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I'm trying to solve something, which requires me to calculate

$ E((\frac 1 n \sum^n_1e_i^2)^2) $ and I don't understand how to do it. Note that $e_i$ is a white noise.

From what I've tried, I got

$\frac 1 {n^2}E((\sum^n_1e_i^2)^2) $ but this seems complicated to calculate. Should I use the fact that $ \frac 1 n \sum^n_1e_i^2 $ is the expected value of the variance of $e_i$ ?

thank you

  • 0
    What is a white noise? – 2017-01-21
  • 0
    0 expected value, constant variance, iid – 2017-01-21
  • 0
    They are independent. – 2017-01-21

1 Answers 1

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$$ E\left[\left(\frac1n\sum_{i=1}^ne_i^2\right)^2\right] =\frac1{n^2}E\left[\sum_{i=1}^n \sum_{j=1}^n e_i^2e_j^2\right]=\frac1{n^2}\sum_{i=1}^n \sum_{j=1}^nE[e_i^2e_j^2]=$$ $$=\frac1{n^2}\left[\sum_{i=1}^nE[e_i^4]+{\sum_{i=1}^n\sum_{i=1}^n}_{\{i\not =j\}}E[e_1^2e_j^2]\right]=\frac1{n}\left[E[e_1^4]+(n-1)\sigma^4\right].$$

wher $\sigma^2$ is the common variance. Without learning more about $E[e_i^4]$ we cannot go further.