2. TRANSITIVITY IN EVIDENCE-IP: SOC VERSUS NDC

There are counterexamples to the thesis that evidence-IP is transitive. It will help in terms of understanding the rationale behind my appeal to screening-off to understand why there are counterexamples to the thesis that evidence-IP is transitive.

Consider the following theorem of the probability calculus:

$$\matrix{ {\Pr (H_2 \vert E) - \Pr (H_2 )} \hfill & = \hfill & {\left[ {\Pr (H_1 \vert E) - \Pr (H_1 )} \right]\left[ {\Pr (H_2 \vert H_1 ) - \Pr (H_2 )} \right]} \hfill & + \hfill \cr {} \hfill & {} \hfill & {\left[ {\Pr (\neg H_1 \vert E) - \Pr (\neg H_1 )} \right]\left[ {\Pr (H_2 \vert \neg H_1 ) - \Pr (H_2 )} \right]} \hfill & + \hfill \cr {} \hfill & {} \hfill & {\Pr (H_1 \vert E)\left[ {\Pr (H_2 \vert H_1 \& E) - \Pr (H_2 \vert H_1 )} \right]} \hfill & + \hfill \cr {} \hfill & {} \hfill & {\Pr (\neg H_1 \vert E)\left[ {\Pr (H_2 \vert \neg H_1 \& E) - \Pr (H_2 \vert \neg H_1 )} \right]} \hfill & {} \hfill \cr} $$

This theorem is due to Shogenji (Reference ShogenjiForthcoming) and thus shall hereafter be referred to as “Shogenji's theorem”. Let the first addend on the right be “A1”, the second addend on the right be “A2”, the third addend on the right be “A3”, and the fourth addend on the right be “A4”. Suppose that E is evidence-IP for H ₁ and that H ₁ is evidence-IP for H ₂. Then each of A1 and A2 is positive and thus A1 + A2 is positive. It does not follow, though, that A1 + A2 + A3 + A4 is positive, because it could be that A3 + A4 is negative and greater than or equal to in absolute value A1 + A2. All counterexamples to the thesis that evidence-IP is transitive are cases where A1 + A2 is positive but A3 + A4 is negative and greater than or equal to in absolute value A1 + A2.

Suppose, to illustrate, that a card is randomly drawn from a standard and well-shuffled deck of cards. Let E be the proposition that the card drawn is not a Heart, H ₁ be the proposition that the card drawn is a Diamond, and H ₂ be the proposition that the card drawn is a Red. Then, since E is evidence-IP for H ₁ and H ₁ is evidence-IP for H ₂, A1 + A2 is positive:

$$\left[ {\displaystyle{1 \over 3} - \displaystyle{1 \over 4}} \right]\left[ {1 - \displaystyle{1 \over 2}} \right] + \left[ {\displaystyle{2 \over 3} - \displaystyle{3 \over 4}} \right]\left[ {\displaystyle{1 \over 3} - \displaystyle{1 \over 2}} \right] = \displaystyle{1 \over {18}}$$

A3 + A4, however, is negative and greater than or equal to in absolute value A1 + A2:

$$\left[ {\displaystyle{1 \over 3}} \right]\left[ {1 - 1} \right] + \left[ {\displaystyle{2 \over 3}} \right]\left[ {0 - \displaystyle{1 \over 3}} \right] = - \displaystyle{2 \over 9}$$

It follows that A1 + A2 + A3 + A4 is negative and that so too is Pr(H ₂ | E) – Pr(H ₂). This case is thus a counterexample to the thesis that evidence-IP is transitive.Footnote ⁷

It is clear from Shogenji's theorem than any condition that is sufficient for transitivity in evidence-IP is a condition that guarantees that, when E is evidence-IP for H ₁ and H ₁ is evidence-IP for H ₂, it is not the case that A3 + A4 is negative and greater than or equal to in absolute value A1 + A2. There is simply no way around it.

Now consider the condition:

This is the screening-off condition in EEE_SOC and TIP_SOC. SOC guarantees that each of A3 and A4 is non-negative and thus guarantees that, when E is evidence-IP for H ₁ and H ₁ is evidence-IP for H ₂, it is not the case that A3 + A4 is negative and greater than or equal to in absolute value A1 + A2. It follows that TIP_SOC holds without exception and that so too does EEE_SOC.

