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

Given

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

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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    @Raskolnikov: It is still very easy to do this from conditional probability point of view, unless you can give the complete formalized argument. And if so, I would accept your answer. :D2011-02-13

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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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    @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 $P(⋅)/P(X)$.2011-02-13
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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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    oops.. i forgot to see that..2011-02-12