Understanding $E(\frac{1}{n}\sum_{i=1}^n{X_i})=\frac{1}{n}nEX_1=\mu$ in point estimates

Question

Let $X_1, ..., X_n$ be a sample of a distribution with expected value $\mu$ and dispersion $\sigma^2$. Then $E\bar{X}=\mu$.

Now the prove goes like this:

$E\bar{X}=E(\frac{1}{n}\sum_{i=1}^n{X_i})=\frac{1}{n}nEX_1=\mu$

My question is why $E(\frac{1}{n}\sum_{i=1}^n{X_i})=\frac{1}{n}nEX_1=\mu$ ?

Answer 1

$$E\left(\frac{1}{n}\sum_{i=1}^n X_i\right) = \frac1nE\left(\sum_{i=1}^n X_i\right) = \frac{1}{n}\sum_{i=1}^nE\left( X_i\right)$$

by linearity (i.e. $E(\alpha X)=\alpha E(X)$ and $E(X+Y)=E(X)+E(Y)$).

Then, use the fact that $X_i$ are equally distributed, which means $E(X_1)=E(X_2)=\cdots=E(X_n)$