It is important to note that neither conjunct of SOC is sufficient for transitivity in evidence-IP. Return to the card case above. E is evidence-IP for H ₁, H ₁ is evidence-IP for H ₂, and the first conjunct of SOC holds (because each probability equals 1). But it is not the case that E is evidence-IP for H ₂. Hence the first conjunct of SOC is not sufficient for transitivity in evidence-IP. Now let E be the proposition that the card drawn is a Club, the Ace of Spades, or the Ace of Hearts, H ₁ be the proposition that the card drawn is not a Spade, and H ₂ be the proposition that the card drawn is a Red. Then E is evidence-IP for H ₁, H ₁ is evidence-IP for H ₂, and the second conjunct of SOC holds (because each probability equals 0). But, as Pr(H ₂ | E) = 1/15 < 1/2 = Pr(H ₂), it is not the case that E is evidence-IP for H ₂. Hence the second conjunct of SOC is not sufficient for transitivity in evidence-IP.

T&C suggest that the second conjunct of SOC be dropped and that the first conjunct be replaced by a no-defeaters condition:

This is the no-defeaters condition in EEE_NDC. NDC is weaker than the first conjunct of SOC on the assumption that H ₁ is evidence-IP for H ₂. Any case where H ₁ is evidence-IP for H ₂ and the first conjunct of SOC holds is a case where Pr(H ₂ | H ₁ & E) ≥ Pr(H ₂ | H ₁) > Pr(H ₂) and thus is a case where NDC holds. But some cases where H ₁ is evidence-IP for H ₂ and NDC holds are cases where Pr(H ₂ | H ₁) > Pr(H ₂ | H ₁ & E) > Pr(H ₂) and thus are cases where the first conjunct of SOC does not hold. Is NDC sufficient for transitivity in evidence-IP?

The answer, clearly, is no. NDC is weaker than the first conjunct of SOC on the assumption that H ₁ is evidence-IP for H ₂ and so holds in all cases where the first conjunct of SOC holds and H ₁ is evidence-IP for H ₂. But, as noted above, there are cases where E is evidence-IP for H ₁, H ₁ is evidence-IP for H ₂, the first conjunct of SOC holds, and yet it is not the case that E is evidence-IP for H ₂. All such cases are counterexamples to the thesis that NDC is sufficient for transitivity in evidence-IP.

T&C acknowledge that my appeal to SOC works. They claim, though, that it works by “brute force”. They want to know why it works. This in effect is their first objection to my appeal to SOC. It should be clear, at this point, why my appeal to SOC works:

It should also be clear why an appeal to NDC would not work:

There is simply no getting around the point that any condition that is sufficient for transitivity in evidence-IP is a condition that guarantees that, when E is evidence-IP for H ₁ and H ₁ is evidence-IP for H ₂, it is not the case that A3 + A4 is negative and greater than or equal to in absolute value A1 + A2.

There is more. Consider what can be called the “Trivial Condition”:

This condition, as with SOC, is sufficient for transitivity in evidence-IP. But TC is not at all interesting. TC is impotent to help explain in a given case why E confirms H ₂ and it is impotent to help in verifying in a given case that E confirms H ₂. SOC is different. It can help explain in a given case why E confirms H ₂ and it can help in verifying in a given case, for example, the case of John and the departmental meeting, that E confirms H ₂.Footnote ^8, Footnote ⁹

What about T&C's charge that SOC is not necessary for transitivity in evidence-IP? How exactly do T&C mean for this charge to be understood?

One answer is that they mean for it to be understood as the charge that SOC is not necessary for the relation of “being evidence-IP for” to be transitive, that is, for it to be such that any case where E is evidence-IP for H ₁ and H ₁ is evidence-IP for H ₂ is a case where E is evidence-IP for H ₂. If this answer is right, then T&C's charge is clearly correct. This can be seen by considering what can be called the “Weaker Condition”:

This condition is weaker than SOC. Yet it is sufficient for transitivity in evidence-IP.

It turns out, though, that T&C mean for their charge to be understood differently. Their worry is that there are cases where E is evidence-IP for H ₁, H ₁ is evidence-IP for H ₂, SOC does not hold, and yet E is evidence-IP for H ₂.Footnote ¹⁰ They want a condition C such that (i) C is sufficient for transitivity in evidence-IP and (ii) if E is evidence-IP for H ₁, H ₁ is evidence-IP for H ₂, and C does not hold, then E is not evidence-IP for H ₂.

It is true that SOC is not a condition of that sort. But note that the same is true of NDC. This can be seen by appeal to a variant of the card cases above. Let E be the proposition that the card drawn is a Two or a Three, H ₁ be the proposition that the card drawn is a Three or a Four, and H ₂ be the proposition that the card drawn is a Two or a Four. Pr(H ₁ | E) = 1/2 > 2/13 = Pr(H ₁), so E is evidence-IP for H ₁. Pr(H ₂ | H ₁) = 1/2 > 2/13 = Pr(H ₂), so H ₁ is evidence-IP for H ₂. Pr(H ₂ | H ₁ & E) = 0 < 2/13 = Pr(H ₂), so NDC does not hold. But Pr(H ₂ | E) = 1/2 > 2/13 = Pr(H ₂), so E is evidence-IP for H ₂.

