What is the prior corresponding to a weighted sum of pdfs? Recently, I have read a paper (Bayesian Bridge Regression
, H. Mallick, 2018) which states that since
$$
\frac{\lambda^{1/\alpha}}{2\Gamma(1+1/\alpha)}e^{-\lambda|\beta|^\alpha}=\int_{u>|\beta|^\alpha}\frac{\lambda^{1+1/\alpha}}{2u^{1/\alpha}\Gamma(1+1/\alpha)}u^{(1+1/\alpha)-1}e^{-\lambda u}du
$$
then hierarchical representation is as follows
$$
\beta|u\sim \text{Uniform}(-u^{1/\alpha},u^{1/\alpha})
$$
$$
u\sim \text{Gamma}(1+1/\alpha,\lambda)
$$
My question: Suppose that instead
$$
\frac{\lambda^{1/\alpha}}{2\Gamma(1+1/\alpha)}e^{-\lambda|\beta|^\alpha}=\int_{u>|\beta|^\alpha}\frac{\lambda^{1+1/\alpha}}{2u^{1/\alpha}\Gamma(1+1/\alpha)}(c_1 f(u)+c_2 g(u))du
$$
where $f$ and $g$ are two known pdfs and $c_1,c_2$ some constants. Then, how would we write the hierarchical representation? and are there any problems with this kind of representation?
 A: The integral representation
$$\int_{u>|\beta|^\alpha}\frac{\lambda^{1+1/\alpha}}{2u^{1/\alpha}\Gamma(1+1/\alpha)}(c_1 f(u)+c_2 g(u))\,\text du$$
involves unnecessary terms $\lambda^{1+1/\alpha}$  and $\Gamma(1+1/\alpha)$ and should rather be written as
$$\int\frac{\mathbb I_{u>|\beta|^\alpha}}{2u^{1/\alpha}}\times[c_1 f(u)+c_2 g(u)]\,\text du\quad \text{with}\quad c_1+c_2=1$$
meaning that
\begin{align}
\beta|u&\sim \text{Uniform}(-u^{1/\alpha},u^{1/\alpha})\\
u&\sim c_1 f(u)+c_2 g(u)
\end{align}
a mixture of two distributions. Which can be rewritten as a three level hierarchical model:
\begin{align}
\beta|u&\sim \text{Uniform}(-u^{1/\alpha},u^{1/\alpha})\\
u|z=1&\sim f(u)\qquad u|z=2\sim g(u)\\
z &\sim c_1^{\mathbb I_{z=1}}c_2^{\mathbb I_{z=2}}
\end{align}
A trivial mixture that produces the same Gamma marginal on $\beta$ is
$$c_1 f(u) + c_2 g(u) = \frac{\lambda^{1+1/\alpha}}{\Gamma(1+1/\alpha)}u^{(1+1/\alpha)-1}e^{-\lambda u}\mathbb I_{u<\mu} + \frac{\lambda^{1+1/\alpha}}{\Gamma(1+1/\alpha)}u^{(1+1/\alpha)-1}e^{-\lambda u}\mathbb I_{u>\mu}$$
for an arbitrary $\mu$.
