Bayesian statistics and Basis for continous functions

Question

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}

Answer 1

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

Answer 2

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.