How to show that P(A ∩ B) ≤ P(A) ≤ P(A ∪ B)?

Question

Assume that A, B are two events on a probability space.

Show that $P(A \cap B) \leq P(A) \leq P(A \cup B).$

Answer 1

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 $$\Bbb P(E_2) = \underbrace{\Bbb P(E_2\setminus E_1)}_{\geq 0} + \Bbb P(E_1) \geq \Bbb P(E_1).$$

Answer 2

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.

Answer 3

You can deduce it by using:

• $P(A\cap B)=P(A)P(B\mid A)=P(B)P(A\mid B)$
• $P(A\cup B)=P(A)+P(B)-P(A\cap B)$

Then (since $P(B\mid A)\leq 1$) $$P(A\cap B)=P(A)P(B\mid A)\leq P(A)$$

Also since $P(B)\geq P(B)P(A\mid B)$ we have: $$P(A)\leq P(A)+P(B)-P(A\cap B)=P(A\cup B)$$

Combining these two inequalities we get: $$P(A\cap B)\leq P(A)\leq P(A\cup B)$$

Answer 4

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.