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Thread: anyone help?

  1. #1
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    Exclamation anyone help?

    A random variable R has probability density,

    p_R(x) = (x^m / m!)e^-x if x >= 0; and 0 otherwise

    where m is a positive integer.


    (a) Show that

    P(0 <= R <= 2(m + 1)) = 1 - P(|R - (m + 1)| > m + 1)


    (b) Use Chebyshev's inequality and part (a) to prove that

    P(0 <= R <= 2(m + 1)) > m/m + 1
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  2. #2
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    Quote Originally Posted by bogazichili View Post
    A random variable R has probability density,

    p_R(x) = (x^m / m!)e^-x if x >= 0; and 0 otherwise

    where m is a positive integer.


    (a) Show that

    P(0 <= R <= 2(m + 1)) = 1 - P(|R - (m + 1)| > m + 1)


    (b) Use Chebyshev's inequality and part (a) to prove that

    P(0 <= R <= 2(m + 1)) > m/m + 1
    (a) Note that $\displaystyle 0 \leq R \leq 2(m + 1) \Rightarrow -(m + 1) \leq R - (m + 1) \leq m + 1 \Rightarrow | R - (m + 1) | \leq m + 1$.


    (b) Note that $\displaystyle R$ follows a gamma distribution with $\displaystyle \alpha = m + 1$ and $\displaystyle \beta = 1$.

    Therefore:

    $\displaystyle \mu = \alpha \beta = m + 1$

    $\displaystyle \sigma^2 = \alpha \beta^2 = m + 1 \Rightarrow \sigma = \sqrt{m + 1}$.


    So $\displaystyle | R - (m + 1) | = | R - \mu |$ and $\displaystyle m + 1 = \sqrt{m + 1} \, \sigma$.

    So apply Chebyshev's inequality to $\displaystyle \Pr( | R - \mu | \geq \sqrt{m + 1} \, \sigma)$.
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