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Math Help - Chebyshev's inequality

  1. #1
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    Chebyshev's inequality

    I guess this is not a difficult problem, but I can not find the function for which to use Chebyshev's inequality...

    Assume X is a random variable with property that \mathbb{E}[X]=0 \;\;and\;\;\mathbb{E}[X^2]=\sigma^2. Use Chebyshev's inequality for the random variable Y=(X+c)^2,c>0 to prove that
    \mathbb{P}(X>x)\le\frac{\sigma^2}{x^2+\sigma^2} for x>0.

    Thanks.
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  2. #2
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    Quote Originally Posted by Veve View Post
    I guess this is not a difficult problem, but I can not find the function for which to use Chebyshev's inequality...

    Assume X is a random variable with property that \mathbb{E}[X]=0 \;\;and\;\;\mathbb{E}[X^2]=\sigma^2. Use Chebyshev's inequality for the random variable Y=(X+c)^2,c>0 to prove that
    \mathbb{P}(X>x)\le\frac{\sigma^2}{x^2+\sigma^2} for x>0.

    Thanks.
    You have, using Chebyshev's inequality, for x,c>0, P(X>x)=P(X+c>x+c)\leq P((X+c)^2>(x+c)^2)\leq \frac{E[(X+c)^2]}{(x+c)^2}. Now, expand the square inside the expectation; then the right-hand side is a function of c, while the left-hand side isn't. The "best" (sharpest) inequality is the one when the right-hand side is lowest, so you have to choose c which minimizes the right-hand side. This resumes to a standard function study (derivative, etc.) to find this minimum.
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