Below is a material about the prior disribution for the proportions.

The appropriate prior distribution for the parameter $\theta$ of a Bernoulli or binomial distribution is one of the oldest problems in statistics. ... We denote $\phi=logit(\theta)$.

an (improper) uniform prior on $\phi$ is formally equivalent to the (improper) Beta(0,0) distribution on the $\theta$ scale, i.e., $p(\theta)\propto \theta^{-1}(1-\theta)^{-1}$:the code below illustrates the effect of the bounding the range for $\phi$ and hence making these distributions proper.

The code in WinBUGS is:

  phi   ~ dunif(-5, 5)
  logit(theta) <- phi

The empirical distributions (based on 100,000 samples) corresponding to the priors is shown in Fig.1. I am not sure what "the effect of the bounding the range" is reflected?

  • 1
    $\begingroup$ In most cases you know more about the problem that such priors would indicate, hence the popularity of Gaussian priors for log odds or $\beta$ priors for probabilities. You can also use a truncated distribution, e.g., a Gaussian prior for the logit, truncated to logits that create the needed truncation in the probabilities. $\endgroup$ Jun 29, 2019 at 11:31

1 Answer 1


Using the improper uniform prior, the parameter $\phi$ can take any value in $(-\infty, \infty)$. A logit transform $\theta = \textrm{logit}^{-1}(\phi)$ would make $\theta \in (0, 1)$ with the density $f(\theta) \propto \theta^{-1} (1-\theta)^{-1}$, which is improper as well because $$\int_{0}^{1} \theta^{-1} (1-\theta)^{-1} d\theta = \infty.$$

If we restrict $\phi$ to the range $(-5, 5)$ then $f(\phi)$ is not improper. We still have $f(\theta) \propto \theta^{-1} (1-\theta)^{-1}$ but now the range for $\theta$ is $(\textrm{logit}^{-1}(-5), \textrm{logit}^{-1}(5)) \approx (0.993, 0.007)$, which is a proper distribution because $$\int_{\textrm{logit}^{-1}(-5)}^{\textrm{logit}^{-1}(5)}\theta^{-1} (1-\theta)^{-1} d\theta < \infty.$$

The purpose of this is to approximate the improper distributions with close-enough proper distributions.


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