Proving Expectations at Infinity in a Paper: Tips and Tricks

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The discussion revolves around proving an expectation formula related to the convergence of sums involving random variables as n approaches infinity. The key equation presented is E{(Σ(α^i e^2(n-i-1)))(Σ(α^i e^2(n-i-1)))} = (2α(E(e_∞^2))^2)/((1-α)^2(1+α)) + E(e_∞^4)/(1-α^2). Initially, the poster sought help due to difficulties in deriving this result from a referenced paper. After some exchanges, the poster confirmed that they successfully proved the equation. The discussion highlights the importance of defining variables clearly for effective communication in mathematical proofs.
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 e_k.
 
mathman said:
More information is needed. Specifically: define e_k.
Thank you all for your comments. I finally proved it.
 
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