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Math Help - Chebyshev's inequality: "It is obvious ..."

  1. #1
    Member courteous's Avatar
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    Unhappy Chebyshev's inequality: "It is obvious ..."

    Having trouble internalizing the last sentence from my textbook's snippets:
    For a binomially distributed random variable X, it can be shown, that the following holds:
    Var(X)=np(1-p)=npq (16)
    [...]
    Chebyshev: t>0 \rightarrow P\big(E(X)-t < X < E(X)+t\big)\geq 1-\frac{D(X)}{t^2}
    [...]
    From (16) it is obvious, that the probability of a complementary event, i.e. the random variable X having value which differs from E(X) for at least t, is at most \frac{Var(X)}{t^2}.
    I would like to understand it visually (e.g.: probability distribution graph, E(X) is at MAX., now look at the E(X)\pm t points, ...).


    BTW, I've also found a similar thread, but could not work out the proof for Theorem 4 from .pdf mr fantastic provided:
     Pr(|R-Ex(R)| \geq x) = Pr\big((R-Ex(R))^2 \geq x^2) \leq \frac{Ex\big((R-Ex(R))^2}{x^2} = \frac{Var[R]}{x^2}. Why does the \leq hold?

    PS: Please, feel free to correct any grammatical mistake you find.
    Last edited by courteous; December 25th 2009 at 01:02 AM. Reason: colored
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  2. #2
    MHF Contributor matheagle's Avatar
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    The inequality holds because you'rer splitting the integral/sum into two parts.
    You throw one of those away and you obtain a lower bound on the other.
    It's actually a very crude bound that pertains to all distributions hence it's not close in many situations.
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  3. #3
    Member courteous's Avatar
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    Red face Still can't internalize

    I still don't intuitively know why
    Pr\Big(\text{that some random variable }X\text{ differs at least }t \text{ from }E(X)\Big) \leq \frac{Var(X)}{t^2}.
    I've put red color in original post.

    It bothers me, because it says it is obvious from the fact that Var(X)=npq, so I should be able to intuit.
    Wikipedia doesn't help the intuition (and I don't know anything about measure theory).
    Last edited by courteous; January 28th 2010 at 05:22 AM.
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