Show that the correlation coeffcient of $x_{i}-\overline{x}$ and $x_{j}-\overline{x}$ is $-(n-1)^{-1}$

Question

Suppose that the second moment exists, and $x_{1},...,x_{n}$ is a group of sample, show that the correlation coeffcient of $x_{i}-\overline{x}$ and $x_{j}-\overline{x}$ is $-(n-1)^{-1}$, here $\overline{x}=\frac{x_{1}+...+x_{n}}{n}$, I know for a group of sample, these $x_{i}$ are independent with each other and they share the same distribution, I do some calculation and here is what I get, and then I don't know what to do

Answer 1

\begin{align} & x_1 - \frac{x_1+x_2+x_3+\cdots+x_n} n \\[8pt] = {} & \left( 1 - \frac 1n\right)x_1+ \frac{x_2 +x_3 + \cdots +x_n} n \\[8pt] \text{So } & \operatorname{var}\left( x_1 - \frac{x_1+x_2+x_3+\cdots+x_n} n \right) \\[8pt] = {} & \left( 1 - \frac 1 n\right)^2 \operatorname{var}(x_1) + \frac 1 {n^2}\operatorname{var} (x_2) +\cdots + \frac 1 {n^2} \operatorname{var}(x_n) \end{align} and all of these variances are the same number. $\require{cancel}$ \begin{align} & \operatorname{cov}\left(x_1 - \frac{x_1+\cdots+x_n}n,\, x_2-\frac{x_1+\cdots + x_n} n \right) \\[8pt] = {} & \cancelto0{\operatorname{cov}(x_1,x_2)} - \operatorname{cov}\left(x_1, \frac{x_1+\cdots+x_n} n\right) \\ & {} - \operatorname{cov}\left(x_2,\frac{x_1+\cdots+x_n} n\right) + \operatorname{var}\left( \frac{x_1+\cdots+x_n} n \right) \end{align} And you seem to know what to do from there.

Another way is to recall that $$ \left( x_1- \frac{x_1+\cdots+x_n} n \right) + \cdots + \left( x_n - \frac{x_1+\cdots +x_n} n \right) = 0 $$ and so the variance of the above must be $0.$ Therefore \begin{align} 0 = {} & \sum_{k\,=\,1}^n \operatorname{var}\left(x_k-\frac{x_1+\cdots+x_n} n \right) \\ & {} - 2\sum_{i,j\,:\,i\,<\, j} \operatorname{cov}\left(x_i - \frac{x_1+\cdots+x_n}n,\, x_j-\frac{x_j-\cdots + x_n} n \right) \\[8pt] = {} & n \operatorname{var}\left(x_k-\frac{x_1+\cdots+x_n} n \right) \\ & {} - 2 \binom n 2 \operatorname{cov}\left(x_1 - \frac{x_1+\cdots+x_n}n,\, x_2-\frac{x_j-\cdots + x_n} n \right) \end{align} and then from the value of that variance deduce the value of the covariances.