Probability Question And, Or, and Bivariates

Question

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:

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

Answer 1

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 $\mathbb P(|A+B| > c) \le \mathbb P(|A|>c/2) + \mathbb P(|B|>c/2).$