If $A,B,C$ are events on a probability space then $P(A\cap B\cap C)\geq P(A)+P(B)+P(C)-2$

Question

The Problem: Show that if $A,B,C$ are events on a probability space $(\Omega,\mathcal F,P)$, then $$P(A\cap B\cap C)\geq P(A)+P(B)+P(C)-2.$$

My Thoughts: It follows from the inclusion-exclusion formula that $$P(A\cup B\cup C)=P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C),$$ and since $(A\cup B\cup C)\subset\Omega$, monotonicity of the probability measure, yields $P(A\cup B\cup C)\leq P(\Omega)=1$, so $$1\geq P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C).$$ Now observe that $A\cap B=(A\cap B\cap C^\complement)\cup(A\cap B\cap C)$ is a union of disjoint events, and so is $A\cap C=(A\cap B^\complement\cap C)\cup(A\cap B\cap C).$ Hence, finite additivity yields $$P(A\cap B)=P(A\cap B\cap C^\complement)+P(A\cap B\cap C),$$ $$P(A\cap C)=P(A\cap B^\complement\cap C)+P(A\cap B\cap C).$$ Finally, we note that $(A\cap B\cap C^\complement)\cap(A\cap B^\complement\cap C)\cap(B\cap C)=\varnothing$, so that finite additivity and monotonicity of the probability measure yield \begin{equation}\begin{split} &\quad\,\,P(A\cap B\cap C^\complement)+P(A\cap B^\complement\cap C)+P(B\cap C)\\ &=P((A\cap B\cap C^\complement)\cup(A\cap B^\complement\cap C)\cup(B\cap C))\\ &\leq P(\Omega)\\ &=1. \end{split}\end{equation} Putting all the above information together and noting that $P(G)\geq0$ for all events $G$ we see that $$P(A\cap B\cap C)\geq P(A)+P(B)+P(C)-2.$$

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Do you agree with the proof presented above? Any feedback is much appreciated. Thank you for your time.

Answer 1

Your solution is correct except for one slight detail. Where you say

Finally, we note that $(A\cap B \cap C^c)\cap (A\cap B^c\cap C)\cap(B\cap C)=\varnothing$, so that finite additivity and monotonicity of the probability measure yield...

you actually need the stronger (but true) statement that these events are pairwise disjoint.

There are several simpler ways to do this. I would have gone with

$$P(A\cap B\cap C)=P(A)-P(A\cap B^c)-P(A\cap B\cap C^c)\\ \geq P(A)-P(B^c)-P(C^c)\\ =P(A)+P(B)+P(C)-2.$$

Answer 2

The answer is rather straightforward: $$P(A)+P(B \cap C)-P(A\cap (B \cap C))=P(A \cup (B \cap C)) \leq 1$$ $$P(B)+P(C)-P(B\cap C)=P(B \cup C) \leq 1$$ Summing both sides of the above inequalities yields the desired result.

Answer 3

We have: $$P(A\cap B\cap C)=P(A\cup B\cup C)+\sum_{cyc}(P(A\cap B)-P(A))=$$ $$=P(A\cup B\cup C)+\sum_{cyc}(P(A)+P(B)-P(A\cup B)-P(A))=$$ $$=P(A\cup B\cup C)+\sum_{cyc}P(A)-\sum_{cyc}P(A\cup B)\geq\sum_{cyc}P(A)-2$$ because $$P(A\cup B\cup C)+2\geq\sum_{cyc}P(A\cup B).$$

Answer 4

It is not difficult to verify that: $$\mathbf1_{A\cap B\cap C}=\mathbf1_A\mathbf1_B\mathbf1_C\geq\mathbf1_A+\mathbf1_B+\mathbf1_C-2$$ Now take expectation on both sides and conclude that:$$P(A\cap B\cap C)\geq P(A)+P(B)+P(C)-2$$