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I got a little stuck on a simple proof of the following probability identity.


$\mathbb{P}(A^c \cap B^c)=1-\mathbb{P}(A)-\mathbb{P}(B)+\mathbb{P}(A\cap B)$

how to prove for any set $X$,

$\mathbb{P}(X \cap A^c \cap B^c)=\mathbb{P}(X)-\mathbb{P}(X\cap A)-\mathbb{P}(X\cap B)+\mathbb{P}(X\cap A\cap B)$

Looks very intuitive; just replace the whole space by $X$. But how to prove it simply and rigorously? Thanks.

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Why would you want to deduce the latter from the former? The latter can be proven the same way as the former. – Rasmus Feb 12 '11 at 20:35
@Rasmus: oh, I want to know the details about the proof. I thought derivation from the former is the only way to get the latter. Thanks. – Qiang Li Feb 12 '11 at 21:34
If you have seen conditional probabilities, your intuition can be easily formalized. – Raskolnikov Feb 12 '11 at 22:56
@Raskolnikov: how to argue with conditional probability? – Qiang Li Feb 12 '11 at 23:14
Just think about it and remember that $\mathbb{P}(A \cap X) = \mathbb{P}(A | X) \mathbb{P}(X)$. – Raskolnikov Feb 12 '11 at 23:16

Write $X$ as a disjoint union: $$ X=(X\cap A\cap B)\sqcup (X\cap A^c\cap B)\sqcup (X\cap A\cap B^c)\sqcup (X\cap A^c\cap B^c). $$ This gives: $$ P(X)=P(X\cap A\cap B)+P(X\cap A^c\cap B)+P(X\cap A\cap B^c)+P(X\cap A^c\cap B^c). $$ Hence $$ P(X\cap A^c\cap B^c)=P(X)-P(X\cap A\cap B)-P(X\cap A^c\cap B)-P(X\cap A\cap B^c). $$ Now the result follows from $P(X\cap A\cap B)+P(X\cap A^c\cap B)=P(X\cap B)$ and $P(X\cap A\cap B)+P(X\cap A\cap B^c)=P(X\cap A)$.

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actually, the reason I asked the question above is that I wanted to extend the inclusion-exclusion identity. your method seems nice to this problem, but not in the general case, am i correct? – Qiang Li Feb 12 '11 at 22:06
@Qiang Li: If you mean the general form of the inclusion-exclusion identity, then the same reasoning will work. Essentially, your generalisation just consists in considering the subspace $X$ of your probability space. X is itself a probability space with measure $P(⋅)/P(X)$. – Rasmus Feb 13 '11 at 9:41

An easier way would be:

$X \cap A^c \cap B^c = X - (A \cup B)$.

Therefore, $P(X - (A \cup B)) \\= P(X) - P(X \cap (A\cup B)) \\= P(X) - P((X\cap A)\cup(X\cap B)) \\= P(X) - [P(X\cap A) + P(X\cap B) - P(X\cap A\cap B)]$

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Just follow your intuition: if $\mathbb{P}(X) > 0$ the formula $\mathbb{P}_X(E) = \dfrac{\mathbb{P}(E \cap X)}{\mathbb{P}(X)}$ defines a new (conditionnal) probability on $\Omega$. For this probability, you know that:

$$\mathbb{P}_X(A^c \cap B^c)=1-\mathbb{P}_X(A)-\mathbb{P}_X(B)+\mathbb{P}_X(A\cap B). $$

Multiplying by $\mathbb{P}(X)$ gives what you want: $$ \mathbb{P}(X \cap A^c \cap B^c)=\mathbb{P}(X)-\mathbb{P}(X\cap A)-\mathbb{P}(X\cap B)+\mathbb{P}(X\cap A\cap B). $$

Notice that the identity is trivial when $\mathbb{P}(X)=0$, since every term is $0$.

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so.. i am using the latter to derive the former..

  1. write P(X) as $\mathbb{P}(X\cap (A U A^c))$ , equal to $P(X \cap A)+P(X \cap A^c)$, cancel some terms

  2. then expand $P(X \cap A^c)$ to $P(X \cap A^c \cap (B U B^c))$, which is equal to $P(X \cap A^c \cap B)+ P(X\cap A^c \cap B^c)$, cancel the terms to get the answer.

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Hint: Like the cap command for intersections there is the cup command for unions. – Rasmus Feb 12 '11 at 22:07
the latter is supposed to be an extension of the former. just take $X=\Omega$, you get the former. – Qiang Li Feb 12 '11 at 22:08
oops.. i forgot to see that.. – adobriyal Feb 12 '11 at 22:22

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