There are conditions of the sort that T&C want. One is TC. But this condition, as noted above, is not at all interesting. A better condition is WC. This condition, unlike TC, is distinct from the condition that E is evidence-IP for H ₂. There are cases where WC holds but, since in part E is not evidence-IP for H ₁ or H ₁ is not evidence-IP for H ₂, it is not the case that E is evidence-IP for H ₂.

I see no need to choose between SOC and WC or between theses involving SOC such as TIP_SOC and EEE_SOC and similar theses involving WC. They all have their place.

I turn now to EEE_NDC and some variants of it.

3. A COUNTEREXAMPLE TO T&C'S EEE THESIS

Consider the following variant of EEE_NDC:

EEE*_NDC is a special case of:

TIP_NDC, though, is false, for it is not the case that NDC is sufficient for transitivity in evidence-IP. Does it follow that EEE*_NDC too is false?

The answer, it turns out, is negative. It could be that TIP_NDC is false but nonetheless holds without exception in special cases where H ₁ is a proposition about evidence. Consider, for example, the thesis:

EEE**_NDC holds without exception if it is assumed both that all subjects have some propositions (with non-extreme unconditional probabilities) as evidence and that if a subject has a proposition as evidence, then she also has any implication (with a non-extreme unconditional probability) of that proposition as evidence. For, if those things are assumed, then there are no cases where the first condition in EEE**_NDC’s antecedent holds. This is easy to see. Suppose that S has E* as evidence. Then, since E* entails E* ∨ H ₂, it follows that S has that disjunction as evidence. Given this, and given that E* ∨ H ₂ is entailed by H ₂, it follows that E* ∨ H ₂ is evidence-IP for H ₂ (assuming that H ₂ has a non-extreme unconditional probability). Hence S has some evidence-IP for H ₂. So, though EEE**_NDC is a special case of TIP_NDC, and though the latter is false, it is not the case that the former is false.Footnote ^11, Footnote ¹² Hence the mere fact that TIP_NDC is false leaves it open that it is not the case that EEE*_NDC, which is a special case of TIP_NDC, is false.

Consider now a different version of the case of John and the departmental meeting. Each of John's credences is either high or not high, either in a proposition for which his total evidence is evidence-HP or not in a proposition for which his total evidence is evidence-HP, and either in a true proposition or not in a true proposition. Suppose that you know that:

1. (a) 3/5 of John's credences are high & in a proposition for which his total evidence is evidence-HP & in a true proposition

2. (b) 1/15 of John's credences are high & in a proposition for which his total evidence is evidence-HP & not in a true proposition

3. (c) 0 of John's credences are high & not in a proposition for which his total evidence is evidence-HP & in a true proposition

4. (d) 2/27 of John's credences are high & not in a proposition for which his total evidence is evidence-HP & not in a true proposition

5. (e) 0 of John's credences are not high & in a proposition for which his total evidence is evidence-HP & in a true proposition

6. (f) 0 of John's credences are not high & in a proposition for which his total evidence is evidence-HP & not in a true proposition

7. (g) 1/4 of John's credences are not high & not in a proposition for which his total evidence is evidence-HP & in a true proposition

8. (h) 1/108 of John's credences are not high & not in a proposition for which his total evidence is evidence-HP & not in a true proposition

(Note that 3/5 + 1/15 + 0 + 2/27 + 0 + 0 + 1/4 + 1/108 = 1.) Let E be the proposition that John has a high credence in H ₂, H ₁ be the proposition that John's total evidence is evidence-HP for H ₂, and H ₂ be the proposition that the departmental meeting is scheduled for 3:30 today. Suppose that H ₂ is one of the propositions in which John has a credence, and that, from your perspective, John's credence in H ₂ is a random member of the set of propositions in which he has a credence. Suppose, further, and consistent with all this, that your probability for H ₂ is 0.85. Then you should have the following probability distribution:

It follows that:

1. (i) Pr(H ₁ | E) = 9/10 > 2/3 = Pr(H ₁)

2. (j) Pr(H ₂ | H ₁) = 9/10 > 85/100 = Pr(H ₂)

3. (k) Pr(H ₂ | H ₁ & E) = 9/10 > 85/100 = Pr(H ₂)

4. (l) Pr(H ₂ | E) = 81/100 < 85/100 = Pr(H ₂)

So, E is evidence-IP for H ₁ where H ₁ is the proposition that John's total evidence is evidence-HP for H ₂, H ₁ is evidence-IP for H ₂, and NDC holds, and yet it is not the case that E is evidence-IP for H ₂. Hence EEE*_NDC, as with TIP_NDC, is false.

