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Let $X_i \quad$ be independent identically distributed Random Variables s.t. $E|X|^q < \infty$ ( some $q \in \mathbb{N} )$. When defining

$Y_i:= (X_i-\bar{X}_n)^q$

why is it true that $Y_i$ are iid for $i \in (1,..,n)$ ? Does subtracting the sample mean not introduce interdependence ?

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    thanks a lot guys, I m not there yet, but the fog is lifting I think. By the weak law of large number we can say, that $(X_i - \hat{X}_n) \to (X_i - EX)$ Once this is established I can start working with $Y_i$ as iid Random Variables right ? (i.e. after I have taken the limit. )2012-01-27

2 Answers 2

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In the case where $q=1$ and $X_i\sim N(\mu,\sigma^2)$, then although the $Y_i$ are (negatively) correlated, the sum $\sum_{i=1}^n Y_i^2$ is distributed as if it were the sum of the squares of $n-1$ (not $n$) independent random variables distributed as $N(0,\sigma^2)$ (with $0$, not $\mu$). And in fact, it is the sum of $n-1$ (not $n$) independent random variables distributed as $N(0,\sigma^2)$ (with $0$, not $\mu$). Those random variables are linear combinations of the $X_i$.

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Who told you they are independent? Except in trivial cases, they are dependent, because $Y_1 + \ldots + Y_n = 0$.

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    thanks for you help ! unfortunately I had a typo though, so I m still trying to understand the question that I ment to ask originally. My bad !2012-01-26