ the mapping $R_\omega[\phi] : \mathbb{R}^3 \rightarrow \mathbb{R}^3$ :

$$R_\omega[\phi](y):=\phi(y+w)=\phi(\tau_yw), \quad y \in\mathbb{R}^3$$ where $\tau_y :\Omega \rightarrow \Omega$ is the shift application (which is measurable in this case).

Let $\varphi : \Omega \rightarrow\mathbb{R^3}$ a smooth function and $w \in \Omega$, we define the stochastic gradient by $$\bar{\nabla}(\varphi)(\omega):=\nabla (t \mapsto \varphi(t+w))(0).$$ where $\nabla$ is the regular spatial gradient. Using this definition, we define the partial derivative $\bar{\partial}_i$.

the integration by parts

Let $u,v$ smooths functions from $\Omega$ to $\mathbb{R}^3$. According to the text I'm reading, we have the following integration by parts : $$\mathbb{E} \left[ \bar{\partial}_iu v\right]=-\mathbb{E} \left[ u \bar{\partial}_i v\right]$$ The proof starts with the following equality, that I'm strugling to understand : \begin{aligned} \mathbb{E} \left[\bar{\partial}_iu v\right]= \mathbb{E} \left[\int_{K_1} \partial_i R_\omega[u](y) \ R_\omega[v](y) \ \mathrm{d}y\right] \quad \quad (\star) \end{aligned} with $K_1=\left[-\frac{1}{2},\frac{1}{2}\right]^3$. Then, we can use the regular integration by parts formula for the $K_1$ integral, which gives us :

$$- \mathbb{E} \left[\int_{K_1} R_\omega[u](y) \ \partial_i R_\omega[v](y) \ \mathrm{d}y\right]+ \underbrace{\mathbb{E} \left[\int_{ \partial K_1} n_i R_\omega[u](y) \ R_\omega[v](y) \ \mathrm{d}y \right]}_{:=A}$$ and then it is said that $A=0$. The rest of the proof follows easily using $(\star)$ again.

My questions

1. Where does the first equality $(\star)$ comes from ?
2. Why do we have $A=0$ ? In the usual case it is because the test functions have a their support include in a compact but we do not have such hypothesis here.

Any helps or hint are welcomed !

Answer 1

Actually it is necessary to add one hypothesis so that the result holds.

We need to assume that the application shift $\tau_y : \Omega \rightarrow \Omega$ is $\mathbb{P}$ preserving for any vector $y \in \mathbb{R}^3$. Using this stationarity hypothesis, we see that :

$$\mathbb{E}\left[ \bar{\partial}_i v u \right]= \mathbb{E}\left[ \bar{\partial}_i (v \circ \tau_y) \ (u \circ \tau_y) \right], \quad \forall y \in \mathbb{R}^3. $$

Integrating this equality over $y$ on the domain $K_1$ which has a Lebesgue measure equals to 1, we get by Fubini's theorem the equality $(\star)$.

A similar argument on the boundary integral leaves us with $A=\mathbb{E}[uv] \int_{\partial K_1} n_i$, and it is well known that the latter integral worths $0$, hence $(\star \star)$.