Let X be a set, called the “sample space”. Let S be a sigma-algebra on X, i.e., a set of subsets of X closed under complementation and countable unions and containing the empty set. (In cases where X is countable, S will typically be ÃX. In general, X need not be all of ÃX—there may be “non-measurable sets”.) S always contains both the empty set and X. It need not contain anything else.
Fact: If A and B are members of S, so are their union, intersection, and difference (A\B={xÎA|xÏB}).
A member of S is called an “event” or a “measurable set”.
A probability measure P on X (or, more precisely, on <X,S>) is a function P:SàR (real numbers) satisfying the Kolmogorov Axioms:
1. P(E) ³ 0 for every event E
2. P(X)=1.
3. If E1,E2,… are “pairwise disjoint” events (i.e., EkÇEn is empty whenever k is not n), then P(E1ÈE2È…)=P(E1)+P(E2)+….
Proposition 1: P(Æ)=0.
Proof: 1=P(X)=P(ÆÈX)=P(Æ)+P(X)=P(Æ)+1. Therefore P(Æ)=0.
Proposition 2: If A is a subset of B, then P(A)£P(B).
Proof: Let C=B\A. Then P(B)=P(B\A)+P(A). But P(B\A)³0. So P(A)£P(B).
In the case that X is countable, we often define the probability measure by letting S=ÃX, and having some positive real valued function f:X®R, such that f(x)³0 always and f(x1)+… =1, where x1,…, are the members of X. We can then define P(A) as the sum of f(x) as x ranges over A. Fact: The Kolmogorov Axioms will then hold. 1 and 2 are easy. 3 is a consequence of the associativity of addition.
Example: Let X be the possible outcomes of the throw of two dice={<a,b>|a,bÎ{1,…,6}}. Define f(<a,b>)=1/36 for each a, b.
In philosophy we often want to talk about the probability of a proposition. Here’s one way to do this. Consider a set X of “situations”. These might be worlds, or parts of worlds. Suppose we have a set A of propositions, such that for any member p of A, and any situation s in X, we can ask whether p holds in s or not.
For instance, the propositions might be statements about chess positions, such as “The white king is in check”, and then the situations will be chess positions (let’s not worry about historical questions, such as whether the king has moved already or not). Or the propositions might be about the numbers that show up on a die after it is tossed, and then X={1,2,3,4,5,6}.
Now, if we have a probability measure P on X, we can define Xp={sÎX | p holds in s} (my notation): this is “the event of p holding”. Assuming that Xp is a measurable set, then we can define P(p) = P(Xp).
Say that a proposition is impossible relative to X iff it holds in no member of X, and necessary relative to them iff it holds in all of them. Say that p entails q relative to X iff every situation in which p holds is a situation in which q holds. Say that p and q are incompatible relative to X iff their conjunction is impossible relative to X.
Proposition 3: Every impossible proposition has probability zero.
Proposition 4: If p entails q, then P(p) £ P(q).
Proposition 5: If p and q are incompatible, then P(p Ú q)=P(p)+P(q).
One could try to define analogues of the Kolmogorov Axioms for propositions directly if one had the notions of necessity, compatibility and entailment:
Question: In the die case, what is “the probability that the sum of the dice is 4”?
Xp={<1,3>,<2,2>,<3,1>}. P(Xp)=3/36=1/12.
Definition: The events A and B are independent iff P(AÇB)=P(A)P(B).
Definition: The propositions p and q are independent iff Xp and Xq are independent.
Proposition 6: p and q are independent iff P(pÙq)=P(p)P(q).
Proof: Observe that XpÙq=XpÇXq.
Fact: In the two dice case, “first die result is even” and “second die result is odd” are independent. In fact, any statement solely about the first die is independent of any statement solely about the second die.