It is unclear at this point whether the same is true of EEE_NDC. This is because it is unclear at this point how exactly EEE_NDC is to be understood. T&C never specify how degree of evidence-IP is to be measured. They use expressions like “entailing evidence” and “evidence that raises the probability to at least 4/5”. But expressions like those leave it open how exactly evidence-IP is to be measured. Consider the following measures of evidence-IP:

$$e_d (H,E) = \Pr (H \vert E) - \Pr (H)$$

$$e_{KO} (H,E) = \displaystyle{{\Pr (E \vert H) - \Pr (E \vert \neg H)} \over {\Pr (E \vert H) + \Pr (E \vert \neg H)}}$$

The first of these measures is the “difference” measure. The second is the “Kemeny-Oppenheim” measure.Footnote ¹³ Suppose that E entails H and thus is evidence-IP for H (assuming that neither E nor H has an extreme unconditional probability). It follows by e _d that whether the degree of evidence-IP is high or low depends on whether Pr(H) is high or low. If Pr(H) is high, then the degree of evidence-IP is low. If Pr(H) is low, then the degree of evidence-IP is high. It follows by e _KO, in contrast, that the degree of evidence-IP is high, indeed, is maximal at 1, regardless of whether Pr(H) is high or low.

T&C presumably hold that EEE_NDC is true regardless of which of the leading measures of evidence-IP in the literature is assumed (otherwise, presumably, they would have specified how degree of evidence-IP is to be measured). I shall assume e _d and thus shall assume that in the case of John and the departmental meeting the degree to which H ₁ is evidence-IP for H ₂ is 0.05. But nothing of importance for my purposes hinges on this, since all the main points below would hold mutatis mutandis if e _KO or any alternative measure of evidence-IP (in the literature) were assumed instead.

Now, continuing with the second version of the case of John and the departmental meeting, let H* be the proposition that there is a proposition E* such that E* is true and is evidence-IP for H ₂ to degree 0.05. There is no reference to H* in the probability distribution above. It turns out, though, that H* is logically equivalent (in the context) to H ₁. (See Appendix for details.) It follows immediately that:

1. (m) Pr(H* | E) = 9/10 > 2/3 = Pr(H*)

2. (n) Pr(H ₂ | H*) = 9/10 > 85/100 = Pr(H ₂)

3. (o) Pr(H ₂ | H* & E) = 9/10 > 85/100 = Pr(H ₂)

4. (p) Pr(H ₂ | E) = 81/100 < 85/100 = Pr(H ₂)

By (m) it follows that E is evidence-IP for H* to degree 7/30 where H* is the proposition that there is a proposition E* such that E* is true and is evidence-IP for H ₂ to degree 0.05. By (n) it follows that H* is evidence-IP for H ₂ to degree 0.05. By (o) it follows that H* & E is evidence-IP for H ₂ to degree 0.05. By (p) it follows that E is not evidence-IP for H ₂. Hence EEE_NDC, as with EEE*_NDC, is open to counterexample.

T&C close their discussion by giving a gloss of EEE_NDC and writing (where notation has been slightly modified):

They can no longer say that.

Now return to the point from Section 2 that there are counterexamples to the thesis that NDC is sufficient for transitivity in evidence-IP. It turns out that things are different in the special case where E entails H ₁. The following holds without exception:

Suppose that (i)-(iv) in TIP*_NDC all hold. It follows from (iv) that E is logically equivalent to H ₁ & E. Hence, given (iii), it follows that Pr(H ₂ | E) > Pr(H ₂). Should this be welcome news to T&C?

No. Any case where (iv) in TIP*_NDC holds is a case where (v) holds if and only if (iii) holds. It does not matter whether (i) and (ii) hold. This can be seen by noting that the following variant of TIP*_NDC also holds without exception:

TIP*_NDC, as with TIP*_NDC, is a TIP thesis in name only.Footnote ¹⁴

T&C could modify EEE_NDC along the lines of TIP*_NDC. Then it would not be open to counterexample. But, at the same time, it would be an EEE thesis in name only.

A better tack would be to abandon the no-defeaters approach in favor of the screening-off approach. EEE_SOC is not open to counterexample and is not an EEE thesis in name only.