Deciphering conditional probabilities

How would you evaluate the following two probabilities?

  1. P(B | A)
  2. P(A → B)

In words, the first is “the probability that B is true, given that A is true” and the second is “the probability that if A is true, then B is true.” I don’t know about you, but these sound pretty darn similar to me.

But in fact, it turns out that they’re different. In fact, you can prove that P(B | A) is always greater than or equal to P(A → B) (the equality only in the case that P(A) = 1 or P(A → B) = 1). The proof of this is not too difficult, but I’ll leave it to you to figure out.

Conditional probabilities are not the same as probabilities of conditionals. But maybe this isn’t actually too strange. After all, material conditionals don’t do such a great job of capturing what we actually mean when we say “If… then…” For instance, consult your intuitions about the truth of the sentence “If 2 is odd then 2 is even.” This turns out to be true (because any material conditional with a true consequent is true). Similarly, think about the statement “If I am on Mars right now, then string theory is correct.” Again, this turns out to be true if we treat the “If… then…” as a material conditional (since any material conditional with a false antecedent is true).

The problem here is that we actually use “If… then…” clauses in several different ways, the logical structure of which are not well captured by the material implication. A → B is logically equivalent to “A is true or B is false,” which is not always exactly what we mean by “If A then B”. Sometimes “If A then B” means “B, because A.” Other times it means something more like “A gives epistemic support for B.” Still other times, it’s meant counterfactually, as something like “If A were to be the case, then B would be the case.”

So perhaps what we want is some other formula involving A and B that better captures our intuitions about conditional statements, and maybe conditional probabilities are the same as probabilities in these types of formulas.

But as we’re about to prove, this is wrong too. Not only does the material implication not capture the logical structure of conditional probabilities, but neither does any other logical truth function! You can prove a triviality result: that if such a formula exists, then all statements must be independent of one another (in which case conditional probabilities lose their meaning).

The proof:

  1. Suppose that there exists a function Γ(A, B) such that P(A | B) = P(Γ(A, B)).
  2. Then P(A | B & A) = P(Γ | A).
  3. So 1 = P(Γ | A).
  4. Similarly, P(A | B & -A) = (Γ | -A).
  5. So 0 = P(Γ | -A).
  6. P(Γ) = P(Γ | A) P(A) + P(Γ | -A) P(-A).
  7. P(Γ) = 1 * P(A) + 0 * P(-A).
  8. P(Γ) = P(A).
  9. So P(A | B) = P(A).

This is a surprisingly strong result. No matter what your formula Γ is, we can say that either it doesn’t capture the logical structure of the conditional probability P(B | A), or it trivializes it.

We can think of this as saying that the language of first order logic is insufficiently powerful to express the conditionals in conditional probabilities. If you take any first order language and apply probabilities to all its valid sentences, none of those credences will be conditional probabilities. To get conditional probabilities, you have to perform algebraic operations like division on the first order probabilities. This is an important (and unintuitive) thing to keep in mind when trying to map epistemic intuitions to probability theory.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s