1
$\begingroup$

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:

model{
  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?

$\endgroup$
1
  • 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 '19 at 11:31
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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