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I've seen that different authors define the acceptance probability $\alpha$ of the Metropolis-Hastings algorithm with target distribution density $p$ and proposal kernel density $q$ differently; some define $$\alpha(x,y):=\begin{cases}\displaystyle1\wedge\frac{p(y)q(y,x)}{p(x)q(x,y)}&\text{, if }p(x)q(x,y)>0\\\color{red}1&\text{, otherwise}\end{cases}\tag1$$ and others define $$\alpha(x,y):=\begin{cases}\displaystyle1\wedge\frac{p(y)q(y,x)}{p(x)q(x,y)}&\text{, if }p(x)q(x,y)>0\\\color{red}0&\text{, otherwise}.\end{cases}\tag2$$

So, is one of them wrong or does it simply not matter which one we use?

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While the Markov chain $(X_t)$ is producing values such that $p(x_t)=0$ it has clearly not "reached" stationarity, so it should keep moving over the state space. Any way to move around until the support of $p$ is reached is valid, including (1) of course.

However, solution (2) is clearly wrong since, if $p(x)=0$, the chain will never move away from its current value.

Note also that$$p(x)q(x,y)=0$$implies $$p(x)=0\qquad \text{a.s.}$$when $y$ is a realisation from the density $q(x,\cdot)$.

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  • $\begingroup$ You're right. $(2)$ doesn't make sense if $p(x)=0$. I was more thinking about the case $p(x)>0$ and $q(x,y)=0$. Intuitively it would make more sense to me to reject such a problem, but according to $(1)$ we would need to accept it. So, could we use $$\alpha(x,y):=\begin{cases}\displaystyle1\wedge\frac{p(y)q(y,x)}{p(x)q(x,y)}&\text{, if }p(x)q(x,y)>0\\0&\text{, if }p(x)>0\text{ and }q(x,y)=0\\1&\text{, otherwise}\end{cases}\tag3$$ instead? $\endgroup$ – 0xbadf00d Mar 17 '20 at 8:03
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    $\begingroup$ $q(x,y)=0$ is impossible if $Y$ is simulated via this conditional density... $\endgroup$ – Xi'an Mar 17 '20 at 8:05

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