1. ## Markov inequality

Dear All
I have problem with the following inequality

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

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

By Markov inequality

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

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

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

Now if $\displaystyle f$ is strictly increasing function then

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

So I obtained:

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

How can be proved this part.

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

Thanx

2. Hello,

A few remarks, while I'm trying to solve it.
Originally Posted by jovial
Dear All
I have problem with the following inequality

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

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

By Markov inequality

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

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

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

Now if $\displaystyle f$ is strictly increasing function then

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

So I obtained:

$\displaystyle P\{|\xi - E\xi| \ge \epsilon\} \le \frac{Ef(|\xi - E\xi|)}{f(\epsilon)}$
But you're asked for $\displaystyle Ef(\xi-E\xi)$ in the RHS, there is no absolute value.
Although since f is even, $\displaystyle f(|\xi-E\xi|)=f(\xi-E\xi)$.

P.S. : Are you French ?