# iid Random Variables

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. – Davide Giraudo Jan 26 '12 at 18:34
sorry, I had a typo, forgot the power in the definition of $Y_i$, amending post ! – Beltrame Jan 26 '12 at 18:40
If you take $n=2$ and $q$ even then $Y_1=\frac 1{2^q}(X_1-X_2)^q=\frac 1{2^q}(X_2-X_1)^q=Y_2$. – Davide Giraudo Jan 26 '12 at 19:46
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. – Davide Giraudo Jan 26 '12 at 20:13
Sorry I meant $Y_j^{\frac 1q}$ in the last comment. I guess this topic math.stackexchange.com/questions/102455/… is related. – Davide Giraudo Jan 26 '12 at 20:39

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