While reading a paper, i came across the following Expectations:(adsbygoogle = window.adsbygoogle || []).push({});

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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# Expectation at infinity

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