Intuition tells me that $$ p(x\,|\,y) = \int p(x,\theta\,|\,y) \; d\theta$$ by the "law of marginalization", pretty much for any object $\theta$.

I would like to make this statement rigorous, however. For example, let us assume that the random variable $X$ is well defined and absolutely continuous. What minimal conditions are required of $\theta$ in order to make the statement above? I'm thinking

• probability space of $\theta$
• joint measurability of $\theta$ and $x$
• absolute continuity of $\theta$

Also, does the conditioning on y cause any complications?

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###Update 1### Briefly, the motivation for this question is that I want to make the above marginalization statement before defining $\theta$. So I want to say "Don't worry that we have not defined $\theta$ yet, because this is just the rule of marginalization, and it holds for any random variable $\theta$ defined on $(\Omega, S, P)$"

Would it then be sufficient to require that $(\Omega, S, P)$ is also the probability space of $X$, and that the pdf of $Y$ is also absolutely continuous?

Intuition tells me that $$ p(x\,|\,y) = \int p(x,\theta\,|\,y) \; d\theta$$ by the "law of marginalization", pretty much for any object $\theta$.

I would like to make this statement rigorous, however. For example, let us assume that the random variable $X$ is well defined and absolutely continuous. What minimal conditions are required of $\theta$ in order to make the statement above? I'm thinking

• probability space of $\theta$
• joint measurability of $\theta$ and $x$
• absolute continuity of $\theta$

Also, does the conditioning on y cause any complications?

$\phantom{placeholder}$

Update 1

Briefly, the motivation for this question is that I want to make the above marginalization statement before defining $\theta$. So I want to say "Don't worry that we have not defined $\theta$ yet, because this is just the rule of marginalization, and it holds for any random variable $\theta$ defined on $(\Omega, S, P)$"

Would it then be sufficient to require that $(\Omega, S, P)$ is also the probability space of $X$, and that the pdf of $Y$ is also absolutely continuous?

Intuition tells me that $$ p(x\,|\,y) = \int p(x,\theta\,|\,y) \; d\theta$$ by the "law of marginalization", pretty much for any object $\theta$.

I would like to make this statement rigorous, however. For example, let us assume that the random variable $X$ is well defined and absolutely continuous. What minimal conditions are required of $\theta$ in order to make the statement above? I'm thinking

• probability space of $\theta$
• joint measurability of $\theta$ and $x$
• absolute continuity of $\theta$

Also, does the conditioning on y cause any complications?

Intuition tells me that $$ p(x\,|\,y) = \int p(x,\theta\,|\,y) \; d\theta$$ by the "law of marginalization", pretty much for any object $\theta$.

I would like to make this statement rigorous, however. For example, let us assume that the random variable $X$ is well defined and absolutely continuous. What minimal conditions are required of $\theta$ in order to make the statement above? I'm thinking

• probability space of $\theta$
• joint measurability of $\theta$ and $x$
• absolute continuity of $\theta$

Also, does the conditioning on y cause any complications?

$\phantom{placeholder}$

###Update 1### Briefly, the motivation for this question is that I want to make the above marginalization statement before defining $\theta$. So I want to say "Don't worry that we have not defined $\theta$ yet, because this is just the rule of marginalization, and it holds for any random variable $\theta$ defined on $(\Omega, S, P)$"

Would it then be sufficient to require that $(\Omega, S, P)$ is also the probability space of $X$, and that the pdf of $Y$ is also absolutely continuous?