I try to generate random numbers from the conjugate distribution of beta distribution. It is as follows

$$ p(α,β∣a,b,d)∝ \frac{e^{-a \alpha} e^{-b \beta}}{(\beta(\alpha,\beta))^d} \:\:\:\:,\:\:\: \alpha>0,\beta>0$$

where $a>0$, $b>0$ and $d>0$. $\beta(\alpha,\beta)$ is the Beta function. How can I generate samples from the above distribution? Thank you.

  • $\begingroup$ This answer to another question gives you the proper link. $\endgroup$
    – Xi'an
    Aug 29, 2020 at 15:46

1 Answer 1


Here is an excerpt from our book, Introducing Monte Carlo methods with R, indirectly dealing with this case (by importance sampling). The graph of the target shows a smooth and regular shape for the conjugate, meaning a Normal or Student proposal could maybe be of used for accept-reject. An alternative is to use MCMC, eg Gibbs sampling.

Example 3.6. [p.71-75] When considering an observation $x$ from a beta $\mathcal{B}(\alpha,\beta)$ distribution, $$ x\sim \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}\,x^{\alpha-1} (1-x)^{\beta-1}\,\mathbb{I}_{[0,1]}(x), $$ there exists a family of conjugate priors on $(\alpha,\beta)$ of the form $$ \pi(\alpha,\beta)\propto \left\{ \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha) \Gamma(\beta)} \right\}^\lambda\, x_0^{\alpha}y_0^{\beta}\,, $$ where $\lambda,x_0,y_0$ are hyperparameters, since the posterior is then equal to $$ \pi(\alpha,\beta|x)\propto \left\{ \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha) \Gamma(\beta)} \right\}^{\lambda+1}\, [x x_0]^{\alpha}[(1-x)y_0]^{\beta}\,. $$ This family of distributions is intractable if only because of the difficulty of dealing with gamma functions. Simulating directly from $\pi(\alpha,\beta|x)$ is therefore impossible. We thus need to use a substitute distribution $g(\alpha,\beta)$, and we can get a preliminary idea by looking at an image representation of $\pi(\alpha,\beta|x)$. If we take $\lambda=1$, $x_0=y_0=.5$, and $x=.6$, the R code for the conjugate is


leading to the following picture of the target:

enter image description here

The examination of this figure shows that a normal or a Student's $t$ distribution on the pair $(\alpha,\beta)$ could be appropriate. Choosing a Student's $\mathcal{T}(3,\mu,\Sigma)$ distribution with $\mu=(50,45)$ and $$ \Sigma=\left( \begin{matrix}220 &190\\ 190 &180\end{matrix}\right) $$ does produce a reasonable fit. The covariance matrix\idxs{covariance matrix} above was obtained by trial-and-error, modifying the entries until the sample fits well enough:

 x=matrix(rt(2*10^4,3),ncol=2)       #T sample
 E=matrix(c(220,190,190,180),ncol=2) #Scale matrix
 image(aa,bb,post,xlab=expression(alpha),ylab=" ")

If the quantity of interest is the marginal likelihood, as in Bayesian model comparison (Robert, 2001), \begin{eqnarray*} m(x) &=& \int_{\mathbb R^2_+} f(x|\alpha,\beta)\,\pi(\alpha,\beta)\,\text{d}\alpha \text{d}\beta \\ &=& \dfrac{\int_{\mathbb R^2_+} \left\{ \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha) \Gamma(\beta)} \right\}^{\lambda+1}\, [x x_0]^{\alpha}[(1-x)y_0]^{\beta} \,\text{d}\alpha \text{d}\beta} {x(1-x)\,\int_{\mathbb R^2_+} \left\{ \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha) \Gamma(\beta)} \right\}^{\lambda}\, x_0^{\alpha} y_0^{\beta} \,\text{d}\alpha \text{d}\beta}\,, \end{eqnarray*} we need to approximate both integrals and the same $t$ sample can be used for both since the fit is equally reasonable on the prior surface. This approximation \begin{align}\label{eq:margilike} \hat m(x) = \sum_{i=1}^n &\left\{ \frac{\Gamma(\alpha_i+\beta_i)}{\Gamma(\alpha_i) \Gamma(\beta_i)} \right\}^{\lambda+1}\, [x x_0]^{\alpha_i}[(1-x)y_0]^{\beta_i}\big/g(\alpha_i,\beta_i) \bigg/ \nonumber\\ &x(1-x)\sum_{i=1}^n \left\{ \frac{\Gamma(\alpha_i+\beta_i)}{\Gamma(\alpha_i) \Gamma(\beta_i)} \right\}^{\lambda}\, x_0^{\alpha_i}y_0^{\beta_i}\big/g(\alpha_i,\beta_i)\,, \end{align} where $(\alpha_i,\beta_i)_{1\le i\le n}$ are $n$ iid realizations from $g$, is straightforward to implement in {\tt R}:

 f=function(a) exp(2*(lgamma(a[,1]+a[,2])-lgamma(a[,1])
 h=function(a) exp(1*(lgamma(a[,1]+a[,2])-lgamma(a[,1])


 > mean(f(y)/den)/mean(h(y)/den)
 [1] 0.1361185

Our approximation of the marginal likelihood, based on those simulations is thus $0.1361$. Similarly, the posterior expectations of the parameters $\alpha$ and $\beta$ are obtained by

> mean(y[,1]*f(y)/den)/mean(f(y)/den)
[1] 94.08314
> mean(y[,2]*f(y)/den)/mean(f(y)/den)
[1] 80.42832

i.e., are approximately equal to $19.34$ and $16.54$, respectively.


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