As previously explained, I hold that one's sampling method is of crucial importance to determining whether an observed instance of a G-ish F is evidence that all Fs are G. To borrow Blar's example, suppose we're wondering whether there are any extraterrestrials. If God provided me with a

*random*sample of living creatures, and all of them were from Earth, then that would provide me with evidence that all living creatures are from Earth. But I get no such evidence from a sample that is restricted to Earthly creatures to begin with. If I see my neighbour's dog, it is both

*alive*and

*on Earth*, but - contrary to Hempel's R1 principle - this does not provide any evidence whatsoever that all life is on Earth (i.e. there are no aliens).

The argument from exclusion denies this. It begins by noting that there are various rival hypotheses about extraterrestrial life. On one, let's call it H(0), there are

*no*aliens: all living creatures are on Earth. At the other extreme, we have H(1), the claim that

*everything*is an alien, and nothing else but aliens exists. There are infinitely many such hypotheses

**H(p/q)**:

*that the proportion of aliens to objects in the universe is p/q*. [We will see later that this strict definition fails to yield the rough description of H(0) given above.]

We can begin by assigning each of these hypotheses a non-zero probability. Now, when we observe my neighbour's dog (or indeed anything at all that isn't an alien), that allows us to rule out H(1). Consequently, the non-zero probability previously attached to this will get reassigned around all the other H's, including H(0). Each will receive some miniscule, but non-zero, boost. Hence, the argument goes, observing my neighbour's dog

*does*(just slightly) confirm that there are no aliens.

I want to show that this argument fails. I previously showed that it yields a paradox of interpretation, because what conclusions we draw are radically relative upon how we choose to carve up the space of hypotheses. Suppose that, instead of appealing to the proportionally-focused 'H' hypotheses, we instead carved up the problem space in terms of cardinality. That is, we consider the range of hypotheses

**C(n)**:

*that there are exactly n aliens in existence*.

Now, each partition method is exhaustive. It must be the case that exactly one of the H(p/q) hypotheses is correct, and exactly one of the C(n) hypotheses is correct. Each partitions logical space in such a way as to exhaust the possibilities. There is no possible state of affairs that is not covered by exactly one H hypothesis and exactly one C hypothesis. Okay. Now, it seems that H(0) and C(0) are the same hypothesis. The proportion of aliens to other existing objects is zero iff there are zero aliens (but see below). But while observing my neighbour's dog allows us to rule out H(1), and thus slightly confirm H(0), the very same observation does not falsify

*any*of the C(n) hypotheses. For any n, it is possible for there to exist n aliens

*plus*my neighbour's dog. So the C hypotheses are unaffected by the observation. In particular, C(0) is not confirmed.

Thus we have the following inconsistent triad:

1) C(0) = H(0)

2) H(0) is confirmed by evidence E

3) C(0) is not confirmed by evidence E

Perhaps we should reject (1). If it's possible for there to be infinitely many objects, then H(0) could be true when C(0) is false. The proportion of aliens to other objects could be zero even if there are some finite number of aliens. This suggests to me that H's focus on proportions or ratios is misplaced. We're not interested in the relational question of what proportion of objects are aliens. Rather, we're interested in the

*absolute*number of aliens (and, particularly, whether it is non-zero).

This is further highlighted by considering repeated applications of Hempels R1 principle. It claims that

*each*observed co-instantiation of F-properties with G-properties confirms that all F's are G. So if, after observing the dog, I also observe a rabbit, the rabbit provides

*further*evidence that all living creatures are Earthly. How can the argument from exclusion deal with this? Well, it must say that we rule out the hypothesis E(-1) that

*everything is an alien except for one thing*.

There are two things to note here. Firstly, E(-1) is not part of our previous partitions of the problem space. It does not correspond to any C or H hypothesis. Granted, an exhaustive partition of logical space is provided by the hypotheses

**E(-n)**:

*that everything is an alien except for n things*. But this is a bizarre and unhelpful partition. None of the E hypotheses correspond to the desired hypothesis C(0) that there are no aliens. C(0) will be true iff, for some n, E(-n) is true

*and*there are exactly n objects. Depending on how many objects there are, this could potentially end up being

*any*of the E hypotheses. None of them are inconsistent with C(0). So, because they are not

*rival*hypotheses, ruling out E(-1), say, does nothing to confirm C(0).

