Suppose $\mu \in P(\mathbb{R}^d)$ and for each $n$, $T_n:\mathbb{R^d} \rightarrow \mathbb{R^d}$ is such that the pushforward $T_n \# \mu$ has a finite second moment. If $\{T_n \# \mu\}$ converges weakly to $\nu \in P(\mathbb{R}^d)$, could we find $T:\mathbb{R^d} \rightarrow \mathbb{R^d}$ such that $\nu = T_\# \mu$ and the pushforward $T \# \mu$ also has a finite second moment. This problem is very similar to Limit of pushforward measures of random variables is "represented" by a random variable. It would be great if any reference book and paper could be given.
3
-
2$\begingroup$ The maps $T_n$ aren't really doing anything, you can just focus on the $\nu_n=T_{n*}\mu\to\nu$, and of course $\nu$ need not have finite second moment even if the $\nu_n$'s do. For example, take $\nu=c\sum 1/k^2\delta_k$ and $\nu_n$ as a cut off version of this. $\endgroup$– Christian RemlingCommented Feb 6 at 0:44
-
$\begingroup$ In particular, if $\mu$ is atpmless then $T_n\# \mu$ can be any probability measure whatsoever. So yeah, assuming the measures are of the form $T_n\# \mu$ is no restriction at all. $\endgroup$– Nate EldredgeCommented Feb 6 at 3:07
-
$\begingroup$ In order to get the limiting measure $\nu$ to have finite second moment, the natural sufficient condition would be for the measures $T_n\# \mu$ to have uniformly bounded second moments. $\endgroup$– Nate EldredgeCommented Feb 6 at 3:08
Add a comment
|