I have a parameter $\theta$ whose value should lie between $(0,1)$. Therefore, I am assuming the prior distribution of $\theta$ to be a beta distribution with hyper-priors $\alpha$ and $\beta$ ie. $P(\theta|\alpha,\beta)=Beta(\alpha,\beta)$. I have some prior knowledge about $\theta$ from my experiments which I found lies between $(0.45,0.5)$. Is there a way that I can use this prior knowledge of $\theta$ parameter to assign some hyper prior distruibution to $\alpha$ and $\beta$ ?

  • 1
    $\begingroup$ Are you seeking a conjugate prior or are you just after general advice about formulating your prior information into a prior distribution? $\endgroup$
    – Glen_b
    Feb 8, 2015 at 6:53
  • 3
    $\begingroup$ If you think $(0.45,0.55)$ corresponds to a prior credible interval with a certain coverage, like $0.95$, and add that your prior mean is $0.5$, then there exists a single Beta distribution that fits this prior. See, e.g., my former PhD student's paper, [Dupuis (1995)](biomet.oxfordjournals.org/content/82/4/761.full.pdf). $\endgroup$
    – Xi'an
    Feb 8, 2015 at 9:21
  • $\begingroup$ @Glen_b Yes sir, I am seeking what conjugate hyper-prior distribution I should be using for $\alpha$ and $\beta$ if I have some prior information about $\theta$ ? $\endgroup$
    – Spandyie
    Feb 8, 2015 at 18:36
  • $\begingroup$ Take a look here $\endgroup$
    – Glen_b
    Feb 8, 2015 at 21:38

1 Answer 1


Per Xi'an's comment, suppose you want to choose $α$ and $β$ such that the prior probability of $.45 ≤ θ ≤ .5$ is equal to some number $p$; reasonable values of $p$ would be $.5$ and $.95$. I don't know if there's a closed-form solution, but it's easy to get an approximate answer by just plugging in different values of $α$ and $β$.

In R, for example, you could run

optim(c(2, 2), function(v) abs( (pbeta(.5, v[1], v[2]) - pbeta(.45, v[1], v[2])) - p ))

and then look at the value of $par, which will give you $α$ and $β$, respectively. For $p = .5$, I get $α = 130.7, β = 156.8$. For $p = .95$, I get $α = 1367, β = 1571$.


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