Are there any other exhaustive partitions of logical space possible that would support the argument from exclusion? I can't think of any, but it isn't obvious how to prove this. I guess I need to show that the C partition is the appropriate one to make, such that any other partition that yields contradictory results (e.g. the H partition) must be mistaken.

How can I show this? I'm not too sure. It seems fairly self-evident though, really. If we're wondering whether there are no aliens, this corresponds to the claim that the number of aliens is zero, and this contrasts with all the rival C(n) claims that the number of aliens is n. This partition appeals only to the absolute number of aliens -- it is in this sense the most basic and fundamental partition we could make. The R partition appeals to the relative proportion of aliens to other objects, and so is more complex, depending upon both the absolute number of aliens

*and*the absolute number of everything else. Plus it misbehaves terribly if one allows for the possibility of infinitely many objects. Similarly the E partition appeals to both 'everything'

*and*the absolute number of aliens. So both R and E can be redefined in terms of C plus some extra variables. It seems fairly clear that there is no simpler possibility than C. Any other such partition is going to build on it in much the fashion that R and E do.

But that's a fairly rough intuitive account, so if anyone can see how to turn this into a rigorous proof, please let me know!

P.S. Put another way, I need to show that there is no possible partition of logical space such that "there are no aliens" [i.e. C(0)] and "everything is an alien" [i.e. E(0), and perhaps R(1)] are rival hypotheses within this partition. But perhaps this follows simply from the fact that C is an exhaustive partition which contains C(0) but does not contain anything corresponding to E(0)? I don't think it is quite this simple though -- one might always respond that it simply requires a finer-grained partition. Perhaps the better answer is to simply point out that E(0) and C(0) are not mutually exclusive: if

**no objects existed at all**, then it would be true both that there are no aliens

*and*that everything that exists is an alien! Perhaps this suffices for my proof?

Some further thoughts: It might help to rejig the claims [C(0) and E(0)] by appending "and there is at least one object" to each, to ensure that they really are mutually exclusive. Then they're different claims though, so I don't know if that is a legitimate move. Besides, I still don't think it'd be possible to provide the required exhaustive partition of logical space. To fill in the gaps, you'd have to appeal to intermediate hypotheses which say something like: "there are some objects that are aliens and others that aren't". I'll have to think about whether that would work. It certainly seems messier than the cardinal partition [C] I proposed. And it wouldn't save Hempel in cases where you observe more than one object (and so "should" get more than one boost to confirmation).

ReplyDeleteAn earlier thought that I forgot to include in the post: Let T be the most precise partition possible. So each hypothesis T(w) would provide an exhaustive description of exactly one possible world w.

ReplyDeleteThere are many possible worlds w consistent with the claim C(0) that there are no aliens. Thus there are (just as) many hypotheses T(w) consistent with this claim. In fact, C(0) simply amounts to the claim that one of these alienless worlds w obtains, i.e. that one of these T(w) hypotheses is true, where 'w' ranges over all those possible worlds where no aliens exist.

Now, suppose I observe a dog. What hypotheses does it allow us to rule out? Well, obviously enough, we can now exclude all those hypotheses T(k) which did not posit the existence of any dogs. This will rule out some w worlds, but also some non-w worlds. For example, let us say that possible world 997 contains seven aliens and nothing else, whereas possible world 998 contains seven sheep and nothing else. Both get ruled out by the observation of a dog. While excluding T(997) (a non-w world) might seem to support our claim C(0) that a w-world obtains, the exclusion of T(998) yields the opposite result. After all, 998 is a world where there are no aliens, so C(0) would be true if T(998) had been.

Thus, while observing non-alien objects will rule out a wide range of T hypotheses, the impact will be spread across T hypotheses both consistent with, and inconsistent with, C(0). Thus it will have no predictable influence on C(0)'s degree of confirmation.

But T is the most precise and thus fundamental partition possible. All other partitions involve combining various claims T(k) together (e.g. C(0) = the disjunction of all T(w) hypotheses, where 'w' ranges over the possible worlds where no aliens exist).

So if the partition T does not support the argument from exclusion, then there is no possible partition that can support the argument. But T does not support the argument. Therefore, no possible partition does.

Convinced?

One problem with your argument, Richard, is that increasing the probability of one hypothesis does not require ruling out the possibility of some other hypothesis from the same partition. It only requires decreasing the probability of some other hypothesis from the same partition.

