Proof of Inequality (10)

About Posts which Tagged by 'Probability'

Let $X$ be a random variable.  If $\mathscr{E}(X^2)=1$ and $\mathscr{E}|X|\geq a>0$, then $$\mathscr{P}\{|X|\geq\lambda a\}\geq(1-\lambda)^2a^2\mbox{  for }0\leq\lambda\leq1.$$

$\bullet$ Proof.

By H$\ddot{o}$lder's inequality, let $p=q=2$ such that $\frac{1}{p}+\frac{1}{q}=1$ and define the indicator function $I\{|X|\geq\lambda a\}=1$, if $|X|\geq\lambda a$, and 0, otherwise.  We have $$\mathscr{E}\left(|X|\,I\{|X|\geq\lambda a\}\right)\leq\mathscr{E}\left(X^2\right)^{\frac{1}{2}}\mathscr{E}\left(I\{|X|\geq\lambda a\}\right)^{\frac{1}{2}}=\mathscr{P}\{|X|\geq\lambda a\}^{\frac{1}{2}}.$$On the other hand, $$\begin{array}{rl}\mathscr{E}\left(|X|\,I\{|X|\geq\lambda a\}\right)&=\mathscr{E}\left(|X|(1-I\{|X|<\lambda a\})\right)\\  &=\mathscr{E}|X|-\int_{|X|<\lambda a}|X|\,d\mathscr{P}\\  &\geq\mathscr{E}|X|-\lambda a\int_{|X|<\lambda a}\,d\mathscr{P}\\ &\geq\mathscr{E}|X|-\lambda a\int_\Omega\,d\mathscr{P}\\ &\geq a-\lambda a=(1-\lambda)a \end{array}$$by assumption.  Thus we have $$\mathscr{P}\{|X|\geq\lambda a\}^{\frac{1}{2}}\geq\mathscr{E}\left(|X|\,I\{|X|\geq\lambda a\}\right)\geq(1-\lambda)a,$$that is, $$\mathscr{P}\{|X|\geq\lambda a\}\geq(1-\lambda)^2a^2.$$

$\Box$