Let $X \sim P$ for some distribution $P$ and $Y \sim Q(\cdot | X)$ for some distribution $Q$. Assume that $P$ has a density $p$, with respect to some ground measure $\mu$ and $Q(\cdot|X)$ has a density $q(\cdot | X)$ with respect to the same $\mu$ for every $X$. How do I express the likelihood function for a point $y$ in terms of the densities $q(\cdot | X)$ and $p$?
What I've tried
Let $G$ be the marginal distribution of $Y$. Assume $G$ has density $g$. I want to express $g(y)$ in terms of $q(y | X)$ and $p(y)$. My understanding is that this reduces to:
$$\frac{d\mathbb E [ Q(\cdot | X)]}{d\mu}=\frac{d\mathbb E [ Q(\cdot | X)]}{dP}\frac{dP}{d\mu}$$
But I don't know how to proceed from here. It does not make sense to me to exchange the order of the expectation and the Radon-Nikodym derivative (without justification). Any help would be appreciated.
Suppose that $P$ and $Q$ are both measures on $S$, where $S$ is some measurable space. Then \begin{align*} G(A)&= \mathbb{E}[Q(A \: | \: X)] \\ &= \int_S Q(A \: | \: x) \: P(dx) \\ &= \int_S \int_A Q(dy \: | \: x) P(dx) \\ &= \int_S \int_A q(y \: | \: x)p(x) \mu(dy)\mu(dx) \\ &= \int_A \int_S q(y \: | \: x)p(x) \mu(dx) \mu(dy) \quad \quad\text{(using Tonellis theorem.)} \end{align*} It immediately follows that $G$ is absolutely continuous with respect to $\mu$ with density $$\frac{dG}{d\mu}(y) = \int_S q(y \: | \: x)p(x) \mu(dx).$$