In Metropolis-Hastings Markov chain Monte Carlo, the proposal distribution can be anything including the Gaussian (according to the Wikipedia).

Q: What's the motivation for using anything other than Gaussian? Gaussian works, it's easy to evaluate, it's fast and everybody understands it. Why would I consider anything else?

Q: Since the proposal distribution can be anything, can I use an uniform distribution?


1 Answer 1


A1: Indeed the Gaussian distribution is probably the most used proposal distribution primarily due to ease of use. However, one might want to use other proposal distributions for the following reason

  1. Heavy Tails: The Gaussian distribution has light tails. This means that $N(x_{t-1}, \sigma^2)$ will possibly only suggest values between $(x_{t-1} - 3\sigma, x_{t-1} + 3\sigma)$. But a $t$ distribution has heavier tails, and thus can propose values which are farther away. This ensures that the resulting Markov chain explores the state space more freely, and possibly reduces autocorrelation. The plot below shows the $N(0,1)$ compared to the $t_1$. You see how the $t$ will likely propose more values farther from 0.

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  1. Restricted Space: The Gaussian distribution is defined on all reals. If the distribution you are sampling from is lets say only defined on the positives or on $(0,1)$, then the Gaussian will likely propose values for which the the target density is 0. Such values are then immediately rejected, and the Markov chain does not move from its current spot. This is essentially wasting a draw from the Markov chain. Instead, if you are on the positives, you could use a Gamma distribution and on $(0,1)$ you could use a Beta.
  2. Multiple Modes: When the target distribution is multi-modal, a Gaussian proposal will likely lead to the Markov chain getting stuck near one mode. This is in part due to the light tails of the Gaussian. Thus, instead, people use gradient based proposals, or a mixture of Gaussians as a proposal.

You can find more discussion here and here.

A2: Yes you can use a Uniform distribution as long as the support for the uniform distribution is bounded (since if the support is unbounded the Uniform distribution is improper as it integrates to $\infty$). So a Uniform on $(x_{t-1} - c, x_{t-1} + c)$.

  • 1
    $\begingroup$ Can you clarify the meaning of "the space is bounded" in A2 (note that the space where the target parameter lies does not need to be bounded as long as the chain can move everywhere, possibly requiring multiple steps). Also, is there a typo in that one endpoint has $x_{t-1}$ while another has $x_{t+1}$? $\endgroup$ Apr 15, 2016 at 14:33
  • $\begingroup$ @JuhoKokkala Fixed the typo, thanks for pointing that out. The uniform distribution has to be defined on a bounded space otherwise it doesn't integrate to 1 (and integrates to $\infty$). $\endgroup$ Apr 15, 2016 at 14:36
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    $\begingroup$ @Greenparker: you need to clarify further what you mean by "the space". The support of the target distribution may be unbounded, while the support of a uniform proposal is bounded but the corresponding uniform proposal may still produce an irreducible Markov chain on the whole space. $\endgroup$
    – Xi'an
    Apr 19, 2016 at 9:12
  • $\begingroup$ 1. sigma is a parameter of choice so this argument is invalid. 2. If you are discussing random walk MH (as 1. indicates) this will only be a problem at the boundary. $\endgroup$
    – Hunaphu
    Apr 19, 2016 at 16:36
  • $\begingroup$ @Hunaphu $\sigma$ changes the variance of the proposal, but not the structure of the tails. And can you expand on what you mean by "this will only be a problem at the boundary"? $\endgroup$ Apr 19, 2016 at 16:43

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