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For two random variables $X$ and $Y$ taking values in $\mathbb{R}^m$, the convex distance $d_c$ is defined as

$$d_c(X,Y) = \sup_{h} \lvert \operatorname{E}(h(X)) - \operatorname{E}(h(Y)) \rvert,$$

where the supremum is taken over all indicator functions of measurable convex subsets of $\mathbb{R}^m$.

For $m=1$, it is easy to see that $d_c$ coincides with the Kolmogorov distance whenever $X$ and $Y$ are continuous, i.e. we have that $d_c(X,Y)= \sup_{x \in \mathbb{R}} \lvert F_X(x) -F_Y(x) \rvert$, where $F_X$ and $F_Y$ denote the cumulative distribution functions of $X$ and $Y$, respectively.

In particular, if $m=1$, we have that if a sequence $(X_n)$ of continuous real-valued random variables converges to another continuous random variable $Y$ in distribution, then $d_c(X_n,Y) \to 0$ as $n \to \infty$. Does this implication continue to hold if $m \geq 2$?

I could neither find a reference, nor a proof of this myself.

EDIT

If continuity of the random variables is defined as having non-atomic distributions, the implication is false (see the counterexample by Iosif Pinelis below).

If continuity is defined as having CDFs which are absolutely continuous with respect to the Lebesgue measure, the question is still open.

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  • $\begingroup$ This doesn't seem right, even for $m=1$. Convergence in Kolmogorov distance is stronger than convergence in distribution, not weaker. Take for instance deterministic $X_n = 1/n$ and $Y=0$. And they don't converge in convex distance either, by taking $h = 1_{\{0\}}$ or $1_{[-1,0]}$. Is there a mistake, or assumptions missing? $\endgroup$ – Nate Eldredge Jan 31 at 12:16
  • $\begingroup$ @NateEldredge Thany you much for your comment, I somehow forgot to add the crucial assumption that the random variables are continuous. $\endgroup$ – r_faszanatas Jan 31 at 12:35
  • $\begingroup$ So, in the second-to-last paragraph, you want $Y$ to be continuous as well as $X_n$? $\endgroup$ – Nate Eldredge Jan 31 at 12:51
  • $\begingroup$ @NateEldredge Yes, corrected this as well, now the question should be well-posed. $\endgroup$ – r_faszanatas Jan 31 at 12:58
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    $\begingroup$ The other way is absolute continuity with respect to the Lebesgue measure. As illustrated by Iosif's answer, this is not equivalent. $\endgroup$ – Yoav Kallus Jan 31 at 17:32
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The answer is no. For instance, let $Y$ be uniformly distributed on the unit sphere in $\mathbb R^m$ (not the ball, but the sphere) and, for each natural $n$, let $X_n:=(1+1/n)Y$. Then the distribution of each of the random vectors $Y,X_1,X_2,\dots$ is non-atomic and hence, in particular, the values of this distribution on any singleton set are $0$; also, the cumulative distribution function of this distribution is continuous.

Moreover, $X_n\to Y$ in distribution, but $d_c(X_n,Y)=1$ for all $n$.

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    $\begingroup$ Thanks a lot for this counterexample. Given the comments above: Can you add to your answer that continuity is defined the way I commented (cumulative distribution function is just continuous)? Also, it would be nice to settle the question for the better definition of absolute continuity with respect to the Lebesgue measure. $\endgroup$ – r_faszanatas Jan 31 at 22:12
  • $\begingroup$ @r_faszanatas : A better definition of the continuity of a random vector is that its distribution be non-atomic (which in the one-dimensional case coincides with the usual definition of the continuity of a random variable). The distributions of $X_n$ and $Y$ in the counterexample are indeed non-atomic. $\endgroup$ – Iosif Pinelis Feb 2 at 1:03
  • $\begingroup$ As always, I learned a lot from this question, thanks again. So now the question is, if the implication holds in case at least the limiting random variable has a CDF which is absolutely continuous with respect to the Lebesgue measure. $\endgroup$ – r_faszanatas Feb 3 at 16:15
  • $\begingroup$ @r_faszanatas : I think the answer is affirmative if $Y$ is absolutely continuous. To keep things in good order, though, I think this additional question should be posted separately. $\endgroup$ – Iosif Pinelis Feb 3 at 18:26
  • $\begingroup$ just asked this as another question, see mathoverflow.net/questions/351870/… $\endgroup$ – r_faszanatas Feb 3 at 21:16

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