Covariance between centered and scaled normal entries of a random vector

Question

From this post and following a tip by @Ian in the comments.

If $X_1,\dots,X_n \sim \text{ i.i.d. } N(\mu,\sigma^2)$ with $\displaystyle \bar X= \frac{\sum_{i=1}^n X_i}{n},$ the covariance of the random variables $Y_i=\frac{X_i-\bar X}{\sigma}$ (which I supposed can be read as centered and scaled entries of a random vector) is $-1/n.$

I see that

$$\begin{align} \small \operatorname{cov}\left(Y_i,Y_j\right) = {} & \small \operatorname{cov} \left(\frac{(X_i-\bar X)}{\sigma},\frac{(X_j-\bar X)}{\sigma}\right)\\[2ex] = {} &\small\frac{1}{\sigma^2}E\big[(X_i-\bar X)(X_j-\bar X)\big] - E\big[X_i-\bar X\big]E\big[X_j-\bar X\big]\\[2ex] = {} & \small \frac{1}{\sigma^2}E\big[(X_i-\bar X)(X_j-\bar X)\big]\\[2ex] = {} & \small \frac{1}{\sigma^2}E[(X_i-\mu+\mu-\overline{X})(X_j-\mu+\mu-\overline{X})]\\[2ex] = {} &\small \frac{1}{\sigma^2}\left( E[(X_i-\mu)(X_j-\mu)]+E[(X_i-\mu)(\mu-\overline{X})]+E[(X_j-\mu)(\mu-\overline{X})]+E[(\mu-\overline{X})^2]\right)\\[2ex] = {} &\small\frac{1}{\sigma^2}\left( E[(X_i-\mu)(\mu-\overline{X})]+E[(X_j-\mu)(\mu-\overline{X})]+E[(\mu-\overline{X})^2]\right) \end{align}$$

The last term $E[(\mu-\overline{X})^2=\operatorname{var}(\bar X)=\frac{\sigma^2}{n}$ as derived here.

Does, then

$$E[(X_i-\mu)(\mu-\overline{X})]+E[(X_j-\mu)(\mu-\overline{X})] \overset{?}= \frac{-2\sigma^2}{n} \text{ ?}$$

Answer 1

$\newcommand{\c}{\operatorname{cov}}$You seem to be doing it the hard way.

\begin{align} & \c(X_i - \overline X, X_j - \overline X) \\[10pt] = {} & \c(X_i,X_j) - \c(X_i,\overline X) - \c(\overline X, X_j) + \c(\overline X,\overline X) \\[10pt] = {} & 0 - \c\left( X_i, \frac {X_1+\cdots+X_n} n \right) - \c\left( \frac {X_1+\cdots+X_n} n, X_j \right) + \operatorname{var}(\overline X) \\[10pt] = {} & - \frac{\sigma^2} n - \frac{\sigma^2} n + \frac{\sigma^2} n = \frac{-\sigma^2} n. \end{align}