# Acceptance rate for Metropolis-Hastings > 0.5

How come it's possible to get Metropolis-Hastings acceptance rates close to 1 (for example, when exploring a unimodal distribution with a normal proposal distribution with too-small SD), after burn-in is over? I see it in my own MCMC chains but I don't understand how it makes sense. It seems to me that after reaching the summit acceptance rate should stabilize around values that are smaller than 0.5.

• It is unclear what your proposal is. The reasonableness of the acceptance rate depends entirely on the proposal distribution. I'm guessing you are talking about a random walk proposal, but i'm not sure. – jaradniemi Apr 5 '17 at 13:42

The acceptance rate depends largely on the proposal distribution. If it has small variance, the ratio of the probabilities between the current point and the proposal will necessarily always be close to 1, giving a high acceptance chance. This is just because the target probability densities we typically work with are locally Lipschitz (a type of smoothness) at small scales, so the probability of two nearby points is similar (informally).

If your current sample is close to the MAP value, the proposals will have less than one acceptance probability, but it can still be very close to 1.

As a side note, standard practice is to tune the proposal distribution to get around a 0.2-0.25 acceptance rate. See here for a discussion of this.

• Thanks Aaron! I just realized I thought of p(proposed)/(p(current)+p(proposed)) instead of p(proposed)/p(current), as it really is . So when exploring a uniform distribution, acceptance rate should be 1, not 0.5. Cool. Thanks! – TanZor Apr 5 '17 at 6:48
• This answer assumes a random-walk Metropolis algorithm, but this was never stated in the question. If you have an independence proposal with a small variance, you can have acceptance probabilities that are very far from 1. – jaradniemi Apr 5 '17 at 13:44

An easy example of acceptance probability equal to one is when simulating from the exact target: in that case $$\dfrac{\pi(x')q(x',x)}{\pi(x)q(x,x')}=1\qquad\forall x,x'$$ While this sounds like an unrealistic example, a genuine illustration is the Gibbs sampler, which can be interpreted as a sequence of Metropolis-Hastings steps, all with probability one.

A possible reason for your confusion is the potential perception of the Metropolis-Hastings algorithm as an optimisation algorithm. The algorithm spends more iterations on higher target regions but does not aim at the maximum. And while $\pi(x^\text{MAP})\ge\pi(x)$ for all $x$'s, this does not mean values with lower target values are necessarily rejected, since the proposal values $q(x^\text{MAP},x)$ and $q(x,x^\text{MAP})$ also matter.

• +1 for "Metropolis-Hastings is not an optimization algorithm." I was going to post that as an answer myself, but now I don't need to. :) – Ilmari Karonen Apr 5 '17 at 13:07