Proof the E(x+y)=E(x)+E(y) using characteristic function

Question

Let y be independent of x, then it holds that

$$E(x+y)=E(x)+E(y)$$ and also that $$V(x+y)=V(x)+V(y)$$

This blog proves both of these theorems. I'm curious if it is possible to prove the same using moments of the characteristic functions and the fact that $$\phi_{x+y}(t) = \phi_x(t)\phi_y(t) = E[\exp(itX)]E[\exp(itY)]$$

Maybe something along the lines $$E[x+y]=i^{-1}*\frac{d}{dt}(E[\exp(itX)]E[\exp(itY)]) = \dots = E[x]+[y]$$

Answer 1

The relation $\mathbb E[x+y]=\mathbb E[x]+\mathbb E[y]$ holds always as mentioned in the comments.

Proof of $\mathbb V[x+y]=\mathbb V[x]+\mathbb V[y]$:

From $\phi_{x+y}(t)=\phi_x(t)\phi_y(t)$ we get \begin{align} \frac{d}{dt}\phi_{x+y}(t)&=\phi_y(t)\frac{d}{dt}\phi_x(t)+\phi_x(t)\frac{d}{dt}\phi_y(t)\\ \frac{d^2}{dt^2}\phi_{x+y}(t)&=\phi_y(t)\frac{d^2}{dt^2}\phi_x(t)+\phi_x(t)\frac{d^2}{dt^2}\phi_y(t)+2\frac{d}{dt}\phi_x(t)\frac{d}{dt}\phi_y(t)\\ \end{align} and therefore \begin{align} \mathbb E[(x+y)^2]&=-\frac{d^2}{dt^2}\Bigg|_0\phi_{x+y}(t)\\ &= -\phi_y(0)\frac{d^2}{dt^2}\Bigg|_0\phi_x(t)-\phi_x(0)\frac{d^2}{dt^2}\Bigg|_0\phi_y(t)-2\frac{d}{dt}\Bigg|_0\phi_x(t)\frac{d}{dt}\Bigg|_0\phi_y(t)\\[2mm] &=\mathbb E[x^2]+\mathbb E[y^2]+2\,\mathbb E[x]\,\mathbb E[y]\,. \end{align} Therefore, \begin{align} \mathbb V[x+y]&=\mathbb E[(x+y)^2]-\mathbb E[x+y]^2\\ &=\mathbb E[x^2]+\mathbb E[y^2]-\mathbb E[x]^2-\mathbb E[y]^2\\ &=\mathbb V[x]+\mathbb V[y]\,. \end{align}