Suppose that we have a conditional density function $p(y|x;\theta^*)$, where $\theta^*$ represents distribution parameters and are assumed to be deterministic. Is it possible that we write this conditional density as a deterministic function of $x$ and $\theta$ where $\theta$ is a random variable independent of $x$? In other words,
$y|x \sim p(y|x;\theta^*)$
is equivalent to
$y = g(x, \theta)$
$\theta \sim p(\theta)$
Furthermore, is this representation unique?
For example, if $y$ has a Gaussian distribution with mean $x$ and s.d. $\sigma^*$, we can write
$y = x + \sigma,$
where $\sigma$ has a Gaussian distribution with mean zero and s.d. $\sigma^*$. My question might be related to the question discussed here.