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Thread: Markov inequality

  1. #1
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    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
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  2. #2
    Moo
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    Hello,

    A few remarks, while I'm trying to solve it.
    Quote Originally Posted by jovial View Post
    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 ?
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