2
$\begingroup$

Let's say I have a beta distribution parametrized with $\mu, \phi$ such that given a random variable $X$ $$X\sim Beta(\mu,\phi),\quad \mu =\frac{a}{a+b}, \quad \phi = a+b$$

Now let a and b be independent gamma distributions such that $$a\sim Gamma(\alpha, 1),\quad b\sim Gamma(\beta,1)$$

Thus it follows from change of variables and gamma sums that $$\mu\sim Beta(\alpha,\beta)$$ $$\phi\sim Gamma(\alpha+\beta,1)$$

The Probability distribution of $X$ is then $$f(x)= \int_{0}^{\infty}{\int_{0}^{\infty}{B_x(a,b)G_a(\alpha,1)G_b(\beta,1)}dadb}$$

Where $B_x(\cdot)$ is the beta pdf and $G_x(\cdot)$ is the Gamma PDF. Is there a closed form solution to this equation? Or do I have to estimate it numerically?

Edit: Re-expressed the model since my original didnt address $a$ and $b$

$\endgroup$
  • 1
    $\begingroup$ How can $\mu$ be a random variable, while $\theta$ is not, if they are both functions of $a$ and $b$? And if $\mu =\frac{a}{a+b}$ is a random variable, then presumably you really want to say that $a$ and $b$ are random variables ... but you do not provide a model for them. $\endgroup$ – wolfies Mar 7 '16 at 2:37
  • 2
    $\begingroup$ If $a \sim Gamma(\alpha, 1)$ and $b \sim Gamma(\beta, 1)$ independently of $a$ where $\alpha$ and $\beta$ are fixed, then $\mu \sim Beta(\alpha, \beta)$. For that reason, $\phi \sim Gamma(\alpha + \beta, 1)$. Using this you can fix your distribution of $X$. $\endgroup$ – Greenparker Mar 7 '16 at 6:29
  • $\begingroup$ Fixed original post. Yes, I didn't really consider a and b before but your specifications makes much more sense. Given this, is there still a closed-form solution we can derive or do we still have to estimate it numerically? $\endgroup$ – Kevin Pei Mar 7 '16 at 14:22
  • 2
    $\begingroup$ No - I meant dropping $a$ and $b$. Fix $\phi$ as a fixed parameter. And let $\mu$ have some simple distribution. $\endgroup$ – wolfies Mar 7 '16 at 16:23
  • 1
    $\begingroup$ I see. Thank you, I will try that. If you want you can write that as an answer and I'll mark that as answered. $\endgroup$ – Kevin Pei Mar 8 '16 at 13:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.