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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.

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1 Answer 1

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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 $$1 \wedge \dfrac{\pi(x^\star,y^\star,z^\star)}{\pi(x^-,y^-,z^-)}\times \dfrac{p(x^-,y^-,z^-)}{p(x^\star,y^\star,z^\star)}$$

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    $\begingroup$ Well that is simple enough. Thank you. I did not consider conditioning like you did. Thanks for your help, Xi'an! $\endgroup$
    – Oski
    Commented Mar 4, 2022 at 14:50
  • $\begingroup$ If we are allowed to calculate the acceptance probability as a 3 by 1 vector in R (same log likelihood but three different prior values), can we compared it with three random numbers from U~[0,1] and accept part of elements and reject the rest? Is this still a valid MH algorithm? $\endgroup$
    – Keith Lau
    Commented Feb 9, 2023 at 9:26
  • $\begingroup$ @KeithLau: Please ask a new question with all details included as I find your question too vague to reply. $\endgroup$
    – Xi'an
    Commented Feb 9, 2023 at 12:11

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