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 $$P(A) \leq P(A \Delta B) + P(A \cap B) \leq P(A\Delta B) + P(B)$$.
2026-04-03 12:19:08.1775218748
How do we show $P(A) \leq P(A \Delta B) + P(A \cap B) \leq P(A\Delta B) + P(B).$?
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Based on your comments you meant $\lvert{P(A)-P(B)}\rvert \leq P(A\Delta B)$
$A \subseteq (A \Delta B) \cup (A \cap B)$, and the two sets on the RHS are disjoint. So $P(A) \leq P(A \Delta B) + P(A \cap B) \leq P(A\Delta B) + P(B).$
So $P(A) - P(B) \leq P(A\Delta B)$. We can similarly show that $P(B)-P(A) \leq P(A\Delta B)$ and the result follows.