How to show that P(A ∩ B) ≤ P(A) ≤ P(A ∪ B)?
Solution 1
Both inequalities follows directly from the monotonicity of the measure $\mathbb{P}$ and the facts $A \cap B \subset A$ and $A \subset A \cup B$.
For reference: monotonicity of measure under properties.
Solution 2
It follows from the inclusions $A \cap B \subseteq A \subseteq A \cup B$.
Applying $\Bbb P$ gives $$\Bbb P(A \cap B) \leq \Bbb P(A) \leq \Bbb P(A \cup B).$$
To see in general that $E_1 \subseteq E_2$ implies $\Bbb P(E_1) \leq \Bbb P(E_2)$, write the disjoint union $E_2 = (E_2\setminus E_1)\cup E_1$, and use $$\Bbb P(E_2) = \underbrace{\Bbb P(E_2\setminus E_1)}_{\geq 0} + \Bbb P(E_1) \geq \Bbb P(E_1).$$
Solution 3
In general if $E\subset F$ are two events, then $$ P(E)\leq P(F) $$ since $$ P(F)=P(E)+P(F\setminus E) $$ as well as the fact that $P\geq 0$, $F=E\cup(F\setminus E)$ and $P$ is countably additive (and hence finitely additive.
From this result, your inequality follows.
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panda67
Updated on August 19, 2022Comments
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panda67 about 1 year
Assume that A, B are two events on a probability space.
Show that $P(A \cap B) \leq P(A) \leq P(A \cup B).$
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Nate Eldredge almost 6 yearsWhat do you know? What have you tried?
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cansomeonehelpmeout almost 6 years$P(A\cap B)=P(A)P(B\mid A)$, so $P(A\cap B)=P(A)P(B\mid A)\leq P(A)$
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