Suppose $X_1,...,X_n$ are iid as $N(\mu,\sigma^2)$. I'm trying to find the Variance of $X_1-\bar{X}$. Here's what I have:
\begin{align} Var(X_1-\bar{X})&=Var(X_1)+Var(\bar{X})-2Cov(X_1,\bar{X}) \\ &= \sigma^2+\frac{\sigma^2}{n}-2[E(X_1\bar{X})-E(X_1)E(\bar{X})] \\ &= \sigma^2+\frac{\sigma^2}{n}-2[E(X_1\bar{X})-\mu^2] \end{align}
The $E(X_1\bar{X})$ is throwing me off here, as I'm pretty sure $X_1$ is not independent of $\bar{X}$. What am I missing here?
$EX_i\overline {X}=\frac 1 n EX_i \sum_j X_j=\frac 1 n \sum_j EX_iX_j$. Note that $EX_iX_j=EX_i^{2}=\mu^{2}+\sigma ^{2}$ if $i=j$ and $EX_iX_j=\mu^{2}$ if $i \neq j$.