Does Metropolis-Hastings work with the log of the proposal and the density to be sampled from? That is, say we want to sample from a density $\pi(x)$, using a proposal $q(x|x^{old})$, will the Metropolis-Hastings work with $\log(\pi(x))$ and $\log(q(x|x^{old}))$ as well?

When constructing a Gibbs sampler, we often encounter full conditional distributions that are non-conjugate. Techniques to sample from them include ARS,1,2 ARMS,3 and Slice sampling.4 These techniques have the convenient feature that they can take the log of a density to be sampled from. They are technically Metropolis-Hastings samplers, so there are cases where my question will be answered in the affirmative. But is this a general feature of the Metropolis-Hastings algorithm?

The reason for my question is that you often store the log of the density when designing a Gibbs sampler, especially when using an object-oriented language and one of the three samplers mentioned above. If you have $\log( \pi(x) )$ stored, you could take $e^{ \log(\pi(x)) } = \pi(x)$ and use Metropolis-Hastings, but that can create issues with numerical overflow and underflow.

I have been unable to find a reference that explains or provides a proof to this question.


I didn't mean to ask whether I can sample from $\log(\pi(x) )$, since that makes no sense. My question was whether the Metropolis-Hastings can work if I pass it $\log(\pi(x) ) $. That is, is there a way to construct the algorithm that only uses the "better behaved" $\log(\pi(x) ) )$, rather than $\pi(x)$? The latter is usually a product of multiple other densities, and it can get really big really quick. Sums of log-densities are easier to use in algorithms than products of densities.


1Gilks, W., Wild, P.: Adaptive Rejection Sampling for Gibbs sampling, Applied Statistics 41(2), 337–348, 1992

2Gilks, W.: Derivative-free Adaptive rejection sampling for Gibbs sampling, Baysian Statistics 4, Oxford University Press, Oxford, 641–649, Eds: Bernardo, J., Berger, j., Dawid, A., Smith, A., 1992

3Gilks, W., Best, N., Tan, K.: Adaptive Rejection Metropolis Sampling within Gibbs sampling, Applied Statistics 44(4), 455–472, 1995

4Neal, Radford M: Slice sampling, Annals of Statistics 31(3), 705–741, 2003

  • $\begingroup$ Acceptance of MH is $ \min\left(1 , \frac{\pi(X) q(X^{(t-1)}|X) }{\pi(X^{(t-1)}) q(X|X^{(t-1)}) } \right)$. Is this the same as $ \min\left(1 , \frac{\log(\pi(X)) \log(q(X^{(t-1)}|X)) }{\log(\pi(X^{(t-1)})) \log(q(X|X^{(t-1)})) } \right)$? Or at least, will it also converge to $\pi(x)$ as stationary distribution? To me it is not obvious. I can clarify that in the question, if you think it necessary? $\endgroup$
    – dwcoder
    Feb 14, 2015 at 20:49
  • 1
    $\begingroup$ This makes no sense, given that those logs often are negative: what would it mean to "sample" from a negative quantity? There aren't any real issues with under- or over-flow with well written software: you just screen the logarithms before computing the exponentials to make sure they are not too extreme. $\endgroup$
    – whuber
    Feb 14, 2015 at 21:09
  • $\begingroup$ That pretty much answers my question then. I was wondering whether the fact that you could pass $log(x)$ to the Slice and ARMS samplers, since they never use exponentials in there, and I was also wondering how they "sample" from a negative quantity. I thought there was something obvious I was overlooking. $\endgroup$
    – dwcoder
    Feb 14, 2015 at 21:19
  • $\begingroup$ Here is the Slice sampler's code. Everything happens in log terms: cs.toronto.edu/~radford/ftp/slice-R-prog The same is true of the ARS. I was wondering why it worked in log terms, and whether it is a general feature of the MH. But as you said, it makes no sense. $\endgroup$
    – dwcoder
    Feb 14, 2015 at 21:22
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    $\begingroup$ If you look through the code you can see that the only way the log density is used is to compare it to other logarithms. Comparing logarithms is equivalent to comparing their exponentials--but it is not limited to the range of IEEE doubles and doesn't risk over or under flowing. As far as I can see in the code, that is all that's going on. $\endgroup$
    – whuber
    Feb 14, 2015 at 21:48

1 Answer 1


As hinted at by @Tim, the solution was quite simple. A function implementing the Metropolis-Hastings can take $\log(\pi(x) )$ and $\log(q(x|x^{old} ) )$, but then everything will have to happen on a log scale. Let $\alpha$ be the acceptance probability of the Metropolis-Hastings update and $x'$ be the current value of the sampler. Then we propose a new $x$ with acceptance probability of:

$$ \alpha(x | x') = \min\left(1 , \frac{\pi(x) q(x'|x) }{\pi(x') q(x|x') } \right) $$ or in log terms $$ \log\big( \alpha(x | x') \big) = \min\Big( 0 , \log(\pi(x)) + \log(q(x'|x)) - \log(\pi(x')) - \log( q(x|x') ) \Big) \text{.} $$

Then $\log(\alpha)$ can then be used to accept/reject the proposed $x$.

This is a useful interpretation of the Metropolis-hastings that has practical benefits. It is used in this textbook: http://mcmcinirt.stat.cmu.edu/archives/320

That is, an implementation of the Metropolis-Hastings that takes the log of the density to be sampled from, and the log of the proposal. Logs are convenient to use since they are not limited to the range of IEEE doubles, and they don't suffer from numerical over and underflow, as pointed out by @whuber.


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