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I have the following model,

$$ X_i \overset{iid}{\sim} \mathrm{Normal}(0,1), i=1, \dots, n. $$

It is known that the sample variance $\hat{\sigma^2} := \sum_{i=1}^n \frac{(X-\bar{X})^2}{n-1}$ is independent of sample mean, $\bar{X} := \frac{\sum_{i=1}^n X_i}{n}$. But what I wonder is,

Is $\bar{X}^{w} := \sum_{i=1}^n w_i X_i$ also independent of $\hat{\sigma^2}$, where $w_i$'s are given constants such that $\sum w_i = 0$?


My try

I'm trying to show $\bar{X}^{w}$ is a function of $\bar{X}$, but I'm stuck at here.

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Let $n=2$, and $w_1 = 1$ and $w_2 = -1$.

It is easy to have $X_1-\bar X = 0.5X_1 - 0.5X_2$ and $X_2 - \bar X = -0.5X_1 + 0.5X_2$

$$X=\left(\matrix{\bar Y^w\\ Y_1-\bar X\\ Y_2-\bar X }\right) = \left(\matrix{ 1 & -1\\ 0.5 & -0.5 \\ -0.5 & 0.5}\right)\left(\matrix{ Y_1\\Y_2}\right) $$

$$\mathrm{Var}(X) = \left(\matrix{ 1 & -1\\ 0.5 & -0.5 \\ -0.5 & 0.5}\right)\left(\matrix {1 & 0.5 & -0.5\\ -1 &-0.5 & 0.5}\right) = \left(\matrix {2 & 1 & -1\\ -1 & -0.5 & -0.5\\ -1 & -0.5 & 0.5}\right)$$ So $\bar Y^w$ is not independent with $Y_1-\bar X$ and $Y_2-\bar X$. ==> $\bar Y^w$ is not independent with $\hat \sigma^2$

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