Proof variance from mean and overall are not covaried?

Question

Let $X_1, X_2,..., X_n$ be independent and identically distributed random variables having variance $\sigma^2$

Show that $Cov [X_i − \bar{X},X] = 0$

So I get that $E[X_i - \bar{X}] = 0$, so all I'm left with proving is that $E[X_i − \bar{X},X] = 0$.

Here we assume that $X$ is $\sum_{i=1}^n{X_i}$

Answer 1

So you want $C=E((X_j-X/n)X)$, where $X=\sum_{i=1}^nX_i$, where the $X_i$ are iid $N(0,1)$. Well, $C= E(X_jX)-E(X^2)/n = \sigma^2 - n\sigma^2/n = 0$. Because $EX_j X = E X_j^2$ and because $X\sim N(0,n\sigma^2)$.

Answer 2

Use the fact that covariance is bilinear. Indeed, \begin{align*} \text{Cov}(X_i − \bar{X},X) &=\text{Cov}(X_i,X)-\text{Cov}(\bar{X},X)\\ &=\sum_j\text{Cov}(X_i,X_j)-\frac{1}{n}\sum_i\sum_j\text{Cov}(X_j,X_i)\\ &=\sigma^2-\frac{1}{n}(n\sigma^2)=0\tag{1}. \end{align*} where we used the fact that the $X_i$ are independent and hence uncorrelated in (1).