# Proof of an inequality with expectation of random variables

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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you're absolutely right! sorry for the mistake! I corrected it. –  user20869 Dec 11 '11 at 15:07
It's clear that I have to use the first Inequality. But I do not see how. –  user20869 Dec 11 '11 at 15:23

just two small question: why is $E(|h(X)-h(Y)|1\{|X-Y|\ge \epsilon\}1\{(X,Y)\in D(\epsilon)\}) \le E(1\{|X-Y|\ge \epsilon\}1\{(X,Y)\in D(\epsilon)\})$ and why is $E(|h(X)-h(Y)|) \le E(|X-Y|)$ ? And in the last line it should be a $h$ instead of $g$ –  user20869 Dec 11 '11 at 16:06
We have $|h(X)-h(Y)|\leq 1$, and $|h(u)-h(v)|=|\int_u^ve^{-t}dt|\leq |u-v|$. It's indeed $h$, I will correct it. –  Davide Giraudo Dec 11 '11 at 16:12
Ok thanks. But there's still something around I don't get. Why is it true that, $h(X)+h(Y) \le 2min(h(X),h(Y))$. –  user20869 Dec 11 '11 at 17:07
$h$ is decreasing so $h(X)\leq h(\min(X,Y))$. (it's $h(\min)$, not $\min h$) –  Davide Giraudo Dec 11 '11 at 17:09