2
$\begingroup$

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 $\eta$).

$\endgroup$

1 Answer 1

1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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