Absence of evidence in the Bayesian

Here's a little something that maybe you didn't know about induction. Let's say I have evidence B. I can use this evidence B to argue inductively for claim A. Evidence B doesn't prove A, but it does make A more likely. So what happens if I instead have evidence not-B. That is, I've looked, and found that evidence B is absent. Does that make not-A more likely?

In other words, does absence of evidence amount to evidence of absence? Yes. And I can prove it mathematically. [I mention math and thus lose half my readers... Skip the proof section if you must.]

The Proof

Good to read first: Induction and the Bayesian

Bayes' theorem states the following:

P(A|B) is equal to the probability that claim A is true if we find evidence B. P(A) is the "prior" probability that A is true. P(B|A) is the probability of finding evidence B if we know A is true. P(B) is the "prior" probability of finding evidence B. If B is evidence for A, then P(A|B) > P(A). If not-B (written as ~B) is evidence for not-A, then P(~A|~B) > P(~A). Thus we seek to prove the following:

If P(A|B) > P(A), then P(~A|~B) > P(~A)

We will use, in addition to Bayes' theorem, the following identities:

~(~X) = X
P(~X) = 1 - P(X)
P(~X|Y) = 1 - P(X|Y)

This theorem assumes two things. First, none of the prior probabilities can be zero. For example, if P(B) = 0, Bayes' theorem doesn't even make sense anyways, since it divides by zero. Second, it assumes that probability is a good way to model knowledge.*

Proof:

1. Start with Bayes' theorem: P(A|B) = P(B|A)*P(A)/P(B)
2. We're given: P(A|B) > P(A)
3. Combining 1 and 2: P(B|A)*P(A)/P(B) > P(A)
4. Multiply by P(B): P(B|A)*P(A) > P(B)*P(A)
5. Use the identities: [1-P(~B|A)]*P(A) > [1-P(~B)]*P(A)
6. Some algebra: P(~B)*P(A) > P(~B|A)*P(A)
7. Use Bayes' theorem: P(~B)*P(A) > P(A|~B)*P(~B)
8. Use the identities: P(~B)*[1-P(~A)] > [1-P(~A|~B)]*P(~B)
9. Some algebra: P(~B)*P(~A|~B) > P(~A)*P(~B)
10. Divide out by P(~B): P(~A|~B) > P(~A)

QED

*Note: Some people might consider this second assumption questionable. Under certain interpretations, probability is only reliable in analyzing repeatable phenomena, and the universe is not repeatable.

Discussion and Conclusion

What does this mean? It means that if the existence of some evidence supports a claim, then the non-existence of that evidence detracts from the claim. This is a logical necessity in induction. The only assumptions are that we can model our knowledge with probabilities, and that none of the prior probabilities are certain.

This directly contradicts the conventional wisdom that "Absence of evidence is not evidence of absence". So where did this conventional wisdom come from?

There are two justifications I can think of. First, "absence of evidence" might mean that we neither know whether there is or there isn't evidence, because we haven't looked. In that case, the conventional wisdom is true. Second, though I proved that absence of evidence is evidence of absence, I did not prove that it's very good evidence of absence. For example, if I found bigfoot behind a tree, that would provide extremely good evidence for bigfoot, but if I didn't find him behind a tree, that would provide very weak evidence against bigfoot. But it's still evidence, mathematically speaking. I've previously explained this asymmetry as the basis for the concept of "burden of proof".

So, I wasn't lying when I said the Bayesian gives us insight into the inner workings of reason! This is just one of the reasons that math is cool.