Why Evidentialism Sucks (p.2)

The Mathematical Problem for Evidentialism

Ironically enough, mathematics and probability theory pose serious problems for evidentialism. For those whose math is a bit rusty, I recommend reading this quick introduction to Bayes’ theorem before moving forward.

Ready? Good. Here’s some other math stuff you’ll need to know here. E is evidence for hypothesis H if and only if the following is true:

Pr(H|E) > Pr(H)

Where Pr(H) is the prior probability.[1] E is evidence for hypothesis H if it makes the hypothesis more likely than it would have been without E. To oversimplify it a bit, something being evidence for a hypothesis means it makes the hypothesis more likely to be true.

One form of Bayes’ theorem, when comparing hypotheses, goes like this:

Pr(H1|E)

Pr(H2|E)

=

Pr(E|H1)

Pr(E|H2)

×

Pr(H1)

Pr(H2)

For example, one can use hypothesis H and it’s negation ¬H as follows:

Pr(H|E)

Pr(¬H|E)

=

Pr(E|H)

Pr(E|¬H)

×

Pr(H)

Pr(¬H)

Notice the importance Pr(E|H) and Pr(E|¬H) have in changing the likelihood ratio of Pr(H|E) to Pr(¬H|E). As one might suspect from the equation above, for this to get bigger via evidence E:

Pr(H|E)

Pr(¬H|E)

Pr(E|H) must be greater than Pr(E|¬H), and a bit of math shows that for E to be evidence for hypothesis H, this has to be true:

Pr(E|H) > Pr(E|¬H)[2]

Similarly, for something to be evidence against a hypothesis, the following has to be true:

Pr(E|¬H) > Pr(E|H)[3]

These math facts are part of a recipe of doom for evidentialism. Let ES be the set of all memories, sense experiences, and intuitions you now have. Let HB be the hypothesis you are a recently created (say, within the past few years) brain in a vat of chemicals hooked up to a supercomputer that gives you all the memories, sense experiences, and intuitions you now have. Thus, Pr(ES|HB) = 1. You are rational to believe that hypothesis HB is false, but here’s the problem: you cannot cite any evidence against it. Thanks to mathematics, for some would-be piece of evidence E to be evidence against HB, the following has to be true:

Pr(E|¬HB) > Pr(E|HB)

Hypothesis HB however predicts with 100% certainty any memory, sense experience, and intuition you now have, thus Pr(E|HB) = 1, and it is impossible for any evidence E to be such that Pr(E|¬HB) > Pr(E|HB).

This actually teaches us an important lesson: some things are rational to believe without proof or evidence, e.g. the falsity of hypothesis HB. It also teaches us that evidentialism sucks; if evidentialism were true, we could never reject that crazy brain-in-the-vat hypothesis.

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[1] The Pr(H|E) > Pr(H) inequality is oversimplifying it a bit; I haven’t mentioned that e.g. the background knowledge for Pr(H) doesn’t already have E (though I later say E is evidence if “it makes the hypothesis more likely than it would have been without E”), because if our background knowledge for Pr(H) included E then Pr(H|E) = Pr(E) even when E is evidence for H. Hopefully though the general raising-the-probability-of-the-hypothesis idea is apparent.

[2] This holds for when Pr(E) ≠ 0, which is the situation that interests us (presumably, any evidence E we are interested in is the sort that Pr(E) ≠ 0). For math nerds who are interested in the mathematical proof for this, I will first prove a few lemmas.

Lemma (1): (H ∩ E) and (¬H ∩ E) are disjoint (mutually exclusive). We can show that no element in the universe can be a member of both (H ∩ E) and (¬H ∩ E). Let x be an arbitrary element and let’s suppose x is a member of both (H ∩ E) and (¬H ∩ E). With a bit of math logic, we show that there can’t be any x such that x ∈ (H ∩ E) and x ∈ (¬H ∩ E) by assuming there is such an x and deriving an impossibility, like so:

x ∈ (H ∩ E) and x ∈ (¬H ∩ E)
(x ∈ H and x ∈ E) and (x ∈ ¬H and x ∈ E), from (1) and definition of ∩
x ∈ H and x ∈ E and x ∈ ¬H and x ∈ E, from (2)
x ∈ H and x ∈ ¬H, from (3)

Of course, it’s impossible for there to be an element that is a member of a set and its complement, since (¬H ∩ H) = ∅. Thus (H ∩ E) and (¬H ∩ E) are disjoint, i.e. (H ∩ E) and (¬H ∩ E) = ∅.

