Product of independent beta distribution and gamma distribution I'm reading Bayesian Forecasting and Dynamic Models by Mike West and Jeff Harrison. The following conclusion is about the variance discounting in dynamic linear regression model. On page 362, it says,


*

*Dynamic of precision, $\phi_t=\gamma_t\phi_{t-1}/\delta$

*Filtered precision distribution, $\phi_{t-1} \sim Gamma(n_{t-1}/2,d_{t-1}/2) $

*Discounting factor, $\gamma_t \sim Beta(\delta n_{t-1}/2,(1-\delta) n_{t-1}/2) $
$\delta \in (0,1]$. Predicted precision distribution is,
$$\phi_t  \sim Gamma(\delta n_{t-1}/2,\delta d_{t-1}/2) $$
How to derive the predicted precision distribution? 
I tried to apply transformation $U=\phi_t=\gamma_t\phi_{t-1}/\delta$ and auxiliary variable $V=\gamma_t$.
$$f_{U,V}(u,v) = f_{\gamma_t,\phi_{t-1}}(\gamma_t,\phi_{t-
1})|J(u,v)|$$
$J(u,v)$ is the Jacobian w.r.t to $U$ and $V$. But I can't integrate out $V$
 A: To further expand upon why the correct transformation is $U=\gamma_t\phi_{t-1}$ and $V=(1-\gamma_t)\phi_{t-1}$, consider the relation between the Gamma and Beta distributions. Specifically, consider independent random variables $X\sim\Gamma(\alpha,\theta)$ and $Y\sim\text{Gamma}(\beta,\theta)$. Then it follows $X(X+Y)^{-1}\sim\text{Beta}(\alpha,\beta)$. To see this, define the following transformations: $U=X(X+Y)^{-1}$ and auxiliary variable $V=X+Y$. The Jacobian of the inverse transformations $X=UV$ and $Y=V(1-U)$ is then equal to $V$, so it follows the joint distribution of $U$ and $V$ is equal to
\begin{aligned}
f_{U,V}(u,v)&=f_X(uv)f_Y(v(1-v))v\\
&={\frac{1}{\Gamma(\alpha)}(uv)^{\alpha-1}e^{-\theta uv}} \ {\frac{1}{\Gamma(\beta)}(v(1-u))^{\beta-1}e^{-\theta v(1-u)}}v\\
&=\frac{1}{\Gamma(\alpha+\beta)}v^{\alpha+\beta-1}e^{-\theta v}\times \frac{{\Gamma(\alpha+\beta)}}{\Gamma(\alpha)\Gamma(\beta)}u^{\alpha-1}(1-u)^{\beta-1},
\end{aligned}
implying $U$ and $V$ are independent. It follows the marginal of $U$ is distributed $\text{Beta}(\alpha,\beta)$, while $V$ is distributed $\text{Gamma}(\alpha+\beta,\theta)$.
This suggests we instead start with independent random variables $U$ and $V$ as defined above and transform them through $X$ and $Y$ to achieve the desired Gamma marginal distribution for $X$. Taking $\alpha=\delta n_{t-1}$, $\beta=(1-\delta)n_{t-1}$ and $\theta=d_{t-1}$ achieves the desired result.
