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I have a possibly simple question. Let $\{x_i\}_{i=1}^n$ and $\{u_i\}_{i=1}^n$ be two i.i.d. sequences of random variables. I.e. $x_i$ and $u_i$ are both independent and identically distributed over $i$. Then, consider the random variable, \begin{equation} w_i = (x_i - \overline{x})u_i, \end{equation} where $\overline{x} = (1/n) \sum_{i=1}^n x_i$. Is the sequence $\{w_i\}_{i=1}^n$ then an i.i.d. sequence?

Thanks in advance!

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2 Answers 2

No, because all the $w_i$ depend, via $\bar x$, on the all of the $x_i$ and are hence, in general, dependent. As an example, you might consider the case $n=2$, $u_i\equiv 1$, and compute the covariance between $w_i$ and $w_2$.

The $w_i$ are identically distributed.

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So, if I want to determine the limiting distribution of \begin{equation} \sqrt{n}\left( \frac{1}{n} \sum_{i=1}^{n} (x_i-\overline{x}) u_i \right), \end{equation} then I cannot apply the Lindeberg-Levy CLT or the Liapounov CLT. Then, which version of the CLT would be appropriate to use? –  Esben Vibel Dec 22 '12 at 11:16

As a simple counter-example, suppose $n=2$, $X_i = \pm 1$ with equal probabilities and $U_i = \pm 1$ with equal probabilities.

Then $W_i$ can take the values $+ \frac12, -\frac12, 0$ with probabilities $\frac14, \frac14, \frac12$, but you have $\Pr(W_2=0|W_1=0)=1$ while $\Pr(W_2=0|W_1=\frac12)=0$.

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