For any random variable $X$ whose density is $Pr(X=x)=p(x;\theta)$ where $\theta$ is a parameter, its deterministic function representation is $X=f(\theta, \omega)$ where $\omega$ is a random variable whose distribution is independent of $\theta$. The question is that what is the condition of $p(x; \theta)$ such that its deterministic function representation exists?

For example, if $X\sim N(\mu,\sigma^2)$, then $X=\mu+\sigma \omega$ where $\omega$ is a standard normal random variable. If $X \sim Bernoulli(q)$, then $X=1(logit^{-1}(q)>\omega)$ where $\omega$ is a logistic random variable. I don't know if there is a deterministic function representation for a Poisson random variable.

Thanks.

For any random variable $X$ whose density is $\mathbb{P}(X=x)=p(x;\theta),$ where $\theta$ is a parameter, its deterministic function representation is $X=f(\theta, \omega)$ where $\omega$ is a random variable whose distribution is independent of $\theta$.

Question: what is the condition on $p(x; \theta)$ so that its deterministic function representation exists?

For example, if $X\sim N(\mu,\sigma^2)$, then $X=\mu+\sigma \omega$ where $\omega$ is a standard normal random variable. 

If $X \sim \operatorname{Bernoulli}(q)$, then $X=1(\operatorname{logit}^{-1}(q)>\omega)$ where $\omega$ is a logistic random variable. 

I don't know if there is a deterministic function representation for a Poisson random variable.