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Thread: Chebyshev and uniform distribution

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    Chebyshev and uniform distribution

    I'm having a lot of trouble with this question:

    Find $\displaystyle P(|Y-\sigma| \leq 2\sigma)$ for the uniform random variable distribution. Compare with the corresponding probabilistic statement given by Chebyshev's theorem and the empirical rule.

    according to the solution in the book the answer is 1; I have no idea what this number represents.

    The variance of a uniform distribution is $\displaystyle Vr(X) = \sigma ^2 = \frac{(\theta_2-\theta_2)^2}{12}$ so in this case it should be $\displaystyle 4\sigma = \frac{(\theta_2-\theta_2)^2}{12}$. This is where I get stuck, plus I don't think that I'm even going in the right direction. Any help would be appreciated.
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    Quote Originally Posted by lllll View Post
    I'm having a lot of trouble with this question:

    Find $\displaystyle P(|Y-\sigma| \leq 2\sigma)$ for the uniform random variable distribution. Compare with the corresponding probabilistic statement given by Chebyshev's theorem and the empirical rule.

    according to the solution in the book the answer is 1; I have no idea what this number represents.

    The variance of a uniform distribution is $\displaystyle Vr(X) = \sigma ^2 = \frac{(\theta_2-\theta_2)^2}{12}$ so in this case it should be $\displaystyle 4\sigma = \frac{(\theta_2-\theta_2)^2}{12}$. This is where I get stuck, plus I don't think that I'm even going in the right direction. Any help would be appreciated.
    $\displaystyle \Pr(|Y - \sigma| \leq 2 \sigma) = \Pr(-\sigma \leq Y \leq 3 \sigma)$.

    If Y is uniform over [a, b]:

    $\displaystyle \bullet$ $\displaystyle f(y) = \frac{1}{b-a}$ for $\displaystyle a \leq y \leq b$ and zero otherwise.

    $\displaystyle \bullet$ $\displaystyle \sigma = \frac{b-a}{\sqrt{12}}$.


    Note that $\displaystyle - \sigma = \frac{a-b}{\sqrt{12}} < a$ for $\displaystyle b > a$. This means that:

    $\displaystyle \Pr(-\sigma \leq Y \leq 3 \sigma) = \Pr(-\sigma \leq Y \leq a) + \Pr(a \leq Y \leq 3 \sigma) = \Pr(a \leq Y \leq 3 \sigma)$.

    Furthermore:

    $\displaystyle 3 \sigma = \frac{3(b-a)}{\sqrt{12}} < b$ if $\displaystyle a > \frac{-(\sqrt{12} - 3)b}{3}$.

    $\displaystyle 3 \sigma = \frac{3(b-a)}{\sqrt{12}} > b$ if $\displaystyle a < \frac{-(\sqrt{12} - 3)b}{3}$. In this case the required probability is obviously equal to 1.
    Last edited by mr fantastic; Sep 8th 2008 at 08:10 PM.
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