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Math Help - Proof of Expected Value

  1. #1
    Senior Member I-Think's Avatar
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    Proof of Expected Value

    RTP: E[X]=\sum_{i=1}^{\infty} P[X\geq{i}]

    If I expand this algebraically, I can see it
    \sum_{i=1}^{\infty} P[X\geq{i}]= (P[X=1]+P[X=2]+...)+(P[X=2]+P[X=3]+...)+....
    =P[X=1]+2P[X=2]+.....

    But I'm having trouble understanding the proof as given by the text and wikipedia, in particular the interchange of the order of summation

    Given proof
    \sum_{i=1}^{\infty} P[X\geq{i}]= \sum_{i=1}^{\infty}\sum_{j=i}^{\infty} P[X=j]
    Then we interchange the order of summation
    \sum_{i=1}^{\infty} P[X\geq{i}]= \sum_{j=1}^{\infty}\sum_{i=1}^{j} P[X=j]

    This step, I do not understand. Given that i does not appear in the summation of \sum_{i=1}^{j} P[X=j], how am I supposed to expand this?
    Last edited by I-Think; November 17th 2011 at 07:23 AM.
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  2. #2
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    Re: Proof of Expected Value

    Quote Originally Posted by I-Think View Post
    Given proof
    \sum_{i=1}^{\infty} P[X\geq{i}]= \sum_{i=1}^{\infty}\sum_{j=i}^{\infty} P[X=j]
    Then we interchange the order of summation
    \sum_{i=1}^{\infty} P[X\geq{i}]= \sum_{j=1}^{\infty}\color{blue}\sum_{i=1}^{j} P[X=j]
    Look at the bit on blue.
    For each i=1\text{ to }j the term P[X=j] is constant.
    Thus \sum_{i=1}^{j} P[X=j]=j\cdot P[X=j]
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