# 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?

• In your given integral, you are integrating with respect to $\mathbf{u}$ but the range of the integral is given for $\beta$. I don't understand that. Also, the integral is of the form $f(\beta|u)\Pi(u)$. In order to write the hierarchical representation as you asked, you neeed to find $f(\beta|c_1f+c_2g)$ and $\Pi(c_1f+c_2g)$. I don't think you can just plug in any pdf into the hierarchical model without knowing how they are obtained in the first place. also you just omitted the parts of $f(\beta|u)$ and replaced them with $c_1f+c_2g$. Why is that?
– DevD
Jul 2, 2022 at 8:41
• You omitted the parts of $\Pi(u)$, not $f(\beta|u)$
– DevD
Jul 2, 2022 at 8:47
• @DebayanKoley, the integration is on $(|\beta|^\alpha,\infty)$
– gbd
Jul 2, 2022 at 8:50
• The expression will be simply $\int_{u>|\beta|^{\alpha}}\frac{c_1f}{2u^{\frac{1}{\alpha}}}+\frac{c_2g}{2u^{\frac{1}{\alpha}}}$. But the question remains what are the random variables associated with f and g? If $\mathbf{u}$ follows pdf of $c_1f+c_2g$ then we can use their convolution in the resulting hierarchical expression.
– DevD
Jul 2, 2022 at 9:20
• The title question should be edited into What is the prior corresponding to a weighted sum of pdfs? or something like that. The first comment shows a potential confusion into seeking a prior for $x_1f+c_2g$. Jul 2, 2022 at 10:44

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 $$\mu$$.

• If we suppose that $u\sim c_1 Gamma(a_1,b_1)+c_2 Gamma(a_2,b_2)$, then how do we ensure that the integral equals one?
– gbd
Jul 2, 2022 at 9:49
• I mean how do we ensure that the integration of the probability of $u$ equals one.
– gbd
Jul 2, 2022 at 9:55
• could you please explain how \begin{align} 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} works?
– gbd
Jul 2, 2022 at 15:17
• Check the keywords mixtures of distributions mixture models latent variables and EM algorithm Jul 2, 2022 at 22:15