# Changing a conditional probability to a deterministic function

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.

• It is hard to call your $p$ as conditional pdf, because there is no additional random component ($x$ and $\theta^*$ are fixed parameters). – user158565 Jun 20 '19 at 2:55
• " σ has a Gaussian distribution with mean zero and s.d. σ" -- please don't use the same symbol for two completely different things. – Glen_b Jun 20 '19 at 6:06
• @Glen_b Thanks for pointing that out. I changed the s.d. to $\sigma^*$. – KRL Jun 20 '19 at 21:27
• 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. $σ$". – Glen_b Jun 20 '19 at 22:28