# How can I find the acceptance probability for a joint Metropolis-Hastings proposal?

Suppose that I want to generate a proposal $$(x^*,y^*,z^*)$$ with the following:

$$z^*\sim p(z|\alpha,\beta)$$ $$x^*\sim p(x|z^*,\boldsymbol{\gamma}_x)$$ $$y^*\sim p(y|z^*,\boldsymbol{\gamma}_y),$$ where $$\alpha,\beta,\boldsymbol{\gamma}_x,\boldsymbol{\gamma}_y$$ are known hyperparameters. Note the dependence of $$x^*$$ and $$y^*$$ on $$z^*$$. How can I appropriately write the acceptance probability for the proposed Metropolis-Hastings sampling scheme?

My attempt:

I know under the single-proposal case, we can write the acceptance probability $$\alpha$$ for a proposed $$\theta$$ as

$$\alpha=\min\left\{1,\frac{p(\theta^*|\cdot)p(\theta|\theta^*)}{p(\theta|\cdot)p(\theta^*|\theta)}\right\},$$

where $$p(\theta|\cdot)$$ denotes the posterior for $$\theta$$ and $$p(\theta|\theta^*)$$ denotes the transition probability from $$\theta^*$$ to $$\theta$$. This leads me to believe that we can write the joint acceptance probability of the proposed scheme as

$$\alpha=\min\left\{1,\frac{p(z^*,x^*,y^*|\cdot)p(z|z^*)p(x|x^*)p(y|y^*)}{p(z,x,y|\cdot)p(z^*|z)p(x^*|x)p(y^*|y)}\right\}$$

Is there a way to simplify the expression of this acceptance probability further? Thank you.

This is a single-step (independent) proposal, namely generating simultaneously $$(X^\star,Y^\star,Z^\star)$$ from the joint proposal with density $$p(x^\star,y^\star,z^\star)= p(z^\star|\alpha,\beta)p(x^\star|z^\star, \boldsymbol{\gamma}_x)p(y|z^\star,\boldsymbol{\gamma}_y)$$ Therefore the acceptance probability to move from $$(x^-,y^-,z^-)$$ to $$(x^\star,y^\star,z^\star)$$ in this independent Metropolis-Hastings algorithm is $$1 \wedge \dfrac{\pi(x^\star,y^\star,z^\star)}{\pi(x^-,y^-,z^-)}\times \dfrac{p(x^-,y^-,z^-)}{p(x^\star,y^\star,z^\star)}$$