Is the infimum of the Ky Fan metric achieved? - MathOverflow

Is the infimum of the Ky Fan metric achieved?

2010-08-15T22:53:00Z

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:

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

Answer by has2 for Is the infimum of the Ky Fan metric achieved?

2010-09-12T14:47:22Z

Because what follows doesn't fit in a comment, I write it here as an answer; but they are merely comments. After computing the minimizer for several simple distributions, my impression is that the answer to this question is yes, and there will be many minimizers.

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.

Answer by Jon Bannon for Is the infimum of the Ky Fan metric achieved?

2010-10-26T19:58:08Z

This probably helps not at all, but I saw you were interested in Ky-Fan metric, and friends of mine have looked at these in a noncommutative setting in which there are some "extreme value" properties. Maybe there's something useful in there for you: http://arxiv.org/PS_cache/arxiv/pdf/0707/0707.4239v3.pdf