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Consider the following situation:

  • $S, T$ are standard Borel spaces (say $S = [0,1]^k$, $T = [0,1]$ if it is helpful).

  • There is a a random variable $\zeta: \Omega \to S$.

  • $f_n(\zeta) \to^d \eta$, i.e. $f_n(\zeta)$ converges in distribution to some random variable $\eta$.

Will there always exist

  • A random variable $\zeta': \Omega' \to S$ (i.e. possibly with a different base space than $\Omega$), and

  • functions $g_n, g: S \to T$

such that

  • $\zeta$ and $\zeta'$ have the same distribution,

  • for all $n$, $g_n(\zeta')$ has the same distribution as $f_n(\zeta)$, and

  • $g_n(\zeta') \to g(\zeta')\ a.e$, i.e. $g_n(\zeta')$ converges to $g(\zeta')$ almost everywhere (and in particular $g(\zeta')$ has the same distribution as $\eta$).

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1 Answer 1

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This cannot work without additional assumptions, because of the following theorem on weak convergence. Given a probability measure $P$ on $S \times T$, let $\mu = P(\cdot \times T)$ denote the first marginal. If $\mu$ is nonatomic, then there exists a sequence of measurable functions $g_n : S \rightarrow T$ such that, if $P_n$ denotes the image of $\mu$ under the map $S \ni x \mapsto (x,g_n(x)) \in S \times T$, then $P_n$ converges weakly to $P$. As a result, the set joint distributions which are concentrated on the graph of a function is not closed, and it is in fact dense in the set of all joint distributions. In your setting, $g_n(\zeta') \rightarrow g(\zeta')$ a.e. would imply that $(\zeta,f_n(\zeta)) \rightarrow (\zeta,g(\zeta))$ in distribution, and we see that this requires additional assumptions on $f_n$.

This density result is well known in the theory of Young measures, and this book of Valadier (Proposition 8, page 22 of PDF) seems to have a nice short proof in the case of Euclidean spaces. It is likely proven for more general spaces in the 2004 book of Castaing, De Fitte, and Valadier, but I don't have a copy.

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