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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.

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  • $\begingroup$ It is hard to call your $p$ as conditional pdf, because there is no additional random component ($x$ and $\theta^*$ are fixed parameters). $\endgroup$ – user158565 Jun 20 at 2:55
  • $\begingroup$ " σ has a Gaussian distribution with mean zero and s.d. σ" -- please don't use the same symbol for two completely different things. $\endgroup$ – Glen_b Jun 20 at 6:06
  • $\begingroup$ @Glen_b Thanks for pointing that out. I changed the s.d. to $\sigma^*$. $\endgroup$ – KRL Jun 20 at 21:27
  • $\begingroup$ It would have been much better to stick with statistical convention and leave the s.d. as $\sigma$ and change the variable to a more conventional symbol in such a context (like $\varepsilon$ or $\eta$ or $\zeta$ or $\xi$), so that you had something like "For example, if $y$ has a Gaussian distribution with mean $x$ and s.d. $σ$, we can write $y=x+\varepsilon$, where $\varepsilon$ has a Gaussian distribution with mean zero and s.d. $σ$". $\endgroup$ – Glen_b Jun 20 at 22:28

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