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How do we show $P(A) leq P(A Delta B) + P(A cap B) leq P(ADelta B) + P(B).$?

Mathematics Asked by user825841 on December 10, 2020

In some questions I have been going through here, I came across this inequality several times.

$A$ and $B$ are RV. How do we show
$$P(A) leq P(A Delta B) + P(A cap B) leq P(ADelta B) + P(B)$$.

One Answer

Based on your comments you meant $lvert{P(A)-P(B)}rvert leq P(ADelta B)$

$A subseteq (A Delta B) cup (A cap B)$, and the two sets on the RHS are disjoint. So $P(A) leq P(A Delta B) + P(A cap B) leq P(ADelta B) + P(B).$

So $P(A) - P(B) leq P(ADelta B)$. We can similarly show that $P(B)-P(A) leq P(ADelta B)$ and the result follows.

Correct answer by riemleb on December 10, 2020

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