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

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  • $\begingroup$ a trick you can to is to re-parametrize. Let $\theta = e^{-\omega} + 2$, and sample $\omega$ from a non truncated normal. Then $(-\infty, \infty)$ is mapped to $[2, \infty)$ $\endgroup$
    – hakanc
    Mar 4 at 13:59
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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.

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  • $\begingroup$ How did you derive the expressions for $q(\theta|\theta')$ and $q(\theta'|\theta)$? Maybe is simple but I am asking exactly that. $\endgroup$
    – Giorgetto
    Mar 4 at 14:32
  • $\begingroup$ A truncated Normal is a Normal density restricted to a subset like $[2,\infty)$ hence requiring renormalising to integrate to one over that very subset. $\endgroup$
    – Xi'an
    Mar 4 at 14:39
  • $\begingroup$ Could you please show it or at least give me a reference on this. Why do I get the ratio of a pdf and a cdf? Thanks $\endgroup$
    – Giorgetto
    Mar 4 at 14:49
  • $\begingroup$ It is not the ratio of a density and of a cdf, rather the ratio of a density in $\theta^\prime$ and of a constant that happens to depend on $\theta$. Integrate in $\theta^\prime$ to show it has a mass equal to $1$.. $\endgroup$
    – Xi'an
    Mar 4 at 15:21

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