Goodman’s ‘New Riddle’

8/19/2019 Goodman’s ‘New Riddle’

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Nicod & Hempel Hempel & Goodman Goodman & Bayesianism Triviality? Summary References

Goodman’s ‘New Riddle’

Branden Fitelson

Department of PhilosophyGroup in Logic and the Methodology of Science

&

Institute for Cognitive and Brain Sciences

University of California–Berkeley

[email protected]://fitelson.org/

Branden Fitelson Goodman’s ‘New Riddle’ fitelson.org


2/17


The Plan

This is mainly a historical talk (part of an ongoing bookproject on confirmation theory) with the following aims:

First, to place Goodman’s “Grue” paradox in historical

perspective (Keynes, Nicod, Hempel, and friends).

Then, to examine the paradox from both Hempelian and

Bayesian perspectives (with an inductive-logic slant).

Upshot: To try to motivate the following claims:

“Grue” is not necessary for discrediting Hempel’s theory of

confirmation. [This can be done in an elementary way.]

“Grue” is not necessary for establishing underdetermination

cases. [These were known long before Goodman.]

“Grue” is not sufficient to establish Goodman’s desired

“triviality result”. [This fails even for Hempel’s theory.]

“Grand Conclusion” of the talk: Either Goodman’s “New

Riddle” is not new, or it is not much of a riddle (or both).



3/17


Nicod [12], the grandfather of instantial confirmation, said:

Consider the formula or the law: A entails B . How can a particular

proposition, [i.e.] a fact, affect its probability? If this fact consists of the

presence of B in a case of A, it is favourable to the law . . . on the contrary, if

it consists of the absence of B in a case of A, it is unfavourable to this law.

While Nicod is not crystal clear on the logical details of hisaccount of instantial confirmation, three fundamental

aspects of it can be groked from his inchoate remarks:(1) Instantial confirmation is a relation between singular andgeneral propositions/statements (viz., “facts” and “laws”).

(2) Confirmation consists in positive probabilistic relevance, and

disconfirmation consists in negative probabilistic relevance

(for Nicod, the probabilities are Keynesian-inductive [10]).

(3) Universal generalizations are confirmed by their positive

instances and disconfirmed by their negative instances.

Hempel takes Nicod as his point of departure. But, Hempel’s

reconstruction of Nicod is rather uncharitable. Sadly, the

resulting straw man has had profound historical influence.Branden Fitelson Goodman’s ‘New Riddle’ fitelson.org


4/17


Hempel’s reconstruction of Nicod’s instantial principle is(NC0) For all objects x, and for all properties φ and ψ:

x confirms (∀y)(φy ⊃ ψy) iff φx & ψx is true, and

x disconfirms (∀y)(φy ⊃ ψy) iff φx & ∼ψx is true.(NC0) diverges from Nicod’s conception in two key ways:

(NC0) makes the relata of the confirmation relation objects

and statements (as opposed to statements and statements).

(NC0) says nothing about (inductive) probability .

As Hempel explains, (NC0) is quite unattractive (henceuncharitable?), since it violates the equivalence condition

(EQC) If x confirms y , then x confirms anything equivalent to y .

Hempel formulates a general, logical theory of confirmation.

His theory satisfies (EQC) and some other desiderata he lays

down. But, it exhibits some undesirable vestiges of (NC0).E.g., Hempel’s theory implies that, for all objects x, for all(compatible) properties φ and ψ, and for all statements H :

(NC) φx & ψx confirms (∀y)(φy ⊃ ψy) and

φx & ∼ψx disconfirms (∀y)(φy ⊃ ψy).

(M) If φx confirms H , then φx & ψx confirms H .



5/17


(NC) is a more plausible thing to attribute to Nicod. But,

Hempel’s theory is still non-probabilisic. And, (M) is quite

implausible — especially from a Pr-relevance point of view.Interestingly, Hempel himself seems (quite rightly) to have

thought that (M) was (intuitively) false. In his discussion of

the raven paradox, he describes a counterexample to (M)!

Hempel seems to have been unaware of this tension in his

views [1]. My conjecture is that (M) is a carry-over fromHempel’s objectual reconstruction of Nicod. Anyhow, (M) is

responsible for various historical confusions in this area.

E.g., the claim that the raven paradox argument shows that

“a is a white shoe” confirms that all ravens are black.

With that background in place, now we’re ready to look atthe “New Riddle”. There are two parts to the “Riddle”:

Part I: the underdetermination/indistinguishability part.

Part II: the triviality part.

I’ll say something about each part (I think both are crucial).


Ni d & H l H l & G d G d & B i i T i i li ? S R f


6/17


Here is (the all too familiar) Part I of Goodman’s “Riddle”.

