Proving Expectations at Infinity in a Paper: Tips and Tricks

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    Expectation Infinity
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Discussion Overview

The discussion revolves around proving expectations at infinity as presented in a specific paper related to a variable step-size LMS algorithm. Participants are examining a mathematical expression involving expectations of squared terms and their convergence as \( n \) approaches infinity.

Discussion Character

  • Technical explanation
  • Homework-related
  • Debate/contested

Main Points Raised

  • One participant presents an equation involving expectations, stating that \( E\{e^2_{n-i-1}e^2_{n-j-1}\} = E\{e^2_{n-i-1}\}E\{e^2_{n-j-1}\} \) for \( i \neq j \) and seeks proof for a derived expression as \( n \rightarrow \infty \).
  • Another participant expresses difficulty in viewing the LaTeX expressions and questions whether the inquiry is related to homework.
  • A later reply clarifies that the equation is from a paper, not homework, and provides the source of the equation.
  • One participant requests further information, specifically asking for a definition of \( e_k \).
  • In a concluding remark, a participant claims to have proved the expression after the discussion.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the proof of the expression, as there are varying levels of understanding and requests for clarification. The discussion includes both requests for help and claims of resolution.

Contextual Notes

There are limitations regarding the definitions of terms used in the expressions, particularly the variable \( e_k \), which remains undefined in the discussion.

feryee
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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
 
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feryee said:
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

Those latex expressions are not displaying on my browser.

PS Is this homework?
 
PeroK said:
Those latex expressions are not displaying on my browser.

PS Is this homework?
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'')
 
feryee said:
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
More information is needed. Specifically: define [itex]e_k[/itex].
 
mathman said:
More information is needed. Specifically: define [itex]e_k[/itex].
Thank you all for your comments. I finally proved it.
 

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