Exercise
Having two homogeneous and independent Poisson point processes $\Phi_3, \Phi_2$ defined in $\mathbb{R}^2$ with intensities $\lambda_3, \lambda_2$, respectively. Having a Voronoi tessellation with cells centered in the points generated by $\Phi_3$, and defining $V_x(\Phi_3)$ as a Voronoi cell centered in $x\in \Phi_3$. Show that:
$$ \mathbb{E}_{\Phi_3}^o \sum_{z \in \Phi_2 \cap V_o(\Phi_3)} \|z\| = \frac{\lambda_2}{\lambda_3} \mathbb{E}_{\Phi_2}^o \|z_o\| \tag 1 $$
where $z_o$ is the point of the point process $\Phi_2$ nearest to the origin $o$, and $\mathbb{E}_{\Phi_3}^o$ denotes the Palm expectation respect point process $\Phi_3$.
What I've tried
First of all I've expressed the LHS of (1) as $\mathbb{E}_{\Phi_3}^o \sum_{z \in \Phi_2} \|z\| \cdot _{V_o(\Phi_3)}(z) $. The I reformulate that as $\mathbb{E}_{\Phi_3}^o \sum_{z \in \Phi_2}f(z)$ and use Neveu's exchange formula to get:
$$ \frac{\lambda_2}{\lambda_3} \mathbb{E}_{\Phi_2}^o \sum_{t \in \Phi_3} f(-t) = \frac{\lambda_2}{\lambda_3} \mathbb{E}_{\Phi_2}^o \sum_{t \in \Phi_3} \|t\| \cdot _{V_o(\Phi_3)}(-t) \tag 2 $$
Since the Voronoi cell $V_o(\Phi_3)$ only has one point $t \in \Phi_3$ in it, and that is precisely $o$, (2) turns out to be:
$$\frac{\lambda_2}{\lambda_3} \mathbb{E}_{\Phi_2}^o \|o\| $$
but clearly $\|o\|$ is not the same as $z_o$. I think I'm not using correctly Neveu's exchange formula, which I studied at the "Stochastic and Integral Geometry" book of Schneider and Weil (Theorem 3.4.5).
Note
I'm trying to understand the demonstration of the cost formula of the article "Are Heterogeneous Cloud-Based Radio Access Networks Cost Effective?".
