Consider two real-valued random variables $X,Y$ defined on the probability space $(\Omega, \mathcal{F}, P)$. let $\epsilon>0$. Could you help me to show that
$$
P(|X|+|Y|\geq 2\epsilon^2)\leq P(|X|\geq \epsilon^2)+P(|Y|\geq \epsilon^2)
$$
I don't see why. I think this expression is used for example in van der Vaart "Asymptotic Statistic" p.95 in the proof of Theorem 7.2
Best Answer
We have $$\{|X|+|Y|\geqslant 2\varepsilon^2\}\subset\{|X|\geqslant\varepsilon^2\}\cup \{|Y|\geqslant\varepsilon^2\}, $$ which yields \begin{align} \mathbb P(|X|+|Y|\geqslant 2\varepsilon^2) &\leqslant \mathbb P(\{|X|\geqslant\varepsilon^2\}\cup \{|Y|\geqslant\varepsilon^2\})\\ &\leqslant \mathbb P(|X|\geqslant\varepsilon^2)+\mathbb P(|Y|\geqslant\varepsilon^2). \end{align}