Results 1 to 2 of 2

Math Help - Markov inequality

  1. #1
    Newbie
    Joined
    Dec 2010
    Posts
    3

    Markov inequality

    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}

    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)

    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)}

    How can be proved this part.

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

    Thanx
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    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  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}

    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)}

    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 ?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Question relating to Markov's inequality
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: April 16th 2010, 03:35 PM
  2. Markov's Inequality and Random graphs
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: April 7th 2010, 01:55 PM
  3. Markov inequality
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: February 12th 2010, 03:36 PM
  4. [SOLVED] question on markov inequality and functions
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: October 27th 2008, 12:16 AM
  5. Markov's Inequality Problem Help! MIDTERM MONDAY!
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: November 17th 2007, 06:52 PM

Search Tags


/mathhelpforum @mathhelpforum