Iam trying to express $E\left[\left(\sum_{i=1}^{n}X_{i}-\mu_{i}n\right)^{2}\right]$ in terms of Var and Cov. $\mu_{i}=E[X_{i}]$ and $(X_{i})_{i\geq 1}$ stationary.
Expressing the expectation in terms of Var and Cov
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$\begingroup$
probability
probability-theory
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0And more specifically, what did you try? (Unrelated: it seems that $\mu_i$ does not depend on $i$, could you confirm this point?) – 2012-10-24
1 Answers
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Just to be clear, the $\mu_i n$ is not part of the sum, despite the $i$ there. In fact, all $\mu_i$ are equal because of stationarity. If $Y_i = X_i - \mu_i$, you're looking at $E \left[ \left(\sum_i Y_i \right)^2\right]$.
Hint: what are $E[Y_i Y_j]$ and $E[Y_i^2]$ in terms of variance and covariance?
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0Thx for you help. I have found an expression. – 2012-10-24