Now let me introduce another predicate less familiar than “green”. It is the

predicate “grue” and it applies to all things examined before t just in case

they are green but to other things just in case they are blue. Then at time t

we have, for each evidence statement asserting that a given emerald is green,

a parallel evidence statement asserting that that emerald is grue. And the

statements that emerald a is grue, that emerald b is grue . . . will each confirm

the general hypothesis that all emeralds are grue. Thus according to our

definition, the prediction that all emeralds subsequently examined will be

green and the prediction that all will be grue are alike confirmed by evidence

statements describing the same observations. . . . Thus although we are well

aware which of the two incompatible predictions is genuinely confirmed,

they are equally well confirmed according to our present definition. . . .

Next, I will reconstruct this first part of the “New Riddle”,

and evaluate it from Hempelian and Bayesian perspectives.

Then, I will move on to the triviality part of the “Riddle”.


Ni d & H l H l & G d G d & B i i T i i lit ? S R f


7/17


Let Ox x is examined before t, Gx x is green, Ex x

is an emerald, and Gx Ox ≡ Gx (i.e., x is “grue”).

Now, we can state the two salient hypotheses, as follows:

(H 1) All emeralds are green. Formally, H 1 is: (∀x)(Ex ⊃ Gx).

(H 2) All emeralds are grue. Formally, H 2 is: (∀x)(Ex ⊃ Gx).

Even more precisely, H 2 is: (∀x)[Ex ⊃ (Ox ≡ Gx)].

And, here are three salient (instantial) evidential statements:

(E 1) Ea & Ga (E 2) Ea & (Oa ≡ Ga) (E ) Ea & Oa & GaGoodman must be saying that there is some statement

about a that (on Hempel’s theory) confirms both H 1 and H 2.

Neither E 1 nor E 2 does the trick. But, E does . This follows

from (NC) and (M). As such, I will reconstruct Part I of the

“New Riddle” as involving the following two central claims:(6) E confirms both H 1 and H 2.

(6) E confirms both H 1 and H 2 equally .

That is, Part I of the “Riddle” is that H 1 and H 2 are

confirmation-theoretically indistinguishable by E .




8/17


For Bayesians, confirmation is a three-place relation,

defined as follows. E confirms H , relative to background

corpus K iff Pr(H | E & K) > Pr(H | K), for some suitable Pr.

From a Bayesian point of view, (NC) is not generally true[2, 3, 11]. However, a closely related principle is correct:

(NC⊃) For all objects x and all (compatible) properties φ and ψ:

φx ⊃ ψx confirms (∀y)(φy ⊃ ψy), and

φx & ∼ψx disconfirms (∀y)(φy ⊃ ψy).

(NC⊃) makes sense because (∀y)(φy ⊃ ψy) φa ⊃ ψa,

and φa & ∼ψa ∼(∀y)(φy ⊃ ψy). On the other hand,

(NC) is less clear because (∀y)(φy ⊃ ψy) φa & ψa.

Of course, if one assumes (M), then (NC⊃) entails (NC)! But,

(M) is rather obviously false from a Bayesian point of view.

Let Bx x is a black card, let Ax x is the ace of spades,and let Jx x is the jack of clubs. Assuming (K ) that wesample a card a at random from a standard deck, we have:

(7) Ba confirms Aa, relative to K , but

(8) Ba & Ja dis confirms (indeed, refutes !) Aa, relative to K .




9/17


In Hempel’s theory, (6) trades on (M). But, (M) is clearly false

from a Bayesian point of view. So, we’ve seen no good

reason for a Bayesian to accept (6). [In fact, (M) is a reason

to reject Hempel’s theory — no “grue” needed for that !]

Moreover, here’s a Bayesian counterexample to (6) & (6) (a

la Good [2, 3]) such that E confirms H 1 but dis confirms H 2.

(K ) Either: (H 1) there are 1000 green emeralds 900 of which have

been examined before t, no non-green emeralds, and 1 millionother things in the universe, or (H 2) there are 100 green

emeralds that have been examined before t, no green emeralds

that have not been examined before t, 900 non-green emeralds

that have not been examined before t , and 1 million other things.

Now, imagine an urn containing true descriptions of eachobject in the universe, and let E “Ea & Oa & Ga” be

drawn. E confirms H 1 but E dis confirms H 2, relative to K :

Pr(E | H 1 & K) = 900

1001000 >

100

1001000 = Pr(E | H 2 & K)




10/17


Of course, all this shows is that neither (6) nor (6) is a

logical consequence of Bayesian confirmation theory. But, (6)

and (6) are consistent with Bayesian confirmation theory.

As in the literature on the raven paradox [3, 7], we now turn

to the “appropriateness” of our probabilistic assumptions

for modeling the confirmation relations of the “paradox”.

