Branden Fitelson (forthcoming) provides counterexamples to Richard Feldman’s principle that Evidence of Evidence is Evidence (EEE). Here’s the principle in its initial (naïve) form:

(EEE1) If E (non-conclusively) supports the claim that (some subject) S possesses evidence which supports p, then E supports p. (Fitelson forthcoming: 1).

Fitelson’s counterexamples to (EEE) work by presupposing the “positive relevance” (i.e., increase-in-probability) notion of evidential support. In footnote 6 he indicates a more substantive principle of evidential support might be wielded in defending (EEE). In this post I want to explore this possibility, specifically in relation to the notion of propositional justification. Consider the following principle of propositional justification:

S is justified in believing that p iff S’s total evidence sufficiently supports p (Neta 2007: 197).

Though there are many issues that could be raised with this formulation of propositional justification, let’s see if a less demanding iteration of the principle could be used to resist Fitelson’s counterexamples to (EEE). Neta’s principle suggests the following notion of evidential support:

(1) E (evidentially) supports p iff S’s total evidence includes E and S’s total evidence (necessarily) supports p.

The counterexample to (EEE1) involves drawing a card c at random from a deck. All the evidence we are given regarding c is as follows:

(E1) c is a black card.

(E2) c is the ace of spades.

(p) c is an ace.

Imagine a guy named John knows what card c is, and the evidence above constitutes all the facts about the case. This means the following is the case:

(2) E1 supports the claim that John possesses evidence (E2) which supports p.

Positive relevance creates a problem for (EEE1) because (E1) doesn’t raise the probability of (p). (E1) alone is probabilistically irrelevant to (p); so, even though (E1) supports (E2), the second conjunct in (EEE1) is false (i.e., E1 doesn’t support p).

How does the counterexample fare under principle (1) instead of positive relevance? John’s total evidence includes (E1), and John’s total evidence (E1 and E2) necessarily supports (p). (E1) alone doesn’t necessarily support (p), but it also doesn’t support (not-p), and when coupled with (E2) it does necessarily support (p). In fact, (E2) entails (p). John’s total evidence might not sufficiently support (p), but his total evidence does necessarily do so. The next iteration of (EEE) runs as follows:

(EEE2) If E1 supports the claim that S possesses evidence E2 which supports p, then the conjunction of E1 and E2 supports p (Fitelson forthcoming: 2).

This seems like the defense I just gave for (EEE1), assuming (1). Didn’t I just claim the conjunction of (E1) and (E2) supports (p)? If so, then, assuming evidential support principle (1), it looks like the next counterexample will sink (EEE2). However, I think (EEE2) and (1) escape unscathed. Fitelson’s counterexample to (EEE2) is about a guy named Joe:

(E1) Joe has a full head of white hair.

(E2) Joe is over 35 years of age.

(p) Joe is bald.

The example works given the positive relevance principle because the conjunction of (E1 and E2) fails to raise the probability of (p). Being over 35 years of age might raise the probability that one is bald, but having a full head of white hair doesn’t raise the probability one is bald. (E1) supports (not-p), so the conjunction of (E1) and (E2) refutes (p). What about in relation to principle (1) instead of positive relevance?

(E1) is part of the total evidence. But, does the total evidence support (p)? More specifically, is the notion of total evidence equivalent to the conjunction of all the (relevant) evidence? I would argue it is possible to equate the total evidence with the conjunction of all relevant evidence, as I did in defending (EEE1) against a counterexample, but it is not necessarily the case that the total evidence must be regarded as all of the evidence conjoined. There is a probabilistic consideration in favor of this point.

The probabilistic consideration is that the total evidential support for (p) is not determined by simply conjoining the individual probabilities that (E1) and (E2) afford (p). Returning to the card counterexample, the probability that the card is a black card (E1) supports the claim the card is an ace (p) is the probability that: if the card is black, then the card is an ace. Only the ace of clubs and the ace of spades satisfy this condition, so the probability is 2/52 (approx. 4%).

The probability (E2: the card is the ace of spades) supports the claim the card is an ace (p) is 1, as the entailment takes on the maximum value. The fact that (E2) entails (p) means Pr(p|E2) = 1. The amount that the total evidence supports (p) is not simply the product of the probabilities of (p) given (E2) and (p) given (E1). This would yield a probability of Pr(p|2/52 x 52/52) = 104/2704 or about 4%.

A  better estimate is attained through subtraction of the two probabilities. This measures the degree to which one bit of evidence lessens the impact of another bit of evidence on the target proposition. This better approximates the total impact of (E1 and E2) on (p). This makes it the case that Pr(p|52/52 – 2/52) = 50/52 or 96%. The actual probability, though, would be 1 or 100% because the probability the card is black (E1) is swamped by the probability the card is the ace of spades (E2) in relation to the probability the card is an ace (p). As such, (E1) can be disregarded. Again, the correct probabilistic impact of the total evidence on the target proposition is not determined by a conjunction of all the evidence. Let’s apply this back to the second counterexample about Joe’s hair or lack-thereof:

Pr(p|E1) = 0

Pr(p|E2) = .20 (estimate of men over 35 who are bald)

Conjoining the total evidence, again, doesn’t refute the target proposition. Suppose John knows the age of Joe and (2) is true. That is, (E1) supports the claim that John possesses evidence (E2) that supports (p). Does the conjunction of (E1 and E2) refute (p), as Fitelson urges? Multiplying the probabilities, as given above, would yield a probability of 0. It would indeed refute (p). However, this case is dissimilar from the card drawing case because it uses vague terms. Being “bald” is not defined as having “no hair”. Most men who are “bald” still have some hair on their head. Being bald is defined in relation to “male pattern baldness.” This is a progressive condition and, much like the term “a heap”, is wrought with vagueness. The fact that Joe has a full head of white hair (E1) and Joe is over 35 years of age (p) doesn’t make it the case that there is zero probability that Joe is bald. Joe may have hair loss as a result of male pattern baldness, yet to a casual observer (like John) he may appear to have a full head of hair. This is especially the case for someone who has all white hair because the threshold for counting as having a full head of hair is plausibly lower when all of one’s hair is white. Due to vagueness in the terms in (E1) and (p), the fact that Joe is over 35 years of age (E2) is not swamped by (E1).

A better probability estimate is attained by (i) allowing Pr(p|E1) = .05 as a correction on boundary vagueness in the terms, then (ii) subtracting the probabilities to yield the impact of the total evidence on the target proposition: Pr(p|.20 – .05) = .15 or 15%. The total evidence still supports (i.e., does not outright refute) (p) even though the total evidence makes it more likely that (not-p) than (p) is the case.

References

Fitelson, Branden (forthcoming). “Evidence of Evidence is not (Necessarily) Evidence.” Analysis.

Neta, Ram (2007). “Propositional Justification, Evidence, and the Cost of Error.” Philosophical Issues.