ReplyDeleteI think the best argument against R1 is the one that I made in my last comment to the other thread, which I'm not sure if you've seen yet. The key is to abandon interesting examples about ravens or extraterrestrials or whatever and to make everything as precise and rigorous as possible by using a controlled example like the Wason selection task. Suppose that there are four cards on the table, and it is given that each of these four cards has a letter on one side (a-z) and a number on the other side (1-9). On the sides facing you the cards say A, B, 1, and 2, respectively. The hypothesis in question is that every card on the table that C1) has a vowel on its letter-side also C2) has an odd number on its number-side. R1 implies that flipping over the card with a 1 on it could give you evidence in favor of the hypothesis (if it has an A), but that is wrong, as Chris explained in the linked post. We could even make this more rigorous, if we claim that the cards received their labels as follows: first, I wrote the letters and numbers that you see on them: A, B, 1, and 2. Then, for each card I flipped a fair coin, and if it landed heads I put an A or a 1 on the other side (depending if it was a letter-side or a number-side), and if it landed tails I put a B or a 2. Then there is no way that the card that says 1 could give you information about whether the hypothesis is correct. Before you flip any cards, the probability of the hypothesis is 1/4. If you flip the 1 card and see an A, it's still 1/4. If you see a B, it's still 1/4. As far as I can tell, there is no good way out of this counterexample to R1. Your professor might claim that the stipulated information is still in doubt (e.g. the coin was not necessarily fair), deny that probabilities are what matter, or assert that for some reason R1 does not really apply to this case, but those moves do not seem to leave him in a very credible position.

Yeah, I dunno, the artificiality of the Wason task makes it a bit suspicious. I think the advocate of R1 would simply argue that the principle doesn't apply to this case. But it does at least show that R1 is not universally applicable.

ReplyDeleteIntriguingly, some have argued that the common mistakes we make on the Wason task are not really "mistakes" at all. Consider the rule: "If a car has a broken headlight, then it has a broken tail-light." If you want to confirm this rule, you'd do better to check the odd cars with broken tail-lights rather than all the ones that don't, despite the logician's howling objections. Affirming the consequent can be a useful heuristic in some circumstances. (It sounds like this paper applies such considerations to the raven paradox.)

"

One problem with your argument, Richard, is that increasing the probability of one hypothesis does not require ruling out the possibility of some other hypothesis from the same partition. It only requires decreasing the probability of some other hypothesis from the same partition."Why is that a problem for my argument? I don't think observing the dog even allows you to decrease the probability of any unambiguous rival of the "no aliens" hypothesis. Any hypothesis which

isaffected (e.g. the hypothesis that "everything is an alien") is not an unambiguous rival. It is not a wholly distinct option from the same partition.Accepting that R1 is not universally applicable seems like a big blow to its supporters. Once they retreat to R1' (below), they seem to be in a tough position, especially if they need to specify when R1 fails to apply with more detail than "unusual cases like the Wason one."

ReplyDeleteR1': When an object has two attributes C1, C2, it usually constitutes confirming evidence for the hypothesis that every object which has the attribute C1 also has the attribute C2.

What's more, there seem to be cases where such an observation constitutes disconfirming evidence, rather than confirming evidence. I mentioned the black hole example in my other post, and if you want to avoid statements that are trivially true (as you suggest in your post All of None) we could modify it so that the following five events occur in this chronological order (though in this case the example is clearly not historical):

1. Scientists determine that the sun is one star in a galaxy of many others, and that our galaxy is one many different galaxies of stars that are similar in many important respects.

2. Physicists develop a theory that allows for the existence of what are now known as black holes. There is no observational evidence for the existence of black holes.

3. A philosopher makes the following conjecture: every black hole in existence is located within our galaxy. That is, every object that is C1) a black hole is also C2) within our galaxy.

4'. Astrophysicists create a black hole in our galaxy in highly unusual conditions that are unlikely to have ever existed before.

5'. Astronomers observe several naturally occurring black holes in our galaxy in conditions that are often replicated in other galaxies.

The observations under 5' appear to disconfirm the philosopher's hypothesis rather than confirming it as R1 would imply. This would be true even if we modified 4' to 4'', in which the astrophysicists merely observe a black hole in our galaxy under unusual conditions that leave some doubt about whether there are other black holes anywhere.

Yeah, good point.

ReplyDeleteRe-reading Hempel, I note that he insisted we exclude all background information, possibly because of these sorts of problems. It might be that he was grasping towards the idea that random sampling is crucial, without quite knowing how to put it.