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A statistical model is a function $P : \Theta \to \Delta(X)$, where $\Theta$ is a parameter space, and $\Delta(X)$ is the set of probability measures on a state space $X$.

Suppose that $\Theta$ and $X$ are topological spaces, and equip $\Delta(X)$ with the topology of weak convergence of measures. Suppose that the function $P : \Theta \to \Delta(X)$ is continuous, defining a topological statistical model.

An $X$-valued random variable is a measurable function $x : \Omega \to X$ on some abstract probability space $(\Omega, \mathcal F, \mathbb P)$. The push-forward measure $x_* \mathbb P := \mathbb P \circ x^{-1}$ is called the law or marginal of $x$.

If $P : \Theta \to \Delta(X)$ defines a statistical model, can we find an abstract probability space $(\Omega, \mathcal F, \mathbb P)$ and a family of $X$-valued random variables $\{x_\theta\}$ so that $(x_\theta)_* \mathbb P = P_\theta$ for each $\theta \in \Theta$?

Moreover, if $P$ defines a topological statistical model, what regularity properties does this imply about the higher-order function $\theta \mapsto x_\theta$? If so, can the space $L = L(\Omega,X)$ of $X$-valued random variables be equipped with a natural topology so that the function $x : \Theta \to L$ is continuous?

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    $\begingroup$ For the nontopological case, you can simply take $\Omega=X^\Theta$, $\mathcal{F}$ the product $\sigma$-algebry, $\mathbb{P}=\otimes_{\theta\in\Theta}P_\theta$ and $x_\theta$ the projection onto the $\theta$'s factor. $\endgroup$
    – Michael Greinecker
    Commented Nov 16, 2013 at 8:51
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    $\begingroup$ I'm kind of lost: as Michael Greinecker recalled, you can always build a huge probability space in which all your variables live, whatever $\Theta$ is, thanks to the tensor product construction. Then, maybe the "natural" topology you look for just the convergence in distribution of random variables ? In this case, that $x:\Theta\rightarrow L$ is continuous is by definition equivalent to $P:\Theta\rightarrow \Delta(X)$ continuous. But maybe I misunderstood the question. $\endgroup$
    – Adrien Hardy
    Commented Nov 16, 2013 at 9:58
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    $\begingroup$ The tricky part is regularity, then. This is the whole point of Kolmogorov's continuity theorem: the product construction does not guarantee regularity. I have to think about @John Dawkins' answer a little more. $\endgroup$
    – Tom LaGatta
    Commented Nov 16, 2013 at 14:38

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D. Blackwell and L. Dubins [An extension of Skorohod's almost sure representation theorem, Proc. Amer. Math. Soc., vol. 89} (1983) 691–692] have shown that is $X$ is a reasonable space then there is a map $P\to f_P$ of $\Delta(X)$ into $L$ (based on a well-chosen probability space $(\Omega,{\mathcal F}, {\mathbb P})$) such that, for each $P\in\Delta(X)$, (i) the push forward of ${\mathbb P}$ by $f_P$ is $P$ and (ii) $Q\mapsto f_Q(\omega)$ is continuous at $P$, for a.e. $\omega\in\Omega$.

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