Does the beta distribution have a conjugate prior?

Question

I know that the beta distribution is conjugate to the binomial. But what is the conjugate prior of the beta? Thank you.

Answer 1

Yes, it has a conjugate prior in the exponential family. Consider the three parameter family $$ \pi(\alpha, \beta \mid a, b, p) \propto \left\{\frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha)\Gamma(\beta)}\right\}^p \exp\left(a\alpha + b\beta \right). $$ For some values of $(a, b, p)$ this is integrable, although I haven't quite figured out which (I believe $p \ge 0$ and $a < 0, b < 0$ should work - $p = 0$ corresponds to independent exponential distributions so that definitely works, and the conjugate update involves incrementing $p$ so this suggest $p > 0$ works as well).

The problem, and at least part of the reason no one uses it, is that $$ \int_0^\infty \int_0^\infty \left\{\frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha)\Gamma(\beta)}\right\}^p \exp\left(a\alpha + b\beta \right) = ? $$ i.e. the normalizing constant doesn't have a cloed form.

Answer 2

It seems that you already gave up on conjugacy. Just for the record, one thing that I've seen people doing (but don't remember exactly where, sorry) is a reparameterization like this. If $X_1,\dots,X_n$ are conditionally iid, given $\alpha,\beta$, such that $X_i\mid\alpha,\beta\sim\mathrm{Beta}(\alpha,\beta)$, remember that $$ \mathbb{E}[X_i\mid\alpha,\beta]=\frac{\alpha}{\alpha+\beta} =: \mu $$ and $$ \mathbb{Var}[X_i\mid\alpha,\beta] = \frac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)} =: \sigma^2 \, . $$ Hence, you may reparameterize the likelihood in terms of $\mu$ and $\sigma^2$ and use as a prior $$ \sigma^2\mid\mu \sim \mathrm{U}[0,\mu(1-\mu)] \qquad \qquad \mu\sim\mathrm{U}[0,1] \, . $$ Now you're ready to compute the posterior and explore it by your favorite computational method.

Answer 3

In theory there should be a conjugate prior for the beta distribution. This is because

• the beta distribution is one of the exponential family distributions, and
• in theory it should be possible to derive a prior. See, e.g., wikipedia, D Blei's lecture on exponential families.

However the derivation looks difficult, and to quote A Bouchard-Cote's Exponential Families and Conjugate Priors

An important observation to make is that this recipe does not always yields a conjugate prior that is computationally tractable.

Consistent with this, there is no prior for the Beta distribution in D Fink's A Compendium of Conjugate Priors.

Answer 4

Robert and Casella (RC) happen to describe the family of conjugate priors of the beta distribution in Example 3.6 (p 71 - 75) of their book, Introducing Monte Carlo Methods in R, Springer, 2010. However, they quote the result without citing a source.

Added in response to gung's request for details. RC state that for distribution $B(\alpha, \beta)$, the conjugate prior is "... of the form

$$ \pi(\alpha,\beta) \propto \Big\{ \frac{\Gamma(\alpha+\beta)} {\Gamma(\alpha)\Gamma(\beta)} \Big\} ^{\lambda} x_0^{\alpha} y_0^{\beta} $$

where $\{\lambda, x_0, y_0\}$ are hyperparameters, since the posterior is then equal to

$$ \pi(\alpha,\beta \vert x) \propto \Big\{ \frac{\Gamma(\alpha+\beta)} {\Gamma(\alpha)\Gamma(\beta)} \Big\} ^{\lambda} (xx_0)^{\alpha} ((1-x)y_0)^{\beta}." $$

The remainder of the example concerns importance sampling from $\pi(\alpha,\beta \vert x)$ in order to compute the marginal likelihood of $x$.

Answer 5

I do not believe there is a "standard" (i.e., exponential family) distribution that is the conjugate prior for the beta distribution. However, if one does exist it would have to be a bivariate distribution.