A probability inequality about unit second moment

Question

Let $X$ be a r.v. with $E[X^2]=1$ and $E[|X|]\geq a>0$, prove that \begin{equation*} P[|X|\geq\lambda a]\geq(1-\lambda)^2a^2 \end{equation*} holds for $0\leq\lambda\leq 1$. I tried to use the Markov inequality with respect to \begin{equation} P[|X|\leq\lambda a]\leq P[||X|-a|\geq(1-\lambda) a] \end{equation} but failed unfortunately. So what can I do?

Answer 1

Write $E|X| = E[|X| \mathbf{1} \{|X| < \lambda a\}] + E[|X| \mathbf{1} \{|X| \geq \lambda a\}]$

Immediately, $E[|X| \mathbf{1} \{|X| < \lambda a\}] \leq \lambda a$ and from the Cauchy-Schwartz inequality $E[|X| \mathbf{1} \{|X| \geq \lambda a\}] \leq \sqrt{E[X^2]} \cdot \sqrt{E[\mathbf{1} \{|X| \geq \lambda a\}^2]} = 1 \cdot P(|X| \geq \lambda a)$, it follows that

$$a \leq E|X| \leq \lambda a + \sqrt{P(|X| \geq \lambda a)}$$

and the result follows.