In MCMC you calculate a value $\alpha$, which tells you the probability of you accepting or rejecting the current sample.

If you end up rejecting the current sample, you then set $x_{n+1} = x_n$. So a chain could go $0.1,0.11,0.105,0.105,0.105,0.105,0.12,0.11,0.115,0.115$.

In the event of rejecting the current sample, what is the difference between adding the current value $x_n$ again to the chain (as it says to do in the algorithm), and rejecting it in the sense that you don't add anything to the chain (so you don't get repeated consecutive repeated values in the posterior), but do reuse that same value when next sampling.

So instead, the above chain would be $0.1,0.11,0.105,0.12,0.11,0.115$.

Sorry if that was a bit confusing. (Essentially what are the implications of doing what is says in the algorithm vs repeatedly sampling until a point is accepted)? Or is the latter just nonsensical and not producing anything of importance?

  • $\begingroup$ "In MCMC you calculate a value α, which tells you the probability of you accepting or rejecting the current sample" -- this is not true of every MCMC algorithm. Do you mean to specifically refer to a Metropolis-Hastings algorithm? $\endgroup$ – Glen_b May 16 at 8:31

Rejecting until an acceptance without repeating the current value induces a bias in the algorithm since it does not simulate from the proposal but from another distribution. Many questions on this forum address this issue

A simple experiment with a toy target like $\mathcal N(0,1)$ and a toy transition like the Normal random walk $\mathcal N(x,2)$ should show the difference in the outcome. A more advanced study of the chain of accepted values is found in our paper Vanilla Rao-Blackwellisation. The crux of this paper is that the stationary distribution density for the accepted chain is $$\tilde{\pi}(x)\propto\pi(x)\rho(x)=\pi(x)\int \alpha(x,y)q(x,y)\,\text{d}y$$where

  • $\pi(\cdot)$ is the original target density
  • $q(\cdot,\cdot)$ is the proposal density in the Metropolis algorithm
  • $\alpha(\cdot,\cdot)$ is the Metropolis acceptance probability associated with $\pi$ and $q$.

Hence $\rho(x)$ is the average probability of acceptance when starting Metropolis from $x$. This may sound surprising or counter-intuitive but one can show that the transition from one accepted value to the next is endowed with density$$\tilde{q}(x,y)=\dfrac{\alpha(x,y)q(x,y)}{\rho(x)}$$and that $\tilde\pi$ is stationary for this transition.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.