# What is the effect of "bounding the range" for a prior distribution?

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 . I am not sure what "the effect of the bounding the range" is reflected?

• 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. Jun 29 '19 at 11:31

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 $$\int_{\textrm{logit}^{-1}(-5)}^{\textrm{logit}^{-1}(5)}\theta^{-1} (1-\theta)^{-1} d\theta < \infty.$$