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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?

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  • $\begingroup$ isn't this the definition of a marginal distribution? $\endgroup$ Commented Jul 2, 2014 at 13:26
  • $\begingroup$ @CarloBeenakker Almost. Marginalization tells you how to obtain p(x) from p(x,y) for some well-defined multivariate random variable (x,y). But I'm rather asking what is required of y in order for (x,y) to be well defined? E.g. should I define the probability space of y, and define the probability space of (x,y) as the product space of that of x and that of y? Or must y be defined on the same probability space as x? Obviously, I'm not well versed in the foundations of probability theory. I'm also asking about regularity conditions, although that is of less interest. $\endgroup$
    – Patrick
    Commented Jul 2, 2014 at 13:56
  • $\begingroup$ Is $p$ meant to be a density? $\endgroup$ Commented Jul 3, 2014 at 11:13
  • $\begingroup$ Yes, using "function overloading" notation. $\endgroup$
    – Patrick
    Commented Jul 3, 2014 at 14:42

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