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Aysmptotic Approximation

  1. Feb 2, 2012 #1
    Hello,

    I am reading a paper, and the author claimed that in asymptotic sense as M goes to infinite:

    [tex]\sum_{i=1}^M\sum_{l=0}^L|h_i(l)|^2=M[/tex]

    where:

    [tex]\sum_{l=0}^L\mathbb{E}\left\{|h_i(l)|^2\right\}=1[/tex].

    How is that asymptotic follows?

    Thanks in advance
     
  2. jcsd
  3. Feb 3, 2012 #2

    mathman

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    There are undefined symbols. What kind of things are hi(l)? What is the meaning of E?
     
  4. Feb 3, 2012 #3
    E is the expectation, and h are random variables.

    I got it, it is just by using the law of large numbers.

    Thanks
     
  5. Feb 4, 2012 #4

    mathman

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    In the equation cited, what is asymptotic? Ratio -> 1 (true) or difference -> 0 (false)?
     
  6. Feb 5, 2012 #5
    as M goes to infinite.
     
  7. Feb 5, 2012 #6

    mathman

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    I know you mean as M -> ∞, but my question is what is supposed to happening as M -> ∞, expression divided by M -> 1 (true) or expression minus M -> 0 (false)?
     
  8. Feb 5, 2012 #7
    I am sorry, I did not understand you quiet well. Can you say it in different way, please?
     
  9. Feb 5, 2012 #8

    chiro

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    I have the feeling that he is dividing by M.
     
  10. Feb 6, 2012 #9
    If he is dividing by the M the result would be 1 not M.
     
  11. Feb 6, 2012 #10

    mathman

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    Let ∑(M) be the double sum you are talking about. There are two ways of expressing the limit as M -> ∞, {∑(M)}/M -> 1 (true) or ∑(M) - M -> 0 (false).
     
  12. Feb 6, 2012 #11
    [tex]\lim_{M\xrightarrow{}\infty}\frac{1}{M}\sum_{i=1}^M\sum_{l=0}^L|h(l)|^2=1[/tex]
     
  13. Feb 7, 2012 #12

    mathman

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    True. The author is making use of one of the fundamental theorems of probability theory - the law of large numbers.
     
  14. Feb 7, 2012 #13
    Yes, right. Thanks
     
  15. Feb 7, 2012 #14

    mathman

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    There is one caveat: hi independent of hj for i ≠ j.
     
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