Bayesian statistics and Basis for continous functions I was thinking about Bayesian statistics, and one thought bothered me:
In Bayesian statistics, we assume that the pdf $p(x)$ can be described as:
\begin{equation}
p(x)=\int f(x|\theta)g(\theta)d\theta
\end{equation}
usually when $x\in[l,u]$, people choose $f$ to be beta distribution
\begin{equation}
p(x)=\int_l^u f(x|\alpha,\beta)g(\alpha)h(\beta) d\alpha d\beta
\end{equation}
(where $1=\int_{-\infty}^\infty g=\int_{-\infty}^\infty h$ and $h,g\geq 0$)
After that short intro, My question is:
Can we model any continuous function like that ?
In other words: 
If $p: [l,u]\longrightarrow[0,1]$ is a continuous function, 
does that mean that we can find two functions $g,h$ such that
\begin{equation}
p(x)=\int_l^x f(x|\alpha,\beta)g(\alpha)h(\beta)d\alpha d\beta
\end{equation}
 A: Let's take the case of the beta distribution with parameters $\alpha,\beta>0$:
$$
f(x;\alpha,\beta)=\frac{x^{\alpha-1}(1-x)^{\beta-1}}{B(\alpha,\beta)}
\quad\text{where}\quad
B(\alpha,\beta)=\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}.
$$
The question is if a continuous probability distribution $p(x)$ on $[0,1]$ can always be expressed in terms of $f(x;\alpha,\beta)$ with weights $g(\alpha)\cdot h(\beta)$.
There are two problems. The first is if $p(x)$ can be expressed as a combination of $f(x;\alpha,\beta)$ terms at all, i.e. with weights $q(\alpha,\beta)$. The other is if these weights can be split as $g(\alpha)\cdot h(\beta)$.
If $p(x)$ has a zero for some $x\in(0,1)$, e.g. $p(1/2)=0$, then obviously it cannot be expressed as a combination of $f(x;\alpha,\beta)$ as these are all strictly positive on $(0,1)$. Of course, if we require $p(x)>0$, this limitation is avoided, but I suspect there will still be limitations on how small $p(x)$ can be for particular $x$.
The other problem is in terms of splitting the weights as $g(\alpha)\cdot h(\beta)$. I asked about this in a comment since it is a rather different problem, and I wasn't sure if this was really what you were interested in. However, my suspicion is that there will be cases, e.g. on the form
$p(x)=[f(x;n,1)+f(x;1,n)]/2$, for which this splitting cannot be done using non-negative coefficients, but have no proof of this.
A: First, The representation
$p(x)=\int{f(x|\alpha,\beta)g(\alpha)h(\beta)}$
can span the entire $[l,u]$ segement if $g()$ and $h()$ are allowed to be negative.
This can be seen by extending polynomial approximation.
However, if $h,g \geq 0$, the answer is simply no.
$p(x)\equiv 1$ cannot be obtained
