Is it possible to rescale a beta prior to range from, say, $(0.5,1.0)$? For example, say that your likelihood function for some parameter $p$ is binomial and you know that $p \in (0.5,1.0)$ due to some physical constraint, is there a conjugate prior for this model?

  • 5
    $\begingroup$ The truncated Beta with a fixed lower or upper bound like $.5$ remains conjugate: you simply have to write Bayes' formula. $\endgroup$ – Xi'an Sep 28 '15 at 20:42
  • $\begingroup$ You can certainly rescale but it won't be conjugate. Xi'an's suggestion of truncation certainly works. There's also the possibility of mixtures of truncated Betas, which can reasonably emulate a wide variety of densities on the same interval $\endgroup$ – Glen_b -Reinstate Monica Sep 29 '15 at 1:03
  • $\begingroup$ The advantage of truncation over rescaling is that it preserves the interpretation of $p$ as a probability. $\endgroup$ – Neil G Sep 29 '15 at 3:37

In general the support space of the prior-if a subset of the likelihood parameter space-will be the support space of the posterior. For this case write the likelihood

$$L(x| p)= {n\choose{x} }p^{x}(1-p)^{n-x} $$

and the prior as a beta

$$\pi(p| \alpha, \beta) =\frac{\Gamma(\alpha+\beta)p^{\alpha-1}(1-p)^{\beta-1} }{\Gamma(\alpha)\Gamma(\beta)}\mathbb{1}(p\in (a,b))$$

where $\mathbb{1}(A)$ is an indicator function in the set ($A$) and $0\leq a < b \leq 1$.

Now we can drop the irrelevant normalizing constants to see that the posterior will be

$$\pi(p | x, \alpha, \beta) = L(x| p)\times\pi(p| \alpha, \beta) \propto p^{x}(1-p)^{n-x} \times p^{\alpha-1}(1-p)^{\beta-1}\mathbb{1}(p\in (a,b)) \propto p^{\alpha+x-1}(1-p)^{n-x+\beta-1}\mathbb{1}(p\in (a,b)). $$

Now we can calculate the normalizing constant-at least numerically-by using incomplete beta function

$$C = \left(\int_a^b p^{\alpha+x-1}(1-p)^{n-x+\beta-1}dp\right)^{-1}.$$

So the posterior will be

$$\pi(p | x, \alpha, \beta)=C\times p^{\alpha+x-1}(1-p)^{n-x+\beta-1}\mathbb{1}(p\in (a,b))$$

In the question asked here in the post $a=0.5$ and $b=1.0$.


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