Can anyone solve this E[Xi X-bar]? I am trying to solve this expectation and I don't seem to find a logical way to solve it?
Here:
$\mathrm{E}[X_i*\bar{X}]$
$X_i$ is iid. 
 A: \begin{align*}
E[X_i \overline{X}] =& E\left[ X_i \frac{1}{n} \sum_{j=1}^n X_j\right]
\\=& \frac{1}{n}\sum_{j=1}^n E\left[ X_i  X_j\right]
\end{align*}
Now they are independent so we can split up the product of expectations expect for the case when $j=i$ (because a random variable is never independent of itself):
\begin{align*}
E[X_i \overline{X}] =& \frac{1}{n}\sum_{\substack{j=1 \\ j\neq i}}^n E\left[ X_i\right]  E\left[X_j\right] + \frac{1}{n} E[X_i^2].
\end{align*}
Since they are equally distributed then $E[X_j] = E[X_i]$ for all $i,j$ (that is they have the same expected value). So we can for instance take $j=i$ and we can write
\begin{align*}
E[X_i \overline{X}] =& \frac{n-1}{n} E\left[ X_i\right]^2+ \frac{1}{n} E[X_i^2].
\end{align*}
A: As $\bar{X}$ is just $\frac{\sum{X_j}}{n}$ then it reduces to the sum of  $E[X_i*X_j]$ which is easy as $X_i$ is iid. 
A: I know this is an old question, but I think there may be an easier way to do this.
Assuming $(x_1, \cdots, x_n)$ are i.i.d. with mean $\mu$ and variance $\sigma^2$.
Given $$\bar{X} = \frac{\sum_{i=1}^{n}{X_i}}{n},$$ and $$n\bar{X}=\sum_{i=1}^{n}{X_i},$$
we can observe:
$$\mathbf{E}\left[\sum_{i=1}^{n}{X_i \bar{X}}\right] = \mathbf{E}\left[\bar{X} 
 n\bar{X}\right] = n\mathbf{E}\left[\bar{X}^2\right],$$ then using the fact that $$\mathbf{E}\left[\bar{X}^2\right]=\frac{\sigma^2}{n}+\mu^2,$$ the final answer is $$n\left(\frac{\sigma^2}{n}+\mu^2\right)$$.
