Finding the expected value of the sum of $x_i^2$

Question

I'm trying to figure out how to prove $$E\!\left(\bar{x} ^2\right) = \frac{1}{n} \sum_{i=0}^n \left(x_i ^2\right)$$

This is for a bootstrap, with no specification as to the distribution.

I understand that the lefthand side equals $$\frac{1}{n ^2} E\!\left(\left(\sum_{i=0}^n x_i\right)^2\right)$$ but I'm not entirely sure how to break it up from there.

Any tips would be appreciated!

Answer 1

$\bar x = E[x] = \frac 1n \sum_\limits{i=1}^n x_i\\ E[x^2] = \frac 1n \sum_\limits{i=1}^n x_i^2$

$\frac 1n \sum_\limits{i=1}^n (x_i-\bar x)^2 = $$\frac 1n \sum_\limits{i=1}^n (x_i^2- 2\bar xx_i +\bar x^2)\\ \frac 1n \sum_\limits{i=1}^n (x_i^2) - 2\bar x\frac 1n \sum_\limits{i=1}^n (x_i) +\bar x^2\\ E[x^2] - E[x]^2$

For all $i, (x_i - \bar x)^2 \ge 0$

$E[x^2] - E[x]^2 \ge 0$

$E[x^2] = E[x]^2$ if and only if all $x_i = \bar x$ (or all $x_i=x_j$ for all $i,j$)