Vrftsjtryk

I feel like this is a basic question, but I'm having trouble justifying a property. I have the following result (simplified for the post) which I am trying to get an intuitive understanding for:

$P(|A+B| > c) leq P(|A|>c/2) +P(|B|>c/2)$

I believe that this is true but am having trouble working it out.

I can see why $P(|A|>c/2, |B|>c/2) leq P(|A|>c/2) +P(|B|>c/2)$

I guess my question is, is the step I am missing:

$P(|A+B| > c) = P(|A|>c/2, |B|>c/2)$

Some possible steps (there are other routes):

• the triangle inequality gives $|A+B| le |A|+|B|$


• addition gives $|A| le c/2 text { and } |B| le c/2 implies |A|+|B| le c$


• so $|A| le c/2 text { and } |B| le c/2 implies |A+B| le c$


• with the contrapositive $|A+B| > c implies |A|>c/2 text { or } |B|>c/2$
  


• leading to $mathbb P(|A+B| > c) le mathbb P(|A|>c/2 text { or } |B|>c/2)$


• while $mathbb P(|A|>c/2 text { or } |B|>c/2) = mathbb P(|A|>c/2) + mathbb P(|B|>c/2) - mathbb P(|A|>c/2 text { and } |B|>c/2)$ by inclusion-exclusion


• dropping a term gives $mathbb P(|A|>c/2 text { or } |B|>c/2) le mathbb P(|A|>c/2) + mathbb P(|B|>c/2)$


• and combining these gives $mathbb P(|A+B| > c) le mathbb P(|A|>c/2) + mathbb P(|B|>c/2).$