I’m taking a class in Bayesian statistics, and we’re learning about the Metropolis algorithm.

Suppose for simplicity that we just have one parameter we’re trying to estimate: $\theta$. According to the algorithm, if we’re at some $\theta_{current}$, and we generate a new parameter $\theta_{proposed}$ from the proposal distribution, we move to the new parameter with probability $$min \left(1, \frac{prior(\theta_{proposed}) \times likelihood(\theta_{proposed})}{prior(\theta_{current}) \times likelihood(\theta_{current})}\right)$$ In words, we compare the (unnormalized) posterior value for the current parameter with the posterior value for the proposed parameter.

My question is: why make the comparison between current and proposed in the first place? (which is what the title of the question is alluding to, the Markov property). Why not just decide to accept the new parameter based on the value of $${prior(\theta_{proposed}) \times likelihood(\theta_{proposed})}$$ (either by considering it to be a probability or taking some function of it).

One thing I could think of is that we would not be normalizing the posterior, but if we’re always not normalizing, why does it matter? And are there other reasons we need the Markov property?


1 Answer 1


As you know, the denominator takes care of the normalizing constants. It is an integral that is usually not possible to calculate analytically. In the fraction above, it cancels as it does not depend on the parameter $\theta$.

Say you draw your proposal $\theta_{proposal}$ from a proposal density $q$. Since $q$ is different from the posterior you want to infer via MCMC, you cannot accept $\theta_{proposal}$ in each iteration, but only with some acceptance probability that is captured by the ratio of the posterior, evaluated at the current state $\theta_{current}$ and the proposal $\theta_{proposal}$. This is the fraction above.

The underlying idea is that $\theta_{proposal}$ should be accepted with higher probability when the corresponding value of the (unnormalized) posterior is higher than the corresponding value of the (unnormalized) posterior at $\theta_{current}$.

Regarding your 2nd question below: you need a normalized posterior, because you need a posterior distribution. Otherwise you cannot draw inferential statements.

  • $\begingroup$ Thanks for your answer. Regarding your last comment “you need a posterior distribution,” why does it matter? I mean, isn’t the point that we want to move through parameter space in a way that is proportional to the probability of the parameter occurring? Wouldn’t that be accomplished even if we don’t normalize? And actually, couldn’t you normalize your results once you’re done? $\endgroup$ Feb 5, 2022 at 20:43
  • $\begingroup$ The goal of Bayesian statistics is inference of a posterior distribution. It needs to be a distribution since otherwise you could not derive posterior moments or credible intervals that make the Bayesian approach so appealing (because they allow straightforward uncertainty quantification). In different words, if there is no posterior distribution, e.g. if the normalizing constant diverges, there is no Bayesian inference. There is a very good answer here: stats.stackexchange.com/questions/87321/…. $\endgroup$
    – BalaGizeh
    Feb 5, 2022 at 20:53

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