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Expectation at infinity

  1. Jun 28, 2015 #1
    While reading a paper, i came across the following Expectations:

    Given that the ##E\left\{e^2_{n-i-1}e^2_{n-j-1}\right\}=E\left\{e^2_{n-i-1}\right\}E\left\{e^2_{n-j-1}\right\}## for ##i\neq j##.\\

    Then as ##n\rightarrow\infty##

    ##E\left\{\left(\sum\limits_{i=0}^{n-2}\alpha^i e^2_{n-i-1}\right)\left(\sum\limits_{i=0}^{n-2}\alpha^i e^2_{n-i-1}\right)\right\}=\frac{2\alpha\left(E\left( e_\infty^2\right)\right)^2}{(1-\alpha)^2(1+\alpha)}+\frac{E\left( e_\infty^4\right)}{(1-\alpha^2)}##.

    Can you provide me with proof or any hint/help? I tried but couldn't get the same answer.Thanks
     
    Last edited: Jun 28, 2015
  2. jcsd
  3. Jun 28, 2015 #2

    PeroK

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    Those latex expressions are not displaying on my browser.

    PS Is this homework?
     
  4. Jun 28, 2015 #3
    No, This is an equation in a paper.(Eq 17 in ''New Steady-state analysis result for variable step-size LMS algorithm with different noise distributions'')
     
  5. Jun 28, 2015 #4

    mathman

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    More information is needed. Specifically: define [itex]e_k[/itex].
     
  6. Jun 29, 2015 #5
    Thank you all for your comments. I finally proved it.
     
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