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I'm solving some exercises and got stuck. The setting:

let $\varepsilon > 0 $ given, and D(\varepsilon):=\{(x,y) \in \mathbb{R_+}^2\mid |x-y| \ge \varepsilon\mbox{ and }\min(x,y) \le \frac{1}{\varepsilon}\}. Define $ h(x):=\exp{(-x)} $ with domain $ \mathbb{R_+} $. I've proved, that there's a $ \delta > 0 $ (depending on $\varepsilon $ )

$ h\left(\frac{x+y}2\right) \le \frac{h(x)+h(y)}2 - \delta 1_{D(\varepsilon)}(x,y).$

Now how could I use this to show:

$ E(|h(X)-h(Y)|)\le \varepsilon + 2 h\left(\frac{1}{\varepsilon}\right)+ P((X,Y)\in D(\varepsilon))$

where $X,Y$ non negative random variables.

I'm very thankful for any help.

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    It's clear that I have to use the first Inequality. But I do not see how.2011-12-11

1 Answers 1

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Too long to be explained in a comment. We write \begin{align*} E[|h(X)-h(Y)|]&=E[|h(X)-h(Y)|\mathbf 1_{|X-Y|<\varepsilon}]+E[|h(X)-h(Y)|\mathbf 1_{|X-Y|\geq \varepsilon}]\\ &=E[|h(X)-h(Y)|\mathbf 1_{|X-Y|<\varepsilon}]+E[|h(X)-h(Y)|\mathbf 1_{|X-Y|\geq \varepsilon}\mathbf 1_{(X,Y)\in D(\varepsilon)}]\\ &+E[|h(X)-h(Y)|\mathbf 1_{|X-Y|\geq \varepsilon}\mathbf 1_{(X,Y)\notin D(\varepsilon)}]\\ &\leq E[|X-Y|\mathbf 1_{|X-Y|<\varepsilon}]+E[|h(X)-h(Y)|\mathbf 1_{|X-Y|\geq \varepsilon}\mathbf 1_{(X,Y)\in D(\varepsilon)}]\\ &+E[|h(X)-h(Y)|\mathbf 1_{\min(X,Y)\geq \frac 1\varepsilon}]\\ &\leq \varepsilon + E[\mathbf 1_{|X-Y|\geq \varepsilon}\mathbf 1_{(X,Y)\in D(\varepsilon)}]+E[(h(X)+h(Y))\mathbf 1_{\min(X,Y)\geq \frac 1\varepsilon}]\\ &\leq \varepsilon+P((X,Y)\in D(\varepsilon))+2E[h(\min(X,Y))\mathbf 1_{\min(X,Y)\geq \frac 1\varepsilon}] \\ &\leq \varepsilon +P((X,Y)\in D(\varepsilon))+2 h\left(\frac 1\varepsilon\right). \end{align*} But I don't see how to use the previous question.

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    decreasing, that's the point! Thanks a lot! and again sorry for my ignorance!2011-12-11