# Is the infimum of the Ky Fan metric achieved?

Consider the probability space $(\Omega, {\cal B}, \lambda)$ where $\Omega=(0,1)$, ${\cal B}$ is the Borel sets, and $\lambda$ is Lebesgue measure.

For random variables $W,Z$ on this space, we define the Ky Fan metric by

$$\alpha(W,Z) = \inf \lbrace \epsilon > 0: \lambda(|W-Z| \geq \epsilon) \leq \epsilon\rbrace.$$

Convergence in this metric coincides with convergence in probability.

Fix the random variable $X(\omega)=\omega$, so the law of $X$ is Lebesgue measure, that is, ${\cal L}(X)=\lambda$.

Question: For any probability measure $\mu$ on $\mathbb R$, does there exist a random variable $Y$ on $(\Omega, {\cal B}, \lambda)$ with law $\mu$ so that $\alpha(X,Y) = \inf \lbrace \alpha(X,Z) : {\cal L}(Z) = \mu\rbrace$ ?

Notes:

1. By Lemma 3.2 of Cortissoz, the infimum above is $d_P(\lambda,\mu)$: the Lévy-Prohorov distance between the two laws.

2. The infimum is achieved if we allowed to choose both random variables. That is, there exist $X_1$ and $Y_1$ on $(\Omega, {\cal B}, \lambda)$ with ${\cal L}(X_1) = \lambda$, ${\cal L}(Y_1) = \mu$, and $\alpha(X_1,Y_1) = d_P(\lambda,\mu)$. But in my problem, I want to fix the random variable $X$.

3. Why the result may be true: the space $L^0(\Omega, {\cal B}, \lambda)$ is huge. There are lots of random variables with law $\mu$. I can't think of any obstruction to finding such a random variable.

4. Why the result may be false: the space $L^0(\Omega, {\cal B}, \lambda)$ is huge. A compactness argument seems hopeless to me. I can't think of any construction for finding such a random variable.

• How is the fact that you want to fix the rv $X$ relevant? Isn't it true that for any two rvs $X_1$ and $X_2$ with the same law, there is an isomorphism of $(0,1)$ taking $X_1$ to $X_2$? – Ori Gurel-Gurevich Oct 26 '10 at 20:20
• @Ori: I'm not quite sure what kind of isomorphism you have in mind. The random variables $X_1(\omega)=2\omega-\lfloor 2\omega\rfloor$ and $X_2(\omega)=3\omega-\lfloor 3\omega\rfloor$ are both uniform (0,1) random variables, but neither can be written as a function of the other. It's certainly possible that an easy transformation or observation solves the problem. I'd be glad to hear about it! – user6096 Oct 27 '10 at 1:26
• You're right, I didn't understand the question at first. The way I understand it now I would almost say it is not a question in probability as it depends on the representation of the random variable in question. – Ori Gurel-Gurevich Oct 27 '10 at 2:47
• In point 2) are you appealing to Strassen theorem? Don't you need the laws of both X and Y to be tight? – Ngoc Mai Tran Feb 25 '11 at 8:13

Intuitively, it seems possible to build an optimizer as follows: we are given the law $\mu$ and we would like to find a function $f$ such that 1) the distribution of $f$ is $\mu$ and 2) $\alpha(f,X)$ is minimum. Let $\epsilon > 0$ be this minimum. Let $F(x) = \mu( (-\infty, x])$, i.e., $F$ is the distribution function associated with $\mu$. Let $G$ be the inverse function of $F$: $G(x) \doteq \inf\{y: F(y) \ge x \}$. By its definition $G$'s distribution is $\mu$. Draw the graphs of the functions $l(x) = x + \epsilon$ and $u(x) = x -\epsilon$ around the graph of the function $X(x) = x$. To get the minimizer, one cuts the graph of $G$ into $n$ small pieces with lines parallel to the $x$ axis and shifts around the pieces along these lines so that they lie between the graphs of $l$ and $u$ as much as possible. As the number of pieces increase and their size decreases you would expect this to converge to a function that is the desired minimizer. The result will depend on the particulars of this process.
As to non-uniqueness: suppose $f$ is a minimizer. Denote with $E$ the subset of $[0,1]$ over which $f$ differs from $X$ by at least $\epsilon$. The values that $f$ takes over $E$ can be freely permuted without affecting the distribution and the distance between $f$ and $X$. So there will be infinitely many minimizers, when there is one.