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Math Help - Chebyshev's inequality

  1. #1
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    Chebyshev's inequality

    Use Chebyshev's inequality to get an upper bound on the probability P\Big[|X-E[X]|>2\sqrt{Var[X]}\Big], and also calculate the exact probability.

    X is the result of a fair dice toss, 1,2,3,4,5,6, each with a probability of \frac{1}{6}


    I'm not sure what it means by "get an upper bound on the probability."


    This is what I did, and it matches the manual's solution of what the "exact probability" is:

    E[X]=\frac{1+2+3+4+5+6}{6}=\frac{7}{2}

    E[X^2]=\frac{1+4+9+16+25+36}{6}=\frac{91}{6}

    Var[X]=E[X^2]-E[X]^2=\frac{91}{6}-\frac{49}{4}=\frac{35}{12}

    P\left[|X-\frac{7}{2}|>2\sqrt{\frac{35}{12}}\right]= 1-P[0.0843\leq X \leq 6.916]=1-P[1,2,3,4,5,6]=1-1=0

    0\leq \frac{1}{r^2}=0\leq\frac{1}{4}
    Last edited by downthesun01; June 26th 2013 at 11:33 PM.
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  2. #2
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    Re: Chebyshev's inequality

    Hey downthesun01.

    I think what its asking you to do is use the inequality (where k = 2) and to compare with (and state) the exact probability that you have figured out.

    The bound is 1/k^2 which is 1/4 which represents a theoretical upper bound given Chebyshevs Inequality however the true value (i.e. exact) is a lot smaller (i.e. 0) and still remains in the bound (since 0 < 1/4).
    Thanks from downthesun01
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  3. #3
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    Re: Chebyshev's inequality

    Okay, that makes sense now. Thanks
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