I am having some confusion over the Metropolis algorithm. Let $g(x|y)$ be our proposal distribution for the algorithm. For the Metropolis, $g$ must be symmetric (from Wikipedia). In the discrete case, we can think of this as a stochastic matrix. I have also read, that

Transition matrices that are symmetric $P(i,j)=P(j,i)$ always have detailed balance. In these cases, a uniform distribution over the states is an equilibrium distribution.

I assume this is because symmetric matrices are doubly stochastic.

Could someone please clarify the ties between the Metropolis algorithm, Metropolis-Hastings, detailed balance, symmetry and the uniform distribution? More specifically, when does the Metropolis algorithm yield a uniform distribution (or why is it not always uniform since our $g$ is symmetric)?

  • 2
    $\begingroup$ $g$ does not have to be symmetric. The Wikipedia heading refers to the special case where it is, but it is not a requirement. To see why we might not want to use a symmetric proposal, consider a parameter that is bounded, say $\theta > 0$, but with a largeish plausible range, say $0 - 10$, and assume that on the very first draw we proposed and accepted $\theta_1 = 0.000001$. With a symmetric proposal, we'd be probably be spending a very, very long time trapped at the extreme edge of the plausible range. $\endgroup$
    – jbowman
    Mar 4, 2019 at 2:15
  • $\begingroup$ Doesn't $g$ have to be symmetric for Metropolis and then can be asymmetric for Metropolis-Hastings? I'm hoping to focus on the case when $g$ is symmetric under Metropolis. $\endgroup$
    – cpage
    Mar 4, 2019 at 2:58

1 Answer 1


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The case when the Metropolis-Hastings proposal $g$ is symmetric as in $g(x\mid y)=g(y\mid x)$ (and the above excerpt from Metropolis et al. 1953) does not imply that the stationary distribution $\pi$ associated with this proposal and the Metropolis-Hastings algorithm is uniform, but only that the Metropolis-Hastings acceptance ratio $$1\wedge\frac{\pi(x')g(x\mid x')}{\pi(x)g(x\mid x)}=1\wedge\frac{\pi(x')}{\pi(x)}$$ does not depend on $g$. Even though $g$ itself as a Markov kernel is associated with a uniform measure, due to the acceptance step in the Metropolis-Hastings algorithm, the stationary distribution of the algorithm is not the stationary distribution of the proposal. This is the whole idea and appeal of the algorithm that the acceptance probability modifies the stationary distribution of the Markov chain thus generated.

  • $\begingroup$ Suppose our $g$ for the Metropolis is the normal distribution. That distribution is symmetric (or is this a different definition of symmetric), but obviously the normal distribution is not uniform. How can that be since the normal distribution is symmetric and not uniform? $\endgroup$
    – cpage
    Mar 4, 2019 at 16:04
  • $\begingroup$ Additionally, suppose our desired distribution $\pi$ is uniform. Wouldn't $\pi(x')=\pi(x)$ in that case and just cancel out to 1? $\endgroup$
    – cpage
    Mar 4, 2019 at 16:42
  • $\begingroup$ I am afraid you are missing the point: The symmetry in $g(x|y)$ means that $g(x|y)=g(y|x)$. In the special case when $\pi$ is constant and $g$ is symmetric, the acceptance probability is equal to one, which is correct. $\endgroup$
    – Xi'an
    Mar 4, 2019 at 17:13

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