Let X and Y be random variables defined on the probability space $(\Omega, \mathcal{S}, \mathcal{P})$.
I have:
(i) $\mathcal{P}(|X+Y| >\epsilon) \leq \mathcal{P}(X>\epsilon/2) + \mathcal{P}(Y>\epsilon/2) $
(ii) $\mathcal{P}(|X| >\epsilon, |Y|> \epsilon)= c$ implies $ \mathcal{P}(|X+Y|>2\epsilon) =c$
(The real problems are more complicated, I'm just trying to simplify the part I am confused. )
I'm getting caught up in combining and breaking a part these probability measures. I feel like some of these relationships have to do with the triangle inequality. Could use some help understanding the intuition.
(i) Follows from $\{|X+Y| >\epsilon\} \subset \{|X| >\epsilon /2\} \cup \{|Y| >\epsilon /2\} $. (As noted by aleph_two absolute value signs on RHS are necessary).
(ii) $X=-Y=2\epsilon $ is a counterexample.