Two doubts on expectation inequalities

Question

Let's define a probability space $\left(\Omega\text{, }\mathcal{F}\text{, } \mathbb{P}\right)$. Suppose to have two random variables $X$ and $Y$ defined on it.

Two questions:


1. Could I state that $\vert E\{X-Y\}\vert\le E\{\vert X-Y\vert\}$ by Jensen's inequality?

If not, which is the result used to show the above result and how is it applied?;


2. Starting from the fact that $\vert\vert X\vert-\vert Y\vert\vert\le\vert X-Y\vert$, could I end up with $\vert E\{\vert X\vert-\vert Y\vert \}\vert \le E\{\vert X-Y \vert\}$ simply by taking expectations on both sides?
What I mean is the following: if I know that $||X|−|Y||\le|X−Y|$, by monotonicity property of expectation I think that one can get $E\{||X|−|Y||\}\le E\{|X−Y|\}$. And then, since on my book I have the result $|E{|X|−|Y|}|\le E{|X−Y|}$ I am convincing myself that $E\{||X|−|Y||\}=|E\{|X|−|Y|\}|\le E\{|X−Y|\}$, hence that there is some way to prove that $E\{||X|−|Y||\}=|E\{|X|−|Y|\}|$.

If so, why does it hold that $ E\{\vert\vert X\vert-\vert Y\vert\vert\} = \vert E\{\vert X\vert-\vert Y\vert \}\vert $?

Answer 1

1. You can state that and it indeed can be deduced by Jensen's inequality. However, it is more broadly known as the 'triangle inequality for integrals/expectations' and can be shown directly by the face that a Lebesgue-integral is the limit of simple functions, the triangle inequality, and the continuity of the absolute value.

2. That does not follow by taking expectations but by 1. with $$\big \lvert E \big( \vert X \vert - \vert Y \vert \big) \big\rvert \le E\big( \lvert \vert X \vert - \vert Y \vert \rvert \big) \le E \big( \vert X - Y \vert \big) .$$ Following kimchi lover in the comments, I also don't understand your last sentence. The equation you give there is wrong, though.

Answer 2

Here is a simple proof of the inequality in Q1. For any real numbers $x$ and $y$, $$ -|x-y|\le x-y\le |x-y|. $$ Thus, by the monotonicity and linearity of expectations, $$ -\mathsf{E}|X-Y|\le \mathsf{E}[X-Y]\le \mathsf{E}|X-Y|. $$