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I seek to bound the total-variation distance between two probability measures $p_1$ and $p_2$. It is extremely easy to build a parameter space where $p_1$ and $p_2$ are the marginals of some joint probability distribution (see below for details) so it's extremely easy to control the Wasserstein-k distance between the two.

However, what I'm really interested is the total-variation distance between the two. Is there some hope for me there ? Obviously, any sort of bound would involve the standard deviation of the distributions, and would only work for co-measurable $p_1$ and $p_2$. Do you know anything of that kind ?

Details: $p_1$ and $p_2$ are the probability distributions of the position of a particle behaving according to the same stochastic differential equation, at the same time t, and initialized at slightly different positions. Ie, $X_1$ and $X_2$ both respect:

$$ \dot X = - \nabla \phi(X) + 2 dW $$

and their initial positions are very close to one another. And $p_i$ are the marginals of the position of the particle at some time $t$.

It's easy to couple them by considering the processes driven by the same Wiener process dW, but I don't see how to get good total-variation bounds that way

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    $\begingroup$ On an infinite set, no such bound is possible; see Theorem 4 in Gibbs, A. L. and Su, F. E. (2002), On Choosing and Bounding Probability Metrics. International Statistical Review, 70: 419–435 . $\endgroup$
    – Christian Clason
    Commented Jan 31, 2016 at 13:44
  • $\begingroup$ If $X_1$ and $X_2$ both solve the same SDE you've written but are started from distinct initial positions, then the laws (on the path space) of these processes are singular. In this case, there's no (nontrivial) bound on the total variation. If the initial positions are random, and if the law $\mu_1$ of $X_1(0)$ is absolutely continuous to the law $\mu_2$ of $X_2(0)$, then the relative entropy on the path space $H(Law(X_1)|Law(X_2))$ is equal to the relative entropy of the first marginals $H(\mu_1|\mu_2)$. Then use Pinsker's inequality to get a good bound on the total variation. $\endgroup$
    – Dan
    Commented Jan 31, 2016 at 14:59
  • $\begingroup$ You misunderstood me slightly (because I explained poorly \^\^). I mean to bind the total-variation distance between the marginals at time t, and not the distribution of the paths. @ChristianClason : wouldn't the claim in theorem 4 of Gibbs et al not apply only to trivial examples like: $p_1$ is a binomial (only rational values) while $p_2$ is a gaussian. For such an example, the wasserstein can tend to 0, while TV convergence can't happen because these aren't co-measurable. Couldn't a co-measurability condition of some kind save me from such boring counter-examples ? $\endgroup$
    – Guillaume Dehaene
    Commented Jan 31, 2016 at 18:35
  • $\begingroup$ @GuillaumeDehaene That's why I only wrote a comment, not an answer. I'd expect that co-measurability alone would not be sufficient, and you'd need to use more specific properties of the two marginals you wish to compare (i.e., prove the bound for the concrete case you're interested in, rather than apply a general theorem). $\endgroup$
    – Christian Clason
    Commented Jan 31, 2016 at 22:35

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Guillaume, the answer given by Dan contains a response to your question. Indeed, you can use the fact that the TV distance between the distributions of $X_t^{(1)}$ and $X_t^{(2)}$ is smaller than the TV-distance between the distributions of these processes on the path space $C([0,t])$ up to time $t$. Then, you can use the Pinsker inequality in conjunction with the Girsanov formula. More details can be found in https://arxiv.org/pdf/1412.7392v3.pdf (cf, in particular, Eq (17))

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