Random Walk Metropolis: acceptance probability with truncated normal proposal

Question

I want to draw from my target density $p(\theta)$ using Random Walk Metropolis.

$\theta$ has domain $[2, +\infty)$, and I am using as proposal a truncated normal, namely:

$$q(\theta_t') \sim N(\theta_{t-1}, \sigma^2)_{[2, \infty)},$$

where $t$ is the number of iterations. The acceptance probability will be given by:

$$r = \frac{p(\theta'_t)\,q(\theta_{t-1}|\theta'_t)}{p(\theta_{t-1})\,q(\theta'_t|\theta_{t-1})}$$

How can I calculate $$\frac{q(\theta_{t-1}|\theta'_t)}{q(\theta'_t|\theta_{t-1})}?$$

Answer 1

Using a truncated Normal distribution proposal means that it is no longer a random walk Metropolis proposal: the density of $\theta^\prime_t$ given $\theta_{t-1}$ $$q(\theta^\prime|\theta)=\sigma^{-1}\varphi([\theta^\prime-\theta]/\sigma)\mathbb I_{(2,\infty)}(\theta^\prime)\big/\Phi([\theta-2]/\sigma)$$ differs from the density of $\theta^\prime_t$ given $\theta_{t-1}$ $$q(\theta|\theta^\prime)=\sigma^{-1}\varphi([\theta-\theta^\prime]/\sigma)\mathbb I_{(2,\infty)}(\theta)\big/\Phi([\theta^\prime-2]/\sigma)$$ and the ratio of the proposals thus differs from one (1). This is not an issue, apart from the terminology.

Note that another valid proposal is to use the untruncated Normal distribution $$q(\theta^\prime|\theta)=\sigma^{-1}\varphi([\theta^\prime-\theta]/\sigma)$$ and to reject any proposal $\theta^\prime$ less than two (2), replicating the current $\theta$ instead. In this case, the proposal is symmetric and qualifies as a random walk proposal.