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?
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$\begingroup$ What is the $e_n$? $\endgroup$– MichaelCommented Jun 25 at 15:17
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$\begingroup$ @Michael Sorry for the ambiguity. $e_n$ is the $n \times 1$ vector whose entries are all 1. $\endgroup$– fennelCommented Jun 25 at 15:27
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1$\begingroup$ In that case, you can get a simple answer for mean and variance. You can use symmetry to compute $E[Y_i]$ and $E[Y_i^2]$, where I define $$Y_i=\frac{X_i-\overline{X}}{||X-e_n\overline{X}||_2}$$ Notice that $\sum_{i=1}^nY_i$ and $\sum_{i=1}^nY_i^2$ are nice. $\endgroup$– MichaelCommented Jun 25 at 17:29
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$\begingroup$ Strictly speaking, we should define $Y_i=0$ in the case the denominator is zero (which happens with prob 0 if $n\geq 2$ and $X_i$ has a continuous distribution). $\endgroup$– MichaelCommented Jun 25 at 17:36
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1$\begingroup$ Thanks for your answer! I got the answer as you suggested. Using $E[Y_1 + \cdots + Y_m] = 0$ and $E[Y^2_1 + \cdots + Y^2_m] = 1$ with the fact that the $Y_i$ are identically distributed allows to derive $E[Y_i] =0$ and $V[Y_i]=E[Y_i^2]=1/n$ $\endgroup$– fennelCommented Jun 25 at 19:48
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1 Answer
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This just summarizes my comments which led to the answer that fennel gets (also in the comment section). It generalizes to the case when the random variables $(X_1, ..., X_n)$ are exchangeable, meaning the joint distribution is invariant under permutations of the ordering, that is, for all permutations $(\sigma(1), ..., \sigma(n))$ of $\{1, ..., n\}$ and all $(x_1, \ldots, x_n)\in\mathbb{R}^n$ we have $$P[X_1\leq x_1, ..., X_n\leq x_n]=P[X_{\sigma(1)}\leq x_1, ..., X_{\sigma(n)}\leq x_n] $$ Note that i.i.d. implies exchangeable (so exchangeable is more general). Exchangeability implies all $X_i$ have the same distribution as $X_1$. In the following, the distribution of $X_1$ is arbitrary and does not need to have finite moments.
Fix $n$ as a positive integer. For $i \in \{1, ..., n\}$ define $$ Y_i = \left\{\begin{array}{cc} \frac{X_i-\overline{X}}{||X-e_n\overline{X}||_2} & \mbox{if $X-e_n\overline{X}\neq 0$} \\ 0 & \mbox{else} \end{array}\right.$$ where $e_n=(1, 1, ..., 1)$ is the all-1 vector in $\mathbb{R}^n$. Define event $A$ by $$ A = \{X - e_n\overline{X}\neq 0\}$$
Observe that $\{Y_i\}_{i=1}^n$ are random variables that surely satisfy
$$-1 \leq Y_i\leq 1 \quad \forall i \in \{1, ..., n\}$$
$$ \sum_{i=1}^n Y_i=0 \quad, \quad
\sum_{i=1}^n Y_i^2=1_A$$
Boundedness of the $Y_i$ variables ensures the first and second moments are finite. The exchangeable property of $(X_1, ..., X_n)$ means that each $Y_i$ has the same probability distribution and so
$$E[Y_i]=E[Y_1], \quad E[Y_i^2]=E[Y_1^2] \quad \forall i\in\{1, ..., n\}$$
Therefore
$$ \sum_{i=1}^nY_i=0 \implies nE[Y_1]=0 \implies E[Y_1]=0$$
$$\sum_{i=1}^nY_i^2 = 1_A \implies nE[Y_1^2]=P[A]\implies E[Y_1^2]=P[A]/n$$
and since $Var(Y_i)=E[Y_i^2]-0^2 = E[Y_1^2]$ we have
$$\boxed{Var(Y_i) = P[A]/n \quad \forall i \in \{1, ..., n\}}$$
Observe that $$ A^c = \{X_1=X_2=...=X_n\}$$ In the special case when $n\geq 2$ and the $\{X_i\}$ random variables are i.i.d. with a continuous CDF then it can be shown $P[A]=1$, so $Var(Y_i)=1/n$ for all $i \in \{1, ..., n\}$.
