# Thread: Application of Chebyshev's inequality

1. ## Application of Chebyshev's inequality

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

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

I tried like way.

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

Consider,
$|\xi - E\xi|=|\xi - E\xi|.I(|\xi - E\xi| \ge \varepsilon)$
$+|\xi - E\xi|.I(|\xi - E\xi| < \varepsilon)$

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

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

Now since f is nondecreasing function

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

f is non-negative even, so

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

taking expectation, we get

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

$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.

2. There is a quick proof of the first inequality with Markov's inequality
$P(|\xi-E\xi|\geq \epsilon)\leq P(f(\xi-E\xi) \geq f(\epsilon))\leq \frac{E[f(\xi-E\xi)]}{f(\epsilon)}$.

Are you missing something in the second inequality? Something seems wrong. Take f=x^2 and the random variable to be constantly C, then the probability on the RHS is zero for strictly positive epsilon, but $\frac{C^2-\epsilon^2}{C^2}$ can easily be made strictly positive.

Actually i need how to prove this inequality.
$\frac{{Ef(\xi ) - f(\varepsilon )}}{{f(C)}} \le P\left\{ {\left| {\xi - E\xi } \right| \ge \varepsilon } \right\} \le \frac{{Ef(\xi - E\xi )}}{{f(\varepsilon )}}.$.

I have shown one part

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

now i need to prove this

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

4. Originally Posted by amb03
now i need to prove this

$\frac{{Ef(\xi ) - f(\varepsilon )}}{{f(C)}} \le P\left\{ {\left| {\xi - E\xi } \right| \ge \varepsilon } \right\}$.
Originally Posted by Focus
Are you missing something in the second inequality? Something seems wrong. Take f=x^2 and the random variable to be constantly C, then the probability on the RHS is zero for strictly positive epsilon, but $\frac{C^2-\epsilon^2}{C^2}$ can easily be made strictly positive.
See above.

Here is complete question

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

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

In particular if $f(x)=x^2$, then

$\frac{{E{\xi ^2} - {\varepsilon ^2}}}{{{C^2}}} \le P\left\{ {\left| {\xi - E\xi } \right| \ge \varepsilon } \right\} \le \frac{{V\xi }}{{{\varepsilon ^2}}}.$
===================================
I proved this version only

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

Now i need to prove

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

Regards

6. Originally Posted by amb03
Now i need to prove

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

Regards
I understand what you are trying to prove, but this doesn't hold (as my earlier post explains). Take $\xi(w)=C=1$ for a.e. omega and $\epsilon=1/2$ and $f(x)=x^2$. Then you get that $3/4 \leq 0$ which is blatantly not true.