How do you prove Cov $\left( \bar{X} , X_i - \bar{X} \right) = 0$?

Hint: Use properties of the covariance function and independence of $X_1,...,X_n$.

Solution:

Note that $$ Cov(\bar{X},X_i-\bar{X}) = Cov(\bar{X},X_i)-Cov(\bar{X},\bar{X}) = Cov(\bar{X},X_i)-Var(\bar{X}) $$

You know that $Var(\bar{X})=\frac{\sigma^2}{n}$. So you need to find $Cov(\bar{X},X_i)$. But,

$$ Cov(\bar{X},X_i)= Cov(\frac{1}{n}\sum_{j=1}^{n}X_j,X_i) = \frac{1}{n} \sum_{j=1}^{n} Cov(X_j,X_i) $$ Since $X_j$'s are independent it follows that $$Cov(X_j,X_i)=0 \text{ for all j } \neq i $$

So we conclude that $$ Cov(\bar{X},X_i) = \frac{1}{n} Cov(X_i,X_i) = \frac{\sigma^2}{n} $$

It follows that $$Cov(\bar{X},X_i-\bar{X}) = 0$$