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Thread: Binomial distribution

  1. #1
    Oct 2008

    Binomial distribution

    Let $\displaystyle X$ be Binomial $\displaystyle B(p,n)$ with $\displaystyle p>0$ fixed, and $\displaystyle a>0$. How can I show that:

    $\displaystyle P(|\frac{X}{n}-p|>a)\le\frac{\sqrt{p(1-p)}}{a^2n}min(\sqrt{p(1-p)},a\sqrt{n})$
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  2. #2
    MHF Contributor matheagle's Avatar
    Feb 2009
    To be smaller than the min, means it's smaller than both.
    Clearly the first is true by chebyshev's
    That means you must also prove that

    $\displaystyle P\left( \left| {X\over n}-p\right|>a\right)\le {\sqrt{p(1-p)}\over a\sqrt{n}}$

    and that's just the st deviation over a, which is markov.
    Last edited by matheagle; Sep 22nd 2010 at 01:38 PM.
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