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Math Help - Cumulative distrubution function question

  1. #1
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    Cumulative distrubution function question

    Can anyone provide any insight in how I could show the following (attached below)
    Attached Thumbnails Attached Thumbnails Cumulative distrubution function question-q3.jpg  
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  2. #2
    MHF Contributor matheagle's Avatar
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    0\le nP(X>n) =n\int_n^{\infty}f(x)dx =\int_n^{\infty}nf(x)dx\le \int_n^{\infty}xf(x)dx\to 0

    as n\to \infty since the first moment is finite.
    Last edited by matheagle; November 6th 2009 at 10:30 PM.
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  3. #3
    Moo
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    Hello,

    1-F(x)=P(X>n)=P(X\cdot \bold{1}_{X>n}>n)=P(Y_n>n)\leq \frac{E(Y_n)}{n} by Markov's inequality.
    where Y_n=X\cdot \bold{1}_{X>n}
    As n goes to infinity, Y_n obviously goes to 0 (X is almost surely finite since it's integrable)
    and |Y_n| is bounded by |X|, which is integrable.
    So we can apply the dominated convergence theorem, and we have n(1-F(x))\leq E(Y_n) \to 0  \quad \quad \square
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