Dear all

I have problem with the following question, i have solved one inequality like given below. Can anybody help me further for solving this problem

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

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

I tried like way.

Let $\displaystyle \xi$ be a random variable, then $\displaystyle |\xi - E\xi|$ is non-negative

Consider,

$\displaystyle |\xi - E\xi|=|\xi - E\xi|.I(|\xi - E\xi| \ge \varepsilon) $

$\displaystyle +|\xi - E\xi|.I(|\xi - E\xi| < \varepsilon) $

$\displaystyle |\xi - E\xi| \ge |\xi - E\xi|.I(|\xi - E\xi| \ge \varepsilon) $

$\displaystyle |\xi - E\xi| \ge \varepsilon .I(|\xi - E\xi| \ge \varepsilon) $

Now since f is nondecreasing function

$\displaystyle f|\xi - E\xi| \ge f(\varepsilon) .I(|\xi - E\xi| \ge \varepsilon)$

f is non-negative even, so

$\displaystyle f(\xi - E\xi) \ge f(\varepsilon) .I(|\xi - E\xi| \ge \varepsilon)$

taking expectation, we get

$\displaystyle Ef(\xi - E\xi) \ge f(\varepsilon) .P \{ |\xi - E\xi| \ge \varepsilon \} $

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

$

Is this way is correct and can some one help me for second inequality.

Thanks in Advance.