Probability statements (2) Compounds

There is never any difficulty in adding truth-functional connectives to a set of statements, but doing so doesn’t give us any new insight into their character or structure. The better approach is to ask: are there, in this very class of statements itself, already ones that count as conjunctions, disjunctions, and the like? (What is the ‘internal’ logic of this sort of discourse?)

Conjunction

A simple example is (P(A) > x & P(A) < y), which can also be written as: P(A) ε (x, y). But that is the special case, of two statements about the probability of a single proposition. What about such combinations when different propositions are involved?

Theorem: If C is a family of convex sets (finite, countable or uncountable), then the intersection of the members of C is a convex set.

Proof: That the intersection ∩C is convex is trivially so if the intersection is empty, or has just one member. If ∩C has more than one member, consider any two p, p’ of its members: these belong to each member of C, and hence any of their convex combinations are also members of each member of C. Hence all the convex combinations belonging to all members of C, and hence of ∩C, are in ∩C.

This Theorem shows that it is fine to introduce the usual sort of conjunction in to the language, for then the set of measures that satisfy both of two elementary statements will also be convex. So if Q and R are elementary statements then (Q & R) is the statement such that |Q&R| = |Q| ∩ |R|, and this is again an elementary statement.

Disjunction

The same ease is not to be found for disjunction. The union of two convex sets is not in general convex. Anyway, we already know that disjunction does not behave like a truth-function when it comes to probability. In fact, it does not make sense to ask whether p satisfies (P(A) = r or P(A) = s) , as opposed to asking whether it satisfies either (P(A) = r) or satisfies (P(A) = s). At most we can ask whether p satisfies what the two ‘have in common’.

Can we find an operation on elementary statements that has the main characteristics we require of disjunction, in general?

Requirement. The general concept of disjunction of two statements, Q, R, in any kind of language, requires that it must be the logically strongest statement that is implied by both Q and R, and thus itself implies all that is implied by both Q and R.

In the first post I defined entailment for elementary statements. What we should look at therefore is this situation. Suppose that S is a statement S

Q entails S and R entails S

What is the relation that |S|bears to |Q| and |R|?

Theorem. If Q, R, S are elementary statements, Q entails S, and R entails S, then all convex combinations of members p of |Q| and p’ of |R| belong to |S|.

For note that if Q and R entail S then both |Q| and |R| are part of |S|. Therefore, iff is in |Q| and p’ is in |R| then both belong to |S|. Since |S| is convex it will also contain all the convex combinations of p and p’.

The smallest set that fits the Requirement above is therefore the convex hull of the union |Q| ⋃ |R|, that is the set of all convex combinations of members of those two sets. That is the smallest convex set which contains both. Since that is more than just the union, it does not correspond to a truth-functional disjunction. So let’s introduce a special symbol:

Definition. The join of convex sets X and Y is (X ⊕ Y) = {ap +(1-a)p’: a ε [0,1], p in X, p’ in Y}.

That is precisely the convex hull of X ⋃Y. Following upon this we can introduce a statement connective of ‘disjunction’ to the language, which will combine elementary statements into other elementary statements. Without expecting any confusion from this, I will use ⊕ equally for the statement connective and for the operation on convex sets:

| Q ⊕ R| = |Q| ⊕ |R|.

Negation

Really, there is no negation. In a specific case we can make up the negation, but it will typically not be an elementary statement. For example what would be the negation of P(A) = 0.5?

Its contraries are P(A) < 0.5 and P(A) > 0.5. Each of these is an elementary statement. But there is no truth-functional ‘or’ that would combine them into the contradictory of P(A) = 0.5, at least not one that would produce an elementary statement.

The sort of disjunction we do have produces something, but not the contradictory of P(A) = 0.5. In fact,

(P(A) < 0.5) ⊕ (P(A) > 0.5)

is satisfied by the 50/50 convex combination p” of p and p’ which assign 0.25 and 0.75 to A respectively, and p”(A) is 0.5. So we have arrived at a tautology, this disjunction has the same semantic value as |P(A) ε [0,1]|.

The underlying reason is of course that there is no largest convex subset of [0,1] disjoint from [0.5]. The two convex sets disjoint from [0.5] are [0.0.5) and (0.5,1] which are as large as can be, so there is no largest.

New Question: are there other forms of that elementary statements can have? What about other sorts of probability talk, such as about odds or conditional probabilities?

That will be the topic of the next post.