Markov inequality

Hello,

A few remarks, while I'm trying to solve it.

Quote:

Originally Posted by jovial

Dear All
I have problem with the following inequality

Let $f = f(x)$ be a nonnegative even function that is strictly increasing for positive $x$ . Then for a random variable $\xi$ with $|\xi(\omega)|\leq C$ .

$\frac{{Ef(\xi ) - f(\varepsilon )}}{{f(C)}} \le P\left\{ {\left| {\xi - E\xi } \right| \ge \varepsilon } \right\} \le \frac{{Ef(\xi - E\xi )}}{{f(\varepsilon )}}.$

By Markov inequality

$P\{\xi \ge \epsilon\} \le \frac{E\xi}{\epsilon}$

False. It's $P\{|\xi |\ge\epsilon\}\le\frac{E|\xi|}{\epsilon}$

Quote:

I consider $\eta = |\xi - E\xi|$ . Now if we have a function $f: X \rightarrow R^{+}$ then

$P\{f(\eta) \ge f(\epsilon)\} \le \frac{Ef(\eta)}{f(\epsilon)}$
(Is it correct and why)

It's correct :
since f is increasing, $P\{\eta\ge\epsilon\}=P\{f(\eta)\ge f(\epsilon)\}$ and by Markov's inequality, this is $\le \frac{Ef(\eta)}{f(\epsilon)}$

Quote:

Now if $f$ is strictly increasing function then

$P\{\eta \ge \epsilon\}=P\{ f(\eta) \ge f(\epsilon)\} \le \frac{Ef(\eta)}{f(\epsilon)}$

So I obtained:

$P\{|\xi - E\xi| \ge \epsilon\} \le \frac{Ef(|\xi - E\xi|)}{f(\epsilon)}$

But you're asked for $Ef(\xi-E\xi)$ in the RHS, there is no absolute value.
Although since f is even, $f(|\xi-E\xi|)=f(\xi-E\xi)$ .

P.S. : Are you French ? :p