I am confused about how what value should be returned from a uniform distribution when using MCMC simulations.

The proper normal distribution is define as $$ p(\theta) = \left\{ \begin{array}{cc} 1/(b-a) & \text{if} & a < \theta < b \\ 0 & & \mathrm{otherwise} \end{array}\right. $$

However, in the examples from the Python package I am useing (emcee), they instead use the improve definition $$ p(\theta) = \left\{ \begin{array}{cc} 1 & \text{if} & a < \theta < b \\ 0 & & \mathrm{otherwise} \end{array}\right. $$

Now for the parameter estimation itself this makes no obvious difference. Unfortunately I am then using the output from the MCMC algorithm to estimate the marginal likelihood (the evidence). To do this I use the Harmonic mean approximation (which I know is deeply flawed, nevertheless it seems to be doing a reasonable job). In this case the choice of prior makes a significant difference.

Can anyone explain which of these is really appropriate and why? I originally expected the two to reach the same answer, but there is a noticeable difference and it makes me somewhat uncomfortable. Since I am only in the learning stage, I thought it best to ask the stupid questions first.


2 Answers 2


The uniform PDF is as you say:

$$ p(\theta) = \left \{ \begin{array}{ll} 1 / (b-a) \quad &\text{if } a < \theta < b\\ 0 \quad &\text{otherwise.} \end{array} \right. $$

Bayesian methods (MCMC) are usually only concerned with the posterior distributions up to a constant of proportionality, that is:

$$ p(\theta | D) \propto p(\theta) p(D | \theta) $$

for some observed data $D$. Thus you only really need the prior distribution up to a constant of proportionality too, which is why you can use:

$$ p(\theta) \propto \left \{ \begin{array}{ll} 1 \quad &\text{if } a < \theta < b\\ 0 \quad &\text{otherwise.} \end{array} \right. $$

In which sense the prior on $\theta$ only provides upper ($b$) and lower ($a$) bounds on allowable value of $\theta$. In computing terms, the proportional form is faster.

  • $\begingroup$ So if I keep the constant of proportionality, is my result incorrect or just scaled differently? $\endgroup$
    – Greg
    Apr 2, 2015 at 13:19
  • $\begingroup$ @Greg it just does not have a $[0, 1]$ scale so it is not probability proper, but it does not matter if you are not interested in estimating the probabilities. $\endgroup$
    – Tim
    Apr 2, 2015 at 13:32
  • $\begingroup$ @Tim I am interested in calculating the probabilities in the end, but by using the harmonic mean approximation. I think I need to do some more reading by the sounds of it, thanks for the help! $\endgroup$
    – Greg
    Apr 2, 2015 at 13:57
  • $\begingroup$ @Greg how are you calculating the harmonic mean? I've never done it, but I thought you just use the posterior samples from the MCMC (which you have said are the same for both methods) and calculate 1/Likelihood for each sample (radfordneal.wordpress.com/2008/08/17/…) so the 2 methods should be the same. Perhaps you are seeing the problems with using the harmonic mean approximation. $\endgroup$
    – Jeff
    Apr 2, 2015 at 13:58

Bayes theorem states that

$$ p(\theta|D) \propto p(D|\theta) p(\theta) $$

so if $\theta$ is constant, and that is the case in both cases you quote, then it does not have any effect on the estimates. Notice that if you want $p(\theta|D)$ to give you proper probabilities (i.e. to integrate to $1$) then the only thing you have to do is to divide by $\int p(D|\theta) p(\theta) d\theta$, that is constant. See also this thread to learn more on normalizing constant in Bayes theorem.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.