Can anyone solve this E[Xi X-bar]?

Question

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.

Answer 1

\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*}

Answer 2

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.

Answer 3

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 $$n\left(\frac{\sigma^2}{n}+\mu^2\right)$$.