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Thread: Proving Chebyshev's Inequality

  1. #1
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    Proving Chebyshev's Inequality

    Prove Chebyshev's inequality, which states that if $\displaystyle X $ is a random variable with expectation $\displaystyle \mu $ and finite variance $\displaystyle \sigma^2 $, show that $\displaystyle \mathbb{P} (|X - \mu| \geq \varepsilon ) \leq \frac{\sigma^2}{\varepsilon^2} $, for any $\displaystyle \varepsilon > 0 $
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  2. #2
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    Quote Originally Posted by Jimmy_W View Post
    Prove Chebyshev's inequality, which states that if $\displaystyle X $ is a random variable with expectation $\displaystyle \mu $ and finite variance $\displaystyle \sigma^2 $, show that $\displaystyle \mathbb{P} (|X - \mu| \geq \varepsilon ) \leq \frac{\sigma^2}{\varepsilon^2} $, for any $\displaystyle \varepsilon > 0 $
    Have you Googled for a proof ....?
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  3. #3
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    Hi Mr Fantastic,
    I have just realised I left out part of my question , which provides me with a road to go down, but I am really lost so I found that none of the proofs on google really helped me out.

    The question should have been:

    Prove Chebyshev's inequality, which states that if $\displaystyle X $ is a random variable with expectation $\displaystyle \mu $ and finite variance $\displaystyle \sigma^2 $, show that $\displaystyle \mathbb{P} (|X - \mu| \geq \varepsilon ) \leq \frac{\sigma^2}{\varepsilon^2} $, for any $\displaystyle \varepsilon > 0 $
    by first showing that $\displaystyle \mathbb{E}[X - \mu)^2] \geq \mathbb{E}[(X- \mu)^2 I_{\{(X- \mu)^2 \geq \varepsilon^2\}}]. $

    Its this last part that I carelessly left out that has me stumped.
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  4. #4
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    I am really stumped on this one. But the little work that I have been able to do on it is:

    $\displaystyle (X - \mu)^2 \geq \varepsilon^2 \geq \mu - \varepsilon \geq X $
    and $\displaystyle \mu + \varepsilon \leq X $

    and now I am lost
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  5. #5
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    This is what I have put together.
    Could somebody please look over it for me?

    $\displaystyle (X - \mu)^2 \geq \varepsilon^2 \geq \mu - \varepsilon \geq X $
    and $\displaystyle \mu + \varepsilon \leq X $

    So... $\displaystyle (- \infty, \mu - \varepsilon) \cup (\mu + e, \infty) $

    Now
    $\displaystyle \mathbb{E} [(X - \mu)^2] = \int_{-\infty}^{\infty} (X - \mu)^2 \ f(x) \ dx $

    $\displaystyle \geq \ \int (X - \mu)^2 \ f(x) \ dx $

    $\displaystyle \geq \int \varepsilon^2 \ f(x) \ dx $ (because $\displaystyle ((X - \mu)^2 \geq \varepsilon^2) \in (-\infty, \infty) $

    $\displaystyle = \varepsilon^2 \ \int f(x) dx $

    $\displaystyle = \varepsilon^2 \ \mathbb{P}(|X - \mu| \geq \varepsilon $

    therefore

    $\displaystyle \sigma^2 \geq \varepsilon^2 \ \mathbb{P}((|X - \mu| \geq \varepsilon) $

    and

    $\displaystyle \mathbb{P} (|X - \mu| \geq \varepsilon ) \leq \frac{\sigma^2}{\varepsilon^2} $
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