Let $X=(X_1, \ldots, X_n)$ be a random vector whose entries are all independent and identically distributed according to some distribution $f$ with finite moments. Let $\bar{X} = \frac{1}{n}\sum_{i=1}^n X_i$. Can the variance of $$\frac{X_i - \bar{X}}{||X - e_n\bar{X}||_2}$$ be derived as a function of $n$? If not, what assumptions are needed on $f$ to derive it?

Let $X=(X_1, \ldots, X_n)$ be a random vector whose entries are all independent and identically distributed according to some distribution $f$ with finite moments. Let $\bar{X} = \frac{1}{n}\sum_{i=1}^n X_i$ and $e_n = (1, \ldots, 1)$. Can the variance of $$\frac{X_i - \bar{X}}{||X - e_n\bar{X}||_2}$$ be derived as a function of $n$? If not, what assumptions are needed on $f$ to derive it?

Let $(X_1, \ldots, X_n)$ be a random vector whose entries are all independent and identically distributed according to some distribution $f$ with finite moments. Let $\bar{X} = \frac{1}{n}\sum_{i=1}^n X_i$. Can the variance of $$\frac{X_i - \bar{X}}{||X_i - \bar{X}||_2}$$ be derived as a function of $n$? If not, what assumptions are needed on $f$ to be able to derive it?

Let $X=(X_1, \ldots, X_n)$ be a random vector whose entries are all independent and identically distributed according to some distribution $f$ with finite moments. Let $\bar{X} = \frac{1}{n}\sum_{i=1}^n X_i$. Can the variance of $$\frac{X_i - \bar{X}}{||X - e_n\bar{X}||_2}$$ be derived as a function of $n$? If not, what assumptions are needed on $f$ to derive it?