What is the relationship between the spaces $X_1\triangleq \mathscr{P}(C([0,1],\mathbb{R}))$ and $X_2\triangleq C([0,1],\mathscr{P}(\mathbb{R}))$; where $\mathscr{P}(\cdot)$ denotes the Borel probability measures on a space and it is equipped with the totalvariation topology. Specifically, I wonder, is $X_1$ continuously embeded in $X_2$ or the converse?

$\begingroup$ There is a natural "projection" from the space of processes $X_1$ into the space of families of measures $X_2$, but it would require the weak topology on $\mathbb R$ rather than the total variation topology. This projection takes a stochastic process and assigns to it the corresponding family of onedimensional distributions. $\endgroup$ – Mateusz Kwaśnicki Jan 15 at 16:05

1$\begingroup$ Rethinking the question: I bet there is a corresponding embedding $\Phi$ from $X_2$ into $X_1$: take a family of measures $\mu = (\mu_t, t \in [0,1])$ in $X_2$, denote by $F_t(x)$ the corresponding distribution functions, define $x_t(\omega) = F_t^{1}(\omega)$ for $\omega \in [0, 1]$, and set $\Phi(\mu)$ to be the law of the process $x_t$ with $\omega$ distributed uniformly over $[0,1]$. $\endgroup$ – Mateusz Kwaśnicki Jan 15 at 16:12

$\begingroup$ @MateuszKwaśnicki The first point seems very natural bt the second is particularly nice. Have you ever seen this in the literature? $\endgroup$ – Wasserstein's Apprentice Jan 15 at 16:39

$\begingroup$ @MateuszKwaśnicki As your comment perfectly answers the question, I suggest that you post it as an answer (I was planning to write this answer when I saw you had already written it as a comment.) $\endgroup$ – Yuval Peres Jan 15 at 17:08

$\begingroup$ @YuvalPeres: Rethinking again, the construction does not seem to work, really. I just posted an extended version of my comment as an answer. $\endgroup$ – Mateusz Kwaśnicki Jan 15 at 17:48
This is an extended version of my comment above. It is not an answer, or at least: not a positive answer. (Perhaps it is sort of a negative answer?)
There is a (sort of natural) candidate for an embedding $\Phi$ from $X_2$ into $X_1$, defined as follows. Take a family of measures $\mu = (\mu_t, t \in [0,1])$ in $X_2$, and denote by $F_t(x)$ the corresponding distribution functions. Let $F_t^{1}$ be the generalised inverse function of $F_t$, so that if $U$ is a random variable uniformly distributed on $[0, 1]$, then $$ F_t^{1}(U) $$ is a random variable with distribution $\mu_t$. Finally, define a stochastic process $$ x_t = F_t^{1}(U) ,$$ and set $\Phi(\mu)$ to be the law of the process $x_t$ (again with $U$ distributed uniformly over $[0,1]$).
The above construction is natural in the following sense: $x_t$ is the unique process with onedimensional distributions (a.k.a. marginals) $\mu_t$ with the following monotonicity property: if $x_t(\omega_1) \leqslant x_t(\omega_2)$ for some $t$, then $x_t(\omega_1) \leqslant x_t(\omega_2)$ for all $t$.
The above 'embedding' looks nice, but it remains to prove that $\Phi(\mu)$ is indeed in $X_1$. In other words: $x_t$ has almost surely continuous paths. Equivalently: $t \mapsto F_t^{1}(u)$ is continuous for almost every $u \in [0, 1]$. This need not be true, though!
Set $\mu_t = (1  t) \delta_0 + t \delta_1$. Then $x_t = 0$ for $t < U$ and $x_t = 1$ for $t > U$, which means that the path of $x_t$ is almost surely discontinuous!
Clearly, $\mu$ constructed above is not a family of onedimensional distributions of a continuous stochastic process, so in a sense there is no hope to fix the above construction. The only thing one can easily say is that if $t \mapsto \mu_t$ is a function of bounded variation (with respect to the total variation distance), then $x_t$ is almost surely a function of bounded variation: $$ \mathbb E \sum_i x_{t_i}x_{t_{i1}} = \sum_i \\mu_{t_i}  \mu_{t_{i1}}\_{TV} \leqslant \operatorname{TV}(t \mapsto \mu_t) , $$ and hence $$ \mathbb E \operatorname{TV}(t \mapsto x_t) \leqslant \operatorname{TV}(t \mapsto \mu_t) . $$

$\begingroup$ So, it seems to me that the problem is the total variation topology...Do you think it would work if we swap it in for the (relative) weak topology? $\endgroup$ – Wasserstein's Apprentice Jan 15 at 22:14

$\begingroup$ With a weaker topology, things get even worse... For this construction to work, either $X_2$ must be restricted to a smaller class (the infinity Wasserstein distance is a good candidate!), or continuity conditions in $X_1$ have to be relaxed. $\endgroup$ – Mateusz Kwaśnicki Jan 15 at 22:25

$\begingroup$ Though it is tautological, can we say anything about the subset $\tilde{X}_1\subset X_1$ of families of measures which push to a process with $a.s.$continuou paths, when applying $\Phi$? $\endgroup$ – Wasserstein's Apprentice Jan 17 at 10:51

1$\begingroup$ That's a good question. My comment above intended to suggest continuity with respect to the infinity Wasserstein distance, but I have not thought about it carefully. $\endgroup$ – Mateusz Kwaśnicki Jan 17 at 21:04

$\begingroup$ Sure, $t \mapsto x_t'$ is continuous (since $t \mapsto F_t^{1}(u)$ is integrable for all $u$). $\endgroup$ – Mateusz Kwaśnicki Feb 18 at 19:54