Weak symmetrization inequalities

Question

Let $X$ and $X'$ be iid random variables, $X^s=X-X'$ be the symmetrization of $X$, $mX$ be the median of $X$, i.e., the number satisfying $\Pr(X\geqslant mX)\geqslant \frac{1}{2}\leqslant \Pr(X\leqslant mX)$. I want to prove the weak symmetrization inequality: for any $x$, $$ \frac{1}{2}\Pr(|X-mX|\geqslant x)\leqslant\Pr(|X^s|\geqslant x).\qquad (1) $$ I can prove the inequality without absolute value, i.e., $$ \frac{1}{2}\Pr(X-mX\geqslant x)\leqslant\Pr(X^s\geqslant x). \qquad (2) $$ Then, by changing $X$ into $-X$, we have $$ \frac{1}{2}\Pr(-X+mX\geqslant x)\leqslant\Pr(-X^s\geqslant x)=\Pr(X^s\geqslant x).\qquad (3) $$ My question is that: how can we prove eq. $(1)$ by combining eq.'s $(2)$ and $(3)$?

Answer 1

Note that for any random variable $Z$ and positive $t$ the events $\{Z \geq t\}$ and $\{Z \leq -t\}$ are disjoint. Then, using what you have already proved: \begin{align*} \mathbf{P}(|X - mX| \geq t) &= \mathbf{P}(X - mX \geq t) + \mathbf{P}(X - mX \leq -t) \\ &\leq 2 \left( \mathbf{P}(X^s \geq t) + \mathbf{P}(-X^s \geq t) \right) \\ &= 2 \left( \mathbf{P}(X^s \geq t) + \mathbf{P}(X^s \leq -t) \right) \\ &= 2 \left( \mathbf{P}(|X^s| \geq t) \right) \end{align*}