Expected value is a linear operator? Under what conditions is median also a linear operator? I have always taken for granted that expected value is a linear operator. For any random variables $X$ and $Y$: $E(aX + bY) = aE(X) + bE(Y)$. Can anyone point me to a rigorous proof of this?

Also, I know that generally median $Med()$ is not a linear operator, meaning $Med(aX + bY)$ might not be equal to $a Med(X) + b Med(Y)$. Are there absolute criteria / rules when $Med$ is a linear operator, and when it is not?
First question: By definition, on a probability space ($\Omega,\mathcal{F},P$), the expected value of a random variable $X:\Omega\to \mathbb{R}$ is defined as
$$E(X)=\int_\Omega X(\omega) dP(\omega).$$ Note that it is only well-defined if the integral converges absolutely, i.e. $$\int_\Omega |X(\omega)| dP(\omega)&lt;\infty$$

(The integrals above are Lebesgue integrals. If $X$ is discrete, $P$ is the point measure and this integral turns into a sum.)

Therefore, if the expected values of both $X$ and $Y$ exist, then $E(X+Y)$ exists (triangle inequality) and, for $a,b\in \mathbb{R}$, we can use the linearity of the Lebesgue integral to conclude
$$E(aX+bY)=\int_\Omega aX +bYdP=a\int_\Omega X dP+b\int_\Omega YdP=aE(X)+bE(Y).$$
I would venture that expectation being a linear operator is somewhat in the nature of an axiom&mdash;that each generalization to more complex sorts of distributions would have to satisfy this axiom to be considered sensible.