Do the following normalizing constants cancel out in Reversible Jump ratio?

Question

We know that a Strauss Point Process has density

$$p(x_{1}, x_{2},..., x_{K})\propto \prod_{i=1}^{K}\phi(x_{i};\theta)\prod_{1\leq i\leq j \leq K}a^{1(\left | x_{i}-x_{j} \right |\leq \delta)}$$

where $\phi$ is a positive function, and $a\in(0,1)$ and $\delta \in (0,\infty)$.

Suppose that we have the Strauss Point Process defined above for $K$ points, then the normalizing constant is computed as

$$Z_{\theta}=\int \prod_{i=1}^{K}\phi(x_{i};\theta)\prod_{1\leq i\leq j \leq K}a^{1(\left | x_{i}-x_{j} \right |\leq \delta)} dx_{1}...dx_{K}$$

Now let's assume that we have a second Strauss Point Process but this time with $K+1$ points, i.e. the same $K$ points as in the previous point process but we also have one additional $x_{K+1}$.

$$p(x_{1}, x_{2},..., x_{K}, x_{K+1})\propto \prod_{i=1}^{K+1}\phi(x_{i};\theta)\prod_{1\leq i\leq j \leq K+1}a^{1(\left | x_{i}-x_{j} \right |\leq \delta)}$$

This process has normalizing constant

$$Z_{\theta}^{'}=\int \prod_{i=1}^{K+1}\phi(x_{i};\theta)\prod_{1\leq i\leq j \leq K+1}a^{1(\left | x_{i}-x_{j} \right |\leq \delta)} dx_{1}...dx_{K+1}$$

Now if we are interested in calculating a Reversible Jump acceptance ratio, which includes the following ratio

$$\frac{p(x_{1}, x_{2},..., x_{K}, x_{K+1})}{p(x_{1}, x_{2},..., x_{K})} = \frac{\frac{\prod_{i=1}^{K+1}\phi(x_{i};\theta)\prod_{1\leq i\leq j \leq K+1}a^{1(\left | x_{i}-x_{j} \right |\leq \delta)}}{Z_{\theta}^{'}}}{\frac{\prod_{i=1}^{K}\phi(x_{i};\theta)\prod_{1\leq i\leq j \leq K}a^{1(\left | x_{i}-x_{j} \right |\leq \delta)}}{Z_{\theta}}}$$ ideally we would like the normalizing constants $Z_{\theta}$ and $Z_{\theta}^{'}$ to cancel out, but are they actually equal? Since the one depends on $K$ points and the other one on $K+1$. Am I calculating something wrong?

Answer 1

$\require{graphicx}$The full model is not defined in the question, as it should include $K$ in the target. For instance, when $$p(K,x_{1}, x_{2},..., x_{K})\propto \prod_{i=1}^{K}\phi(x_{i};\theta)\prod_{1\leq i\leq j \leq K}a^{1(\left | x_{i}-x_{j} \right |\leq \delta)}$$ with a "flat prior" on $K$, the proportionality constant is $$Z_\theta=\sum_K \int \prod_{i=1}^{K}\phi(x_{i};\theta)\prod_{1\leq i\leq j \leq K}a^{1(\left | x_{i}-x_{j} \right |\leq \delta)} \text dx_{1}...\text dx_{K}$$ and the $K$ dependent normalising constants $$Z^K_\theta = \int \prod_{i=1}^{K}\phi(x_{i};\theta)\prod_{1\leq i\leq j \leq K}a^{1(\left | x_{i}-x_{j} \right |\leq \delta)} \text dx_{1}...\text dx_{K}$$ do not appear in the reversible jump ratio. If in the opposite $$p(K,x_{1}, x_{2},..., x_{K}) \underbrace{=}_{\text{normalised}}\frac{\varrho_K}{Z^K_\theta}\prod_{i=1}^{K}\phi(x_{i};\theta)\prod_{1\leq i\leq j \leq K}a^{1(\left | x_{i}-x_{j} \right |\leq \delta)}$$ with a prior $\pi(K)=\varrho_K$ on $K$, then the $K$ dependent normalising constants $Z^K_\theta$ will be involved in the reversible jump ratio, unless an exchange algorithm à la Murray et al. (2006) is introduced.