Why does $\mathbb{E}(\frac{X_1+...+X_n}{n})=\mathbb{E}(X)$?

Question

Let $X_1$, $X_2$, $...$, be i.i.d. and follow the same distribution as a random variable $X$ that has an expectation $\mathbb{E}(X)$ and a finite variance $\operatorname{Var}(X)$.

Why does it follow that:

$$\mathbb{E}\left(\frac{X_1+\dots+X_n}{n}\right)=\mathbb{E}(X)?$$

This assertion is in a proof of the weak law of large numbers.

Answer 1

Since the expectation is a linear operator, $$E( (X_1+...+X_n)/n ) = E(X_1)/n + ... + E(X_n)/n$$

Since $X_1,..., X_n$ have the distribution of $X$, this simplifies to

$$E(X)/n + ... + E(X)/n=nE(X)/n=E(X)$$