# A question about application of the Ky Fan metric

Q) Let $$\alpha(X,Y) = \text{inf}\{\epsilon\geq 0: P(|X-Y|>\epsilon)\leq\epsilon\}$$ be the Ky Fan metric and let $$\beta(X,Y) = E(|X-Y|/1+|X-Y|)$$. If $$\alpha(X,Y) = a$$, then prove that

$$a^2/(1+a)\leq \beta(X,Y)\leq a+(1-a)a/(1+a)$$ is what Durrett says but as obtained in the answer below, I think the upper bound should be $$a+a[1-\frac{a}{1+a}]$$

I am looking for a starting point. Appreciate a hint. Thanks.

## 1 Answer

It is standard that the infimum in the definition of the Ky-Fan metric is attained, that is $$P(|X-Y|>a)\leq a$$.
By definition of $$a$$, we have $$bb)>b$$.
Also, if $$b\geq a$$, $$P(|X-Y|>b)\leq P(|X-Y|>a)\leq a$$.

Let $$\displaystyle f:x\mapsto \frac{x}{1+x}$$. For the upper bound, \begin{aligned} E(f(|X-Y|) &= \int_0^\infty P(f(|X-Y|)\geq t) dt \\ &= \int_0^{1} P(|X-Y|\geq \frac{t}{1-t}) dt\\ &\leq \int_0^{a/(1+a)} 1 \; dt + \int_{a/(1+a)}^{1} a\; dt \\ &= \frac{a}{1+a} + a\left(1-\frac{a}{1+a}\right)\\ &=a+\frac{a(1-a)}{a+1} \end{aligned}

For the lower bound, consider $$\delta\in (0,a)$$ \begin{aligned} E(f(|X-Y|) &\geq E(f(|X-Y|)1_{|X-Y|>a-\delta})\\ &\geq f(a-\delta) P(|X-Y|>a-\delta)\\ &\geq f(a-\delta)(a-\delta) \end{aligned} Letting $$\delta \to 0$$ yields $$\displaystyle E(f(|X-Y|)\geq \frac{a^2}{1+a}$$, as wanted.

• Thanks. A small correction, in the upper bound, the second term is $\int_{a/a+1}^{1}adt = a/(1+a)$. – manifolded Sep 4 at 10:41
• @manifolded thanks for catching that. If you know that the Ky-Fan metric metrizes convergence in probability, then I guess the inequalities hint that $(X,Y)\mapsto E\left(\frac{|X-Y|}{1+|E-Y|}\right)$ also metrizes convergence in probability. – Gabriel Romon Sep 4 at 10:45
• Neat trick for the lower bound as we know we have to setup $P(|X-Y|>a-\delta)$! Good observation about convergence in probability induced by the metric $E(\frac{|X-Y|}{1+|X-Y|})$ :) – manifolded Sep 4 at 12:14