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Math Help - [SOLVED] question on markov inequality and functions

  1. #1
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    [SOLVED] question on markov inequality and functions

    I want to prove that  Pr( Z \ge a) \le {E(g(Z)) \over g(a)} given that  g(x) \ge 0 \forall x \in\mathbb{R} and that for any
     x_1, x_2 \mbox{ if }x_1 < x_2 \mbox{ then }g(x_1) < g(x_2) .
    Finally I was given the formula
     I_{[z \ge a]} = \left\{\begin{matrix} 1, & \mbox{if }z \ge a \\ 0, & \mbox{otherwise} \end{matrix}\right.
    and the inequality
     g(a)I_{[Z \ge a]} \le g(Z)I_{[Z \ge a]} \le g(Z)
    which I am supposed to use somehow.

    Any help will be appreciated.
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  2. #2
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    Quote Originally Posted by joop View Post
    I want to prove that  Pr( Z \ge a) \le {E(g(Z)) \over g(a)} given that  g(x) \ge 0 \forall x \in\mathbb{R} and that for any  x_1, x_2 \mbox{ if }x_1 < x_2 \mbox{ then }g(x_1) < g(x_2) .
    you've already used the fact that g is increasing to show that:  g(Z) \geq g(a)I_{[Z \ge a]}. thus: \text{E}(g(Z)) \geq \text{E}(g(a)I_{[Z \ge a]})=g(a) \text{E}(I_{[Z \ge a]})=g(a)\text{Pr}(Z \geq a).

    now since g(x) is a positive function, you can divide both sides of the inequality by g(a) to finish the proof.
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  3. #3
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    Thank you very much! I was simply not seeing the equality:

    \text{E}(I_{[Z \ge a]})=\text{Pr}(Z \geq a)
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