# Stochastic integration by parts for random point processes

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 !

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 $$(\star \star)$$.