Hempel [6, 7] explains that the raven paradox is about K ’s

which do not entail that a has (or lacks) any of theproperties involved in the H ’s (i.e., ±ravenhood/± blackness).

Presumably, we should say the same thing about “grue”.

But, it is often assumed (e.g., [13]) that K entails Oa. This

directly contradicts the Hempelian line on “paradoxes”.

Following Hempel, I have not assumed the object inquestion (a) has or lacks any of the properties E , G, or O.

The key question is whether Pr(E | H 2 & K) < Pr(E | H 1 & K)

can be motivated as a reasonable (“appropriate for ‘grue”’)

probabilistic confirmation-theoretic modeling assumption.




11/17


In a recent manuscript, Roger White asks us to think about

what a “grue” (viz., “anti-inductive”) world might be like:

To achieve our goal of making standard inductive methods unreliable in a

world, we are going to have to take a more aggressive . . . approach. Rather

than just assigning properties to objects and leaving it to chance how our

inductivist victims will obtain their data, we must somehow give them a

tendency to obtain misleading data. . . . We put only green emeralds in the

vicinity of the surface of the earth, burying the blue ones deep below, ortucking them away in hard to find places. Then once our victims have

sampled sufficiently many green emeralds that they can inductively predict

that the next one will be green, we cause a blue one to rise to the surface

nearby. . . . Coordinating the . . . show will require either a very smart demon

overseeing it all, or the design of some intricate and peculiar physical laws.

Following this line, it may be reasonable to use models in

which Pr(Oa | H 2 & K) < Pr(Oa | H 1 & K) to evaluate the

salient confirmation relations. If so, this opens the door to

Pr(E |H 2

& K) < Pr(E |H 1

& K). Think inductive Pr [10, 12]!




12/17

p p y y y

Some would-be Bayesians modelers [13] assume something

even stronger than Pr(Oa | K) = 1: that K entails Ea & Oa.

Relative to K ’s that entail both Ea and Oa, both H 1 and H 2

entail E . Then, E must confirm both H 1 and H 2, relative to

K , and Pr(E | H 1 & K) = Pr(E | H 2 & K) = 1. Thus, (6) & (6).

Another (more subtle) tactic used by Bayesians to ensure (6)

& (6) is to argue that K entails that if a’s color is observed

(by S prior to t), then Oa will inevitably be learned (by S ) [9].And, something that’s inevitably learned whenever salient

observations are made can’t (from a Bayesian confirmation

theory point of view) be used to discriminate hypotheses.

Similar arguments have been made concerning observation

selection effects (e.g., Swinburne’s firing squad example [14]).Such arguments would have had little force for people like

Keynes or Nicod, who were talking about logical relations. It

is important here to distinguish confirmation and learning .

[Analogy: entailment vs inference in the case of deduction.]




13/17

p p y y y

Goodman thought that the “new riddle” led (at least in the

context of Hempel’s theory of confirmation) to the triviality

result that everything confirms everything ! Here is Part II:

. . . Moreover, it is clear that if we simply choose an appropriate predicate,

then on the basis of these same observations we shall have equal

confirmation, by our definition, for any prediction whatever about other

emeralds—or indeed about anything else. . . . We are left . . . with the

intolerable result that anything confirms anything.So, Goodman wasn’t merely describing a clever and hard

underdetermination problem. He was saying that his “grue”

problem reduces confirmation theory to utter triviality !

Goodman’s triviality argument fails, even when aimed at

Hempel.1 And, from a Bayesian POV, it’s a non-starter.

Next, I’ll run through the argument in the context of

Hempel’s theory. Then, I’ll make some Bayesian remarks.

1On this score, Hempel seems to have conceded too much to Goodman.

He seems to have accepted the triviality part of Goodman’s argument [8].Branden Fitelson Goodman’s ‘New Riddle’ fitelson.org



14/17

Here’s my best guess as to how this Goodmanian “triviality

argument” against Hempel’s theory is supposed to run [8].

(i ) E confirms both H 1 and H 2.

[This is just (6), which we have already discussed at length.]

(ii ) ∴ E confirms both Gb and Bb.

[where b = “the first emerald examined after t” (assuming

one exists), and G and B are incompatible properties.]

(iii ) ∴ E confirms anything .

(iv ) ∴ Mutatis mutandis , anything confirms anything .

While (i ) is implied by Hempel’s theory, (ii ) cannot be.Hempel’s theory implies the consistency condition [5]:

(CC) If E confirms H 1 and E confirms H 2, then H 1 and H 2 are

logically consistent (provided only that E is self-consistent).

