Static Widom-Rowlinson model In Elena Pulvirenti's slides she introduced a $\textbf{static Widom-Rowlinson model of one species}$. Consider $\Lambda\subset R^2$ with periodic boundary conditions, $\Lambda$ set of particle configurations with
$$\Gamma=\{\gamma\subset \Lambda: N(\gamma)\in \mathbb{N}\}, \, $$
where $N(\gamma)$ is the cardinality of $\gamma$. The halo of a configuration is $h(\gamma)=\cup_{x\in \gamma}B_2(x)$ with radius $2$. Let $H(\gamma)=|h(\gamma)|-N(\gamma)|B_2(0)|$ be the Hamiltonian.
Define the grand-canonical Gibbs measure,
$$(1) \mu(d\gamma)=\frac{z^{N(\gamma)}}{\Xi}e^{-\beta H(\gamma)}\mathbb{Q}(d\gamma) $$
where $\mathbb{Q}$ is Poisson point process with intensity 1 and $\Xi$ is the partition function.
Her result is that the 2-species Widom-Rowlinson model is equivalent to 1-species. $\textbf{The 2-species WR model}$ is two types of particles(blue and red) with configurations $\gamma^B, \gamma^R$. The grand-canonical Gibbs measure:
$$(2) \hat{\mu}(d\gamma^R, d\gamma^B)=\frac{1}{\hat{\Xi}}1_{\{\text{red-blue hard-core}\}}z_R^{N(\gamma^R)}z_B^{N(\gamma^B)}\mathbb{Q}(d\gamma^R)\mathbb{Q}(d\gamma^B)$$
where $1_{\{\text{red-blue hard-core}\}}$ means it is $1$ if $d(\gamma^R, \gamma^B)\geq 1$, otherwise is 0, and $z_R=e^{\beta\lambda_R}$ and $z_B=e^{\beta\lambda_B}$.

$\textbf{My question is why 1-species and 2-species are equivalence?}$ I am confused about that fix the centers of the red discs and integrate over the centers of the blue disc, then:
$$\frac{1}{\hat{\Xi}}\int_{\Gamma} 1_{\{\text{red-blue hard-core}\}}z_R^{N(\gamma^R)}z_B^{N(\gamma^B)}\mathbb{Q}(d\gamma^B)=C \frac{z^{N(\gamma^R)}}{\Xi}e^{-\beta H(\gamma^R)}$$
where $(z_B, z_R)\to (\beta, ze^{\beta V_0})$ and $V_0:=|B_2(0)|$.


 A: So let's think about it this way: if $A$ is an event in the two-type model depending only on red then
\begin{align*}
\mathbb{P}(A) &= \frac{1}{\tilde{\Xi}} \sum_{j,k\geq 0} \frac{z_R^k}{k!} \frac{z_B^j}{j!} \int_{S^k} \int_{S^j} 1_{A} \cdot 1_{RBHC} \,dy\, dx \\
&=\frac{1}{\tilde{\Xi}} \sum_{k\geq0}\frac{z_R^k}{k!} \int_{S^k} 1_A \left(\sum_{j \geq 0} \frac{z_B^j}{j!}\int_{S^j} 1_{RBHC} \,dy \right)\,dx \\
&= \frac{1}{\tilde{\Xi}} \sum_{k\geq0}\frac{z_R^k}{k!} \int_{S^k} 1_A \left(\sum_{j \geq 0} \frac{z_B^j}{j!}\int_{(S \setminus h(\gamma_R))^j}  \,dy \right)\,dx \\
&= \frac{1}{\tilde{\Xi}} \sum_{k\geq0}\frac{z_R^k}{k!} \int_{S^k} 1_A \left(\sum_{j \geq 0} \frac{z_B^j}{j!}(|S| - |h(\gamma_R)|)^j  \right)\,dx \\
&= \frac{1}{\tilde{\Xi}} \sum_{k\geq0}\frac{z_R^k}{k!} \int_{S^k} 1_A \exp(z_B |S| - z_B |h(\gamma_R)|)\,dx \\
&= \frac{C}{\tilde{\Xi}} \sum_{k\geq0}\frac{z_R^k}{k!} \int_{S^k} 1_A e^{- z_B |h(\gamma_R)|)}\,dx \\
&= \frac{C}{\tilde{\Xi}} \sum_{k\geq0}\frac{(z_R e^{-z_BV_0})^k}{k!} \int_{S^k} 1_A e^{- z_B (|h(\gamma_R)| - k V_0))}\,dx \,.
\end{align*}
Doing the change of variables listed completes it.
