# Probability of symmetric difference

How do we prove the following inequality? $$\mathbb{P}[A\triangle B ]\geq \max\left \{ \mathbb{P}[A-B],\mathbb{P}[B-A] \right \}$$ I already proved that $$\mathbb{P}[A\triangle B ]=\mathbb{P}[A]+\mathbb{P}[B]-2\mathbb{P}[A\cap B]$$ $$\left | \mathbb{P}[A]-\mathbb{P}[B] \right |\leq \mathbb{P}[A\triangle B ]$$

• Can you use other inequalities or equalities? Perhaps the definition $P(A-B)=P(A)-P(A\cap B)$? Commented Mar 9, 2018 at 15:28
• Yes yes I can use and I was looking for that kind of probability difference that you posted , I think it would be easy to solve it now . Commented Mar 9, 2018 at 15:29

We have $$P(A-B)=P(A)-P(A\cap B)$$ $$P(B-A)=P(B)-P(B\cap A)$$ Using these we can rewrite the first relation as $$P(A\triangle B)=P(A-B)+P(B-A)$$ Since probabilities are non-negative: $$P(A\triangle B)\ge\max(P(A-B),P(B-A))$$
Hint: Prove that $$P(A\Delta B)=P(A\setminus B)+P(B\setminus A)$$
Therefore, $P(A\Delta B)$ is no less than $P(A\setminus B)$ and no less than $P(B\setminus A)$.
Note that $$A\Delta B=(A\setminus B)\cup (B\setminus A)$$ whence $$(A\setminus B)\subseteq A\Delta B\quad \text{and}\quad (B\setminus A)\subseteq A\Delta B.$$ In particular $$P(A\setminus B)\leq P(A\Delta B) \quad \text{and} \quad P(B\setminus A)\leq P(A\Delta B)$$ so that $$\max\{ P(A\setminus B), P(B\setminus A) \} \leq P(A\Delta B)$$