# Is this a valid Markov Chain Monte Carlo sampler

Let's consider a sampling algorithm which is sampling from the posterior distribution $P(X)$ of a random variable $X$. At a given iteration $t+1$ it proposes a new value $(X_{t+1})$ for $X$ using a symmetric proposal distribution $Q(X_{t+1}|X_t)$ around the old value $X_t$. Due to the symmetry of the proposal distribution, the standard Metropolis Hastings (MH) acceptance rate is as follows: accept $X_{t+1}$ with probability $\alpha=\min(1,P(X_{t+1})/P(X_t))$. Somewhere, I saw a slightly different acceptance rate, it says:

• if $P(X_{t+1})>P(X_t)$ accept $X_{t+1}$ (This part is same as the MH algorithm)
• else sample from the distribution $P=[P(X_t); P(X_{t+1})]/(P(X_t)+P(X_{t+1}))]$ (In this case the sampler will choose $X_{t+1}$ or $X_t$ depending on their posterior probabilities)

I was wondering whether the later is a valid MCMC sampler.

• Would you mind to cite the source of the second approach? – Karel Macek Feb 7 '17 at 10:48
• Why can't you try to prove that the distribution $P$ is stationary with this transition? – Xi'an Feb 14 '17 at 5:56

This alternative version sounds like Barker's (1965) and Hastings' (1970) versions when the acceptance probability is $$\alpha(x,x^\prime) = s(x,x^\prime) \left(1+\dfrac{p(x) q(x^\prime|x)}{p(x^\prime) q(x|x^\prime)}\right)^{-1}$$with $s(x,y)$ is a symmetric function keeping the above between 0 and 1 except that there should be no sure acceptance first, i.e., when $p(x^\prime)>p(x)$.