All that follows from Hempel’s theory is that E confirms both material conditionals : ∼Ob ⊃ Gb and ∼Ob ⊃ Bb.

But, even E & ∼Ob cannot Hempel-confirm both Gb and Bb

[8]. So, Goodman’s “triviality argument” fails at step (ii ).

The step from (ii ) to (iii ) is OK — in Hempel’s theory . . .




15/17

. . . because Hempel’s theory implies the following pair:(&C) If E confirms H 1 and H 2, then E confirms H 1 & H 2.

(SCC) If E confirms H 1, and H 1 entails H 2, then E confirms H 2.

Finally, we can generalize (i )–(iii ) to yield something close to(iv ). For any observation statement E Oa & φa, E

(Hempel-)confirms both (∀x)φx and (∀x)(φx ≡ Ox).

Then, by (i )–(iii ) mutatis mutandis , E confirms anything .

So, while Goodman’s triviality argument against Hempel’s

theory fails at step (ii ), it does have some prima facie

plausibility to it. From a Bayesian perspective, it does not.

First, (i ) is not implied by Bayesian CT. And, even if (i ) were

true, (ii ) would not follow from it. Interestingly, (ii ) could be

true in BCT [since BCT does not imply (CC)!], but (iii ) wouldnot follow, since (&C), (SCC) aren’t implied by BCT either.Using the properties from our counterexample to (M):

∼(CC): Ba confirms both Aa and Ja, which are inconsistent.

∼(&C): However, Ba doesn’t confirm Aa & Ja.

∼(SCC): Nor does Ba confirm (Sa) a is some ace [Aa Sa].




16/17

Main points:

“Grue” is not necessary for discrediting Hempel’s theory of

confirmation. That Hempel’s theory entails (M) is sufficient.

“Grue” is not necessary for establishing underdetermination

cases. Simple curve-fitiing cases with natural kinds will do.

“Grue” is not sufficient to establish Goodman’s desired

“triviality result”. His triviality argument almost works on

Hempel’s theory, but it’s a non-starter for Bayesian theory.

“Grand Conclusion” of the talk:

EITHER

Goodman’s “New Riddle” is not new, because it’s just an

underdetermination/indistinguishability example (thesehad been constructed before without grue-like predicates).

OR

Goodman’s “New Riddle” is not much of a riddle, because

it’s an unsound (or non-starter) triviality argument.




17/17

[1] B. Fitelson, The paradox of confirmation, Philosophy Compass (B. Weatherson

and C. Callender, eds .), Blacwkell (online publication), Oxford, November 2005,

URL: http://dx.doi.org/10.1111/j.1747-9991.2005.00011.x.

[2] I.J. Good, The white shoe is a red herring , BJPS 17 (1967), 322.

[3] , The white shoe qua red herring is pink , BJPS 19 (1968), 156–157.

[4] N. Goodman, Fact, Fiction, and Forecast , Harvard University Press, 1955.

[5] C. Hempel, A purely syntactical definition of confirmation, JSL 8 (1943), 122–143.

[6] , Studies in the logic of confirmation, Mind 54 (1945), 1–26, 97–121.

[7] , The white shoe: no red herring , BJPS 18 (1967), 239–240.

[8] C. Hooker, Goodman, ‘grue’ and Hempel , Phil. of Sci., 35 (1968), 232–247.

[9] J. Joyce, APA Sympoisum comments on an earlier version of this talk, 2006.

[10] J. Keynes, A Treatise on Probability , Macmillan, 1921.

[11] P. Maher, Probability captures the logic of scientific confirmation, Contemporary

Debates in the Philosophy of Science (C. Hitchcock, ed .), Blackwell, 2004.

[12] J. Nicod, The logical problem of induction, (1923) in Geometry and Induction,

University of California Press, 1970.

[13] E. Sober, No model, no inference: A Bayesian primer on the grue problem, in

Grue! The New Riddle of Induction (D. Stalker ed .), Open Court, Chicago, 1994.

[14] , The Design Argument , in The Blackwell Guide to Phil. of Religion, 2004.

[15] R. White, A New Look at the Old Problem of Induction, unpublished manuscript.

URL: http://philosophy.fas.nyu.edu/docs/IO/1180/induction.pdf.

http://dx.doi.org/10.1111/j.1747-9991.2005.00011.xhttp://dx.doi.org/10.1111/j.1747-9991.2005.00011.xhttp://philosophy.fas.nyu.edu/docs/IO/1180/induction.pdfhttp://philosophy.fas.nyu.edu/docs/IO/1180/induction.pdfhttp://philosophy.fas.nyu.edu/docs/IO/1180/induction.pdfhttp://dx.doi.org/10.1111/j.1747-9991.2005.00011.x

Goodman’s ‘New Riddle’

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