We have X and Y, independent, zero mean random variables. Show that:
$$E|X+Y| \geq E|X|$$
What I tried:
I treated them as continuous variables, though it should be true in general. But I could not derive the proof. I found that it should be proven:
$$\int_{-\infty}^{\infty}\int_{-x}^{\infty}(x+y)f(x)g(y)dydx \geq \int_0^{\infty} xf(x)dx$$
where we have the constraints:
$$\int f(x)dx=1$$
$$\int x\cdot f(x)dx = 0$$
$$\int g(y)dy=1$$
$$\int y\cdot g(y)dy = 0$$
We have that
$$|EX| \leq E|X|$$
which is true in general.
$$E(abs(X+Y)| X) \geq |E(X+Y|X)| = |X + E(Y|X)| = |X|$$
take the expected value (and use the tower rule, aka. the law of total expectation):
$$EE(abs(X+Y)|X)\equiv E|X+Y| \geq E|X|$$