3
$\begingroup$

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 $\beta$ ?

$\endgroup$
4
  • 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

2
$\begingroup$

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 $α = 1367, β = 1571$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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