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Math Help - Discrete Random Variable and Expectation

  1. #1
    Senior Member slevvio's Avatar
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    Discrete Random Variable and Expectation

    Let X be a discrete random variable that can only take non-negative integers (i.e.  R_x \subset \mathbb{N} ).

    Show that  \mathbb{E}(X) = \sum_{x=0}^{\infty}\mathbb{P}\{X>x\}.

    I don't know how to work this out and I would appreciate any help. I know that  \sum_{x=0}^{\infty}\mathbb{P}\{X>x\} = \sum_{x=0}^{\infty} 1 - F(x) where F(x) is the CDF but I cannot see how even that gives the expectation. Any help would be appreciated and thanks in advance.
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  2. #2
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    Quote Originally Posted by slevvio View Post
    Let X be a discrete random variable that can only take non-negative integers (i.e.  R_x \subset \mathbb{N} ).

    Show that  \mathbb{E}(X) = \sum_{x=0}^{\infty}\mathbb{P}\{X>x\}.

    I don't know how to work this out and I would appreciate any help. I know that  \sum_{x=0}^{\infty}\mathbb{P}\{X>x\} = \sum_{x=0}^{\infty} 1 - F(x) where F(x) is the CDF but I cannot see how even that gives the expectation. Any help would be appreciated and thanks in advance.
    The simplest proof consists in noting that X=\sum_{k=0}^\infty {\rm 1}_{\{X>k\}}, where {\rm 1}_{\{X>k\}} (called an indicator function) is a random variable that equals 1 if k<X and 0 else. This is just because the X first indicator functions equal 1 and the others equal 0.
    Then, by monotone convergence theorem and linearity of the expectation, E[X]=\sum_{k=0}^\infty E\left[{\rm 1}_{\{X>k\}}\right], and E\left[ {\rm 1}_{\{X>k\}} \right]=P(X>k) as you can check (you integrate 1 when X>k and 0 else...). So this is it.
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