4
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ 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 Nov 16 '13 at 8:51
  • 1
    $\begingroup$ 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 Nov 16 '13 at 9:58
  • 1
    $\begingroup$ 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 Nov 16 '13 at 14:38
4
$\begingroup$

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. $\omega\in\Omega$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.