An alternative measure of scatter. The context:
Let $X$ be a random variable. A familiar measure for its scatter is variance
$$\text{Var}(X)=\mathbb{E} \left( \left( X- \mathbb{E}(X) \right)^2 \right),$$
which has several nice properties. A less common alterative is the "$L_1$-version" of variance
$$\text{MAD}(X)=\mathbb{E} \left( \, \left| X- \mathbb{E}(X) \right| \, \right)$$
that is a lot more difficult to deal with theoretically. However, sometimes when considering samples of observations following a heavy-tailed distribution, this quantity is more useful for applications since it is more robust against extreme values.
The question: Let $X$ and $Y$ be i.i.d. Clearly $$\text{Var}(X+Y) = \text{Var}(X)+ \text{Var}(Y) \geq \text{Var}(X).$$
Do we have a similar inequality for MAD? I.e. is it true that
$$\text{MAD}(X+Y) \geq \text{MAD}(X)?$$
Intuitively, yes, but I'm not sure what to do. Tried Google, but no luck.
 A: There is a much more general result which is true.

Proposition
Let $X$ and $Y$ be independent, real-valued, integrable and centered
  random variables. Let $\varphi$ be a non-negative convex function.
  Then:
$$\mathbb{E} (\varphi (X+Y)) \geq \mathbb{E} (\varphi (X)).$$

Proof
The main tool is a conditional version of Jensen's inequality (e.g. Intégration, probabilités et processus aléatoires, p.148, property f).
Since $Y$ is centered, and $X$, $Y$ are independent,
$$0 = \mathbb{E}(Y) = \mathbb{E}(Y|X).$$
Hence,
$$\varphi (X) = \varphi (\mathbb{E}(X|X)) = \varphi (\mathbb{E}(X+Y|X)).$$
By the conditional version of Jensen's inequality,
$$\varphi (X) \leq \mathbb{E}(\varphi (X+Y)|X).$$
Finally, taking the expectation on both sides,
$$\mathbb{E}(\varphi (X)) \leq \mathbb{E}(\mathbb{E}(\varphi (X+Y)|X)) = \mathbb{E}(\varphi (X+Y)).$$
To get the inequality on the variances, take $\varphi (x) = x^2$. To get the one you want, take $\varphi (x) = |x|$. This works even if $X$ and $Y$ do not share the same distribution.
