Expectation of $\bar X^2$

If all enumerated $X$s are observations from a population with a population $\mu$ and variance $\sigma^2$:

Why is this true?

$$\mathbb E\!\left(\bar X ^2\right) = \frac{\sigma^2}{n}+\mu^2$$

Source.

1 Answer

I think you mean if $X_i$ are iid...

As a hint, recall that $$\text{Var}[\bar X] = E[\bar X^2]-\{E[\bar X]\}^2.\tag a$$

Expand and solve $$E[\bar X] = E\left[\frac{X_1+\dotsb+X_n}{n}\right] \tag b$$ and $$ \text{Var}[\bar X]=\text{Var}\left[\frac{X_1+\dotsb+X_n}{n}\right].\tag c$$ Remember to use independence.

Finally, use (b) and (c) to solve (a) $$E[\bar X^2] = \text{Var}[\bar X]+\{E[\bar X]\}^2.$$