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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.

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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 $b<a\implies P(|X-Y|>b)>b$.
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.

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    $\begingroup$ Thanks. A small correction, in the upper bound, the second term is $\int_{a/a+1}^{1}adt = a/(1+a)$. $\endgroup$ – manifolded Sep 4 at 10:41
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    $\begingroup$ @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. $\endgroup$ – Gabriel Romon Sep 4 at 10:45
  • $\begingroup$ 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 $E(\frac{|X-Y|}{1+|X-Y|})$ :) $\endgroup$ – manifolded Sep 4 at 12:14

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