With this in mind, let ξ be the universal set.

E ∩ ξ = E
⇔ E ∩ (H ∪ ¬H) = E
⇔ (E ∩ H) ∪ (E ∩ ¬H) = E

Lemma (2): Pr(H|E) = 1 − Pr(¬H|E) for Pr(E) ≠ 0.

Proof: since (E ∩ H) ∪ (E ∩ ¬H) and are mutually exclusive as proven in Lemma (1), by the rules of probability:

Pr(E ∩ H) + Pr(E ∩ ¬H) = Pr(E)
⇔ Pr(E) × Pr(H|E) + Pr(E) × Pr(¬H|E) = Pr(E)
⇔ Pr(E) × [Pr(H|E) + Pr(¬H|E)] = Pr(E)
⇔ 1 × [Pr(H|E) + Pr(¬H|E)] = 1
⇔ Pr(H|E) + Pr(¬H|E) = 1
⇔ Pr(H|E) = 1 − Pr(¬H|E)

This holds for all Pr(E) ≠ 0. Thus, Pr(H|E) = 1 − Pr(¬H|E) for Pr(E) ≠ 0.

Lemma (3): If Pr(H|E) > Pr(H), then Pr(¬H|E) < Pr(¬H)

Proof: assume for sake of argument that Pr(H|E) > Pr(H) is true. We can then see how this leads to Pr(¬H|E) < Pr(¬H) using Lemma(2), where Lemma (2) implies that Pr(H|E) = 1 − Pr(¬H|E). If you’ve understood what’s going on thus far, I’ll assume you’re already aware of the well-known probability fact that Pr(¬A) = 1 − Pr(A) and that Pr(A) = 1 − Pr(¬A).

Pr(H|E) > Pr(H)
⇔ 1 − Pr(¬H|E) > 1 − Pr(¬H)
⇔ −Pr(¬H|E) > −Pr(¬H)
⇔ Pr(¬H|E) < Pr(¬H)

The components of the odds form of Bayes’ theorem are these:

posterior
odds

likelihood
ratio

prior
odds

Pr(H|E)

Pr(¬H|E)

Pr(E|H)

Pr(E|¬H)

Pr(H)

Pr(¬H)

Pr(H|E) > Pr(H) entails Pr(¬H|E) < Pr(¬H). If Pr(H|E) > Pr(H) and Pr(¬H|E) < Pr(¬H), then the posterior odds are greater than the prior odds, and the only way for this to happen is if the likelihood ratio is greater than one, which requires Pr(E|H) > Pr(E|¬H). Thus, if Pr(H|E) > Pr(H) then Pr(E|H) > Pr(E|¬H) for all E such that Pr(E) ≠ 0.

[3] To prove this, the task is to show that if Pr(H|E) < Pr(H), then Pr(E|¬H) > Pr(E|H) for all E such that Pr(E) ≠ 0.

One proof for this is quite similar to the one above, and so we can borrow from the same lemmas. Note that Lemma (2) from above entails that Pr(H|E) = 1 − Pr(¬H|E), for Pr(E) ≠ 0. From this we can construct a new lemma.

Lemma (4): Pr(H|E) < Pr(H) entails Pr(¬H|E) > Pr(¬H) for Pr(E) ≠ 0.

Pr(H|E) < Pr(H)
⇔ 1 − Pr(¬H|E) < 1 − Pr(¬H)
⇔ −Pr(¬H|E) < −Pr(¬H)
⇔ Pr(¬H|E) > Pr(¬H)