# Expectation at infinity

1. Jun 28, 2015

### feryee

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. Jun 28, 2015

### PeroK

Those latex expressions are not displaying on my browser.

PS Is this homework?

3. Jun 28, 2015

### feryee

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'')

4. Jun 28, 2015

### mathman

More information is needed. Specifically: define $e_k$.

5. Jun 29, 2015

### feryee

Thank you all for your comments. I finally proved it.