# Embeddings of spaces of probability measures

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 total-variation topology. Specifically, I wonder, is $$X_1$$ continuously embeded in $$X_2$$ or the converse?

• 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 one-dimensional distributions. – Mateusz Kwaśnicki Jan 15 at 16:05
• Re-thinking 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 $[0,1]$. – Mateusz Kwaśnicki Jan 15 at 16:12
• @MateuszKwaśnicki The first point seems very natural bt the second is particularly nice. Have you ever seen this in the literature? – Wasserstein's Apprentice Jan 15 at 16:39
• @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.) – Yuval Peres Jan 15 at 17:08
• @YuvalPeres: Re-thinking again, the construction does not seem to work, really. I just posted an extended version of my comment as an answer. – Mateusz Kwaśnicki Jan 15 at 17:48

## 1 Answer

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 one-dimensional 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 one-dimensional 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: $$\mathbb E \sum_i |x_{t_i}-x_{t_{i-1}}| = \sum_i \|\mu_{t_i} - \mu_{t_{i-1}}\|_{TV} \leqslant \operatorname{TV}(t \mapsto \mu_t) ,$$ and hence $$\mathbb E \operatorname{TV}(t \mapsto x_t) \leqslant \operatorname{TV}(t \mapsto \mu_t) .$$

• 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? – Wasserstein's Apprentice Jan 15 at 22:14
• 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. – Mateusz Kwaśnicki Jan 15 at 22:25
• 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 $\Phi$? – Wasserstein's Apprentice Jan 17 at 10:51
• 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. – Mateusz Kwaśnicki Jan 17 at 21:04
• Sure, $t \mapsto x_t'$ is continuous (since $t \mapsto F_t^{-1}(u)$ is integrable for all $u$). – Mateusz Kwaśnicki Feb 18 at 19:54