
Correlation of samples
Let $\displaystyle X_i, i= 1,..,n,$ be i.i.d. sampes from $\displaystyle N(\mu, \sigma^2).$ Let $\displaystyle \bar{X} = \frac{1}{n}\sum_{i=1}^{n} X_i.$
Prove that $\displaystyle \bar{X}$ and $\displaystyle X_i  \bar{X}$ are uncorrelated for any $\displaystyle i.$

for simplicity let i=1....
Let's obtain the covariance between $\displaystyle \bar X$ and $\displaystyle X_1\bar X$
$\displaystyle Cov(\bar X,X_1\bar X) ={1\over n} Cov\left(X_1+\cdots +X_n,\left(1{1\over n}\right)X_1{1\over n}[X_2+\cdots +X_n]\right)$
Now find the covariance between each pair, taking one term from each set.
ALL we need is for the $\displaystyle Cov(X_i,X_j)=0$ for each $\displaystyle i\ne j$ , we don't need NORMALITY...
$\displaystyle \left({1\over n}\right)\left[\left(1{1\over n}\right)\sigma_1^2{1\over n}\left[\sigma_2^2+\cdots +\sigma_n^2\right]\right)$
Next use the fact that all the variances are equal this becomes zero.
Changing 1 to i is easy, just sum over all the terms that's not i in the second sum.