Recall from last time:
Today: Iversion formula & consequences.
Let \(\Phi\) be a stationary point process on \(\mathbb{R}^d\) with intensity \(\lambda>0\).
The typical cell \(Z\) of \(V_\Phi\) gives the following inversion formula, allowing us to obtain the distribution of \(\Phi\) from the Palm version \(\Phi^0\).
Theorem (Inversion Formula). For all measurable \(f : \mathcal{M}(\mathbb{R}^d) \to \mathbb{R}_{+}\), \[ \mathbb{E}\bigl[f(\Phi)\bigr] = \lambda \,\mathbb{E}\!\left[ \int_{\mathbb{R}^d} f\bigl(t_x \Phi^0\bigr)\, \mathbf{1}_{\{x \in Z\}}\,dx \right], \] where \(Z := V(0,\Phi^0)\) is the typical cell of \(V_{\Phi}\), and \(t_x \mu(B) := \mu(B+x)\).
Some Consequences:
Define \(\tau : \mathbb{R}^d \times \mathcal{M}(\mathbb{R}^d) \to \mathbb{R}^d\) by letting
\(\tau(x,\eta)\) be the point in \(\mathrm{supp}(\eta)\) that is closest to \(x\). Let \(\Phi\) be a stationary point process on \(\mathbb{R}^d\) with intensity \(\lambda>0\).
Also define \(Y := \tau(0,\Phi)\).
Theorem. For all measurable \(h : \mathbb{R}^d \times \mathcal{M}(\mathbb{R}^d) \to \mathbb{R}_{+}\), \[ \mathbb{E}\bigl[h(Y,\Phi)\bigr] = \lambda\,\mathbb{E}\!\left[ \int_{\mathbb{R}^d} h\!\left(-x,\,t_x \Phi^0\right) \mathbf{1}_{\{x \in Z\}}\,dx \right], \] and the inversion formula follows from letting \(h(x,\eta)=f(\eta)\).
Proof. Recall refined Campbell’s theorem together with a change of variables:
\[\begin{equation} \quad \mathbb{E}\!\left[\int_{\mathbb{R}^d} f(x,\Phi)\,\Phi(dx)\right] = \lambda\,\mathbb{E}\!\left[\int_{\mathbb{R}^d} f(-x,\,t_x \Phi^0)\,dx\right]. \end{equation}\]
Now let $ f(x,) := h(x,),_{{(0,)=x}}. $
The LHS of Eq.~[ref] equals:
\[\begin{align*} \mathbb{E}\!\left[ \int_{\mathbb{R}^d} h(x,\Phi)\, \mathbf{1}_{\{\tau(0,\Phi)=x\}}\, \Phi(dx) \right] &= \mathbb{E}\bigl[h(Y,\Phi)\bigr]. \end{align*}\]
The RHS of Eq.~[ref] equals:
\[\begin{align*} \quad \lambda\,\mathbb{E}\!\left[ \int_{\mathbb{R}^d} h(-x,\,t_x \Phi^0)\, \mathbf{1}_{\{\tau(0,t_x \Phi^0)=-x\}} \,dx \right]. \end{align*}\]
We note that \(\tau\) satisfies the following property: \[ \tau(x-y,\,t_y \mu) = \tau(x,\mu)-y. \]
Letting \(y = x\), we have \(\tau(0,t_x\Phi^0)=\tau(x,\Phi^0)-x\). Then [ (0,t_x^0) = -x (x,^0)=0. ]
Therefore,
\[\begin{align*} \text{RHS}&= \lambda\,\mathbb{E}\!\left[ \int_{\mathbb{R}^d} h(-x,\,t_x \Phi^0)\, \mathbf{1}_{\{\tau(x,\Phi^0)=0\}} \,dx \right] \\ &= \lambda\,\mathbb{E}\!\left[ \int_{\mathbb{R}^d} h(-x,\,t_x \Phi^0)\, \mathbf{1}_{\{x \in Z\}} \,dx \right]. \end{align*}\]
\(\square\)
Let \(h(x,\mu) := g(t_x \mu)\).
Then $ h!(-x,,t_x ^0) = g!(t_{-x}t_x ^0) = g(^0), $ and
\[\begin{equation} \mathbb{E}\bigl[g(t_Y \Phi)\bigr]=\lambda\,\mathbb{E}\!\left[\,|Z|\,g(\Phi^0)\right]. \end{equation}\]
This shows that the distribution of \(t_Y \Phi\) is absolutely continuous with respect to the Palm distribution.
Further, letting \(g\) in Eq.~[ref] satisfy $ g() := f(),|Z_{0}()|^{-1}, $ gives \[ \lambda\,\mathbb{E}\!\left[f(\Phi^0)\right] = \mathbb{E}\!\left[ |Z_{0}(t_Y \Phi)|^{-1}\, f(t_Y \Phi) \right]. \]
Since $ Z_{0}(t_Y ) = Z_{0}() - Y, $ we obtain the following.
Proposition. \[ \lambda\,\mathbb{E}\!\left[f(\Phi^0)\right] = \mathbb{E}\!\left[ |Z_{0}(\Phi)|^{-1}\,f(t_Y \Phi) \right]. \]
This gives, for $ f() := |Z_{0}()|^{+1}, $ the more genral formula \[ \lambda\,\mathbb{E}\!\left[|Z|^{\alpha+1}\right] = \mathbb{E}\!\left[|Z_{0}|^{\alpha}\right]. \]