Seeking a More Elegant Proof to an Expectation Inequality

Question

Let $X$ and $Y$ be i.i.d. random variables, and $\mathbb E[|X|]<\infty$, prove that $$\mathbb E[|X+Y|]\geq\mathbb E[|X-Y|].$$

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This question is a re-posting of An expectation inequality. I can prove this with integration. But there must be a more elegant proof via perhaps a symmetry argument. Can someone come up with such a one?

Answer 1

A first observation is that $|x+y|\geqslant|x-y|$ if and only if $xy\geqslant 0$, so defining $$f(x,y):=|x+y|-|x-y|,$$ the positivity of $f$ is linked to that of $xy$. We would like to find a more tractable expression for $f$.

• Assume that $x\gt 0$ and $ y\gt 0$. Then $f(x,y)=x+y-|x-y|=2\min\{x,y\} =\min\{|x|,|y|\}$.
• Since $f(-x,-y)=f(x,y)$ we get $f(x,y)=2\min\{|x|,|y|\}$ if $xy\gt 0$.
• Assume that $x\gt 0$ and $y\lt 0$. Then $x-y\gt 0$, hence $|x-y|=x-y$ $$f(x,y)=|x+y|+y-x=-(x-y-|x-(-y)|)=-2\min\{x,-y\}=-2\min\{|x|,|y| \}.$$
• By symmetry of $f$, we obtain this expression if $x\lt 0$ and $y\gt 0$.

To sum up: for each $(x,y)\in\mathbf R^2$, $$f(x,y)= 2\min\{|x|,|y|\}\left(\mathbf 1\{xy\gt 0\}-\mathbf 1\{xy\lt 0\}\right).$$ Now we compute the expectation: using the fact that the random variable are i.i.d., we have $$\mathbb E[\min\{|X|,|Y|\}\mathbf 1\{XY\gt 0\}]=\int_0^{+\infty}\mu\{X\gt t\}^2+ \mu\{-X\gt t\}^2\mathrm dt \mbox{ and } $$ $$\mathbb E[\min\{|X|,|Y|\}\mathbf 1\{XY\lt 0\}]=2\int_0^{+\infty}\mu\{X\gt t\}\cdot \mu\{-X\gt t\} \mathrm dt. $$ This follows from the equality
$$\mathbb E[Y]=\int_0^{+\infty}\mu\{Y\gt t\}\mathrm dt $$ and the fact that $$\mu\left(\{ \min\{|X|,|Y|\}\gt t \}\cap\{X\gt 0\}\cap\{Y\gt 0\}\right) =\mu\{X\gt t\}\mu\{Y\gt t\} ,$$ $$\mu\left(\{ \min\{|X|,|Y|\}\gt t \}\cap\{X\lt 0\}\cap\{Y\lt 0\}\right) =\mu\{-X\gt t\}\mu\{-Y\gt t\}\mbox{ and } $$ $$\mu\left(\{ \min\{|X|,|Y|\}\gt t \}\cap\{X\lt 0\}\cap\{Y\gt 0\}\right) =\mu\{X\gt t\}\mu\{-Y\gt t\}. $$ We thus infer that $$\mathbb E|X+Y|-\mathbb E|X-Y|=2 \int_0^{\infty}\left(\mu\{X\gt t\}-\mu\{-X\gt t\} \right)^2\mathrm dt. $$ This gives the wanted lower bound, and it shows that the equality is achieved if and only if $X$ is symmetric.