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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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    Is $\bar{X_n}=\frac 1n\sum_{j=1}^nX_j$? In this case, $\sum_{i=1}^nY_i=\sum_{i=1}^nX_i-\frac nn\sum_{i=1}^nX_i=0$, so the random variable $\{Y_i\}_{1\leq i\leq n}$ cannot be independent.2012-01-26
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    sorry, I had a typo, forgot the power in the definition of $Y_i$, amending post !2012-01-26
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    If you take $n=2$ and $q$ even then $Y_1=\frac 1{2^q}(X_1-X_2)^q=\frac 1{2^q}(X_2-X_1)^q=Y_2$.2012-01-26
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    For $q$ odd we have $\sum_{j=1}^ny_j^{\frac 1j}=0$ so the $Y_j$ cannot be independent, except if they are constant.2012-01-26
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    Sorry I meant $Y_j^{\frac 1q}$ in the last comment. I guess this topic http://math.stackexchange.com/questions/102455/alternative-way-of-showing-convergence-of-central-moments is related.2012-01-26
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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

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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