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Math Help - Application of Chebyshev's inequality

  1. #1
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    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 )}}.<br />

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

    Thanks in Advance.
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  2. #2
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    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.
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  3. #3
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    Thanks for your reply.
    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\} .
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  4. #4
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    Quote Originally Posted by amb03 View Post
    now i need to prove this

    \frac{{Ef(\xi ) - f(\varepsilon )}}{{f(C)}} \le P\left\{ {\left| {\xi  - E\xi } \right| \ge \varepsilon } \right\} .
    Quote Originally Posted by Focus View Post
    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.
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  5. #5
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    Again Thanks for your reply and sorry for disturbing again

    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 )}}.<br />

    Now i need to prove

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

    Regards
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  6. #6
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    Quote Originally Posted by amb03 View Post
    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.
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