# Marginalizing probability

Notes for my intro to probability class give a theorem which states the following:

P(A) = P(A ^ B) + P(A ^ ~B)


What would the proof for this look like given the basic probability axioms?

Here's a list of axioms I start off with:

0 <= P(A) <= 1
P(True) = 1
P(False) = 0
P(A v B) = P(A) + P(B) - P(A ^ B)
P(A) = 1 - P(~A)

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From the set-theoretic equations

\eqalign { A &= A \cap 1 \\ &=A \cap (B \cup \sim B) \\ &=(A \cap B) \cup (A \cap \sim B) \\ }

and

\eqalign { (A \cap B) \cap(A \cap \sim B) &= (A \cap A) \cap (B \cap \sim B) = A \cap \varnothing = \varnothing }

we apply the penultimate axiom to obtain

\eqalign { \Pr(A) &= \Pr((A \cap B)\cup (A \cap \sim B)) \\ &= \Pr(A \cap B) + \Pr(A \cap \sim B) - \Pr((A \cap B) \cap(A \cap \sim B)) \\ &= \Pr(A \cap B) + \Pr(A \cap \sim B) - \Pr(\varnothing) }

and then, observing $\Pr(\varnothing)=0$ by the third axiom, we are done.

This nicely illustrates the fact that most probability statements are really about sets (events). That's why Venn diagrams can be so helpful in reasoning about probabilities.

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First prove that $A=(A\cap B)\cup(A\cap B^c)$.

You may just draw a Venn diagram or do it formally, which means that you have to prove that $A\subset (A\cap B)\cup(A\cap B^c)$ and $(A\cap B)\cup(A\cap B^c)\subset A$.

For example, to prove the first inclusion you have to suppose that $\omega\in A$ and conclude that $\omega\in (A\cap B)\cup(A\cap B^c)$.

Now observe that $(A\cap B)\cap(A\cap B^c)=\emptyset$.

Here you may use the same Venn diagram as before or just reason that any $\omega$ in this intersection would be an element of both $B$ and $B^c$, which is not possible (a contradiction).

Finally, use the fact that the probability of the union of two disjoint events is the sum of their probabilities.

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I understand the concept behind it, and I understand the proof with the Venn diagram, but is there a way to prove this using axioms and substitution? –  Kshitij Mehta Mar 1 '12 at 0:36
What axioms are you beginning with? (There is more than one set!) –  whuber Mar 1 '12 at 0:38
Sorry, I'm new to this stuff! I'll edit the question with the axioms I'm given :) –  Kshitij Mehta Mar 1 '12 at 0:40
"Finally, use the fact that the probability of the union of two disjoint events is the sum of their probabilities." This is not one of the "axioms" that the OP is given but needs to be deduced from the two that say P(False) = 0; P(A v B) = P(A) + P(B) - P(A ^ B) In other words, there is more work to be done for the proof "using axioms and substitutions" that the OP wants. –  Dilip Sarwate Mar 1 '12 at 1:56