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I wondered if someone could confirm if this is correct. The probability of accepting a sample from the proposal distribution $q(z)$ is given by the ratio $\frac{\tilde {p}(z)}{kq(z)}$ where $\tilde {p}(z)$ is the unnormalised true posterior and $k$ the scaling factor. This itself is not the acceptance probability (although I thought it was). It seems the acceptance probability is $\frac{1}{k}$ which is rationalised as follows:

$\int \frac{\tilde {p}(z)}{kq(z)} q(z) dz = \frac{1}{k}$ which seems that we take an expectation over this acceptance probability to get an average over all terms? Is this correct?

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Both quantities are acceptance probabilities, assuming of course that the dominating inequality$$\sup_{z\in\mathfrak X}\dfrac{\tilde {p}(z)}{kq(z)}\le 1$$holds.

First, the quantity$$\dfrac{\tilde {p}(z)}{kq(z)}$$is the conditional probability of acceptance given that $z$ is the outcome of a (marginal) simulation from $q(\cdot)$. Defining the indicator random variable $$I=\mathbb I_{U\le {\tilde {p}(z)}/{kq(z)}}$$ when $U\sim \mathfrak U(0,1)$, $I\in\{0,1\}$ and is therefore a Bernoulli random variable. Conditionally on $Z=z$, when $Z\sim q(z)$, $I$ is a Bernoulli random variable$$\mathcal B(\tilde {p}(z)/{kq(z)})$$

Second, marginally, i.e., on its own, $I$ is a Bernoulli random variable with parameter $$p=\mathbb P(I=1) = \int_\mathfrak Z \dfrac{\tilde {p}(z)}{kq(z)} q(z)\text d z=\frac{1}{k}$$ (the last equality assuming both $\tilde p(\cdot)$ and $q(\cdot)$ are probability densities, i.e., are properly normalised). Therefore, $1/k$ is the average and marginal probability of acceptance in the accept-reject algorithm.

Note that, while $Z$ is marginally distributed from $q(\cdot)$, conditional on $I$, we have $$Z|I=1 \sim \tilde p(z) \quad\text{and}\quad Z|I=1 \sim \dfrac{kq(z)-\tilde p(z)}{k-1}$$

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  • $\begingroup$ How can it be a marginal by definition? Wouldn't we require a joint distribution which we then integrate over to get the variable of interest? This seems more like an average as you also pointed out $\endgroup$ Mar 15, 2021 at 14:11
  • $\begingroup$ Would you be able to edit your answer to show this formally? I do not see how $I$ is marginally bernoulli $B(\frac{1}{K})$ $\endgroup$ Mar 15, 2021 at 17:57

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