Confirmation and the ravens paradox 1

Seminar 3: Philosophy of the Sciences

Wednesday, 21 September 2011

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Required readings

Peter Godfrey Smith. Theory and Reality. Section 3.1-3.3 (can be downloaded from HKU library)

Clark Glymour ‘Why I am not a Bayesian’ (on course website)

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Optional readings

Paul Horwich ‘Wittgensteinian Bayesianism’ (on course website)

J. A. Cover and Martin Curd ‘Commentary on confirmation and relevance’, section 5.1, pp 627-638 (on course website)

Hawthorne and Fitelson. ‘The paradox of confirmation’, Philosophy Compass. 2006. pp 93-113 (can be downloaded from HKU library)

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TutorialsTutorials will start on this Friday 23 SeptemberClass 1: 1 PM - 2 PM seminar room 305Class 2: 4 PM – 5 PM seminar room 305

Required reading: ‘The Problem of Induction’, Section I, Chapter 7 of Richard Feldman’s book Epistemology pp 130-141 (on course website)

Required reading and seminar handouts must be brought along to tutorials

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Questions to be addressed

Q1) What is it for evidence E to confirm hypothesis H (that is, what is it for E to be evidence in favour of H)?

Q2) Which propositions confirm which propositions?

Q3) Which propositions do scientists and ordinary people take to confirm which propositions?

Note: Skeptics like Hume might still be interested in Q3

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Instances of Q2

• Does a being a black raven confirm all ravens are black?

• Does a being a white shoe confirm all ravens are black?

• Does there was scratching noises coming from the cupboard last night and the cheese in the cupboard has now disappeared confirm the cheese was eaten by a mouse?

• Do our data concerning changes in temperature and climate confirm the theory of man-made global warming?

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The instantial theory of confirmation (ITC)

For any predicates F and G, and any name a,i) Fa.Ga confirms x(Fx Gx) (All Fs are Gs); andii) Fa.~Ga disconfirms x(Fx Gx). Def: ‘.’ means ‘and’

Problem 1 with ITC: There is no obvious way to extend ITC to deal with plausible cases of confirmation like c) and d) described on slide 6.

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Problem 2: The ravens paradox(Carl Hempel)

Let ‘R’ symbolise ‘is a raven’, and ‘B’ symbolise ‘is black’.(1) By ITC, ~Ba.~Ra confirms x(~Bx ~Rx) (2)x(~Bx ~Rx) is necessarily equivalent to

x(Rx Bx) (3) By (1), (2) and (EQ), ~Ba.~Ra confirms x(Rx

Bx)

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The ravens paradox(cont)

(EQ) If E confirms H1, and H1 is necessarily equivalent to H2, then E confirms H2

(PC) therefore follows from (ITC) and (EQ). But how can a white shoe provide evidence that all ravens are black??

(PC) That a is non-black and non-raven confirms that all ravens are black

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A response to the paradox

Since ‘(xFx).(x(Fx Gx))’ is a better symbolisation of ‘All Fs are Gs’ than ‘x(Fx Gx)’, ITC should be replaced with ITC*.

ITC*) For any predicates F and G, and any name a,i) Fa.Ga confirms (xFx).(x(Fx Gx)); andii) Fa.~Ga disconfirms (xFx).(x(Fx Gx)).

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A problem with this response

Given (SPC), (ITC*) entails (PC).(SPC) If E confirms H1, and H1 entails H2, then E confirms H2.Argument:(1) By ITC*, ~Ba.~Ra confirms (x~Bx).(x(~Bx

~Rx))(2) (x~Bx).(x(~Bx ~Rx)) entails x(Rx Bx) (3) By (1), (2) and (SPC), ~Ba.~Ra confirms x(Rx

Bx) 11

Hempel’s response to the paradox

(PC) is true. It seems false becausea) We falsely think that ‘x(Rx Bx)’ is only about ravens, when in fact it is about all objects, as it is reformulation as ‘x(~Rx v Bx)’ reveals.b) We falsely think that PC is false, because we fail to distinguish it from the false PC*.(PC*) That a is non-black and non-raven confirms that all ravens are black, given the background information that a is a non-raven.

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Two kinds of confirmation relation

3-place: E confirms H relative to background knowledge K2-place: E confirms H absolutelyConnection between them: E confirms H absolutely iff E confirms H relative to no information (or relative to a logical truth T)

ITC is intended as a theory of absolute confirmation

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The hypothetico-deductive theory of confirmation (HDT)

i) E confirms H if E can be divided into two parts, E1 and E2, such that a) E1 does not entail E2, but b) the conjunction of H and E1 does entail E2.

ii) E disconfirms H if E entails ~Hiii) Otherwise E neither confirms or disconfirms

HIn favour of HDT: HDT fits many episodes in the history of science well.

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Problems with HDT

Problem 1: HDT entails PC, and hence faces the ravens paradox

Problem 2: HDT cannot account of confirmation of statistical theories such as the hypothesis that anyone who smokes has a 25% chance of developing lung cancer.

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Problem 3: Irrelevant conjunction

• Suppose evidence E, made up of E1 and E2, is such that i) E1 does not entail E2, but ii) H.E1 does entail E2.

• Then H.S.E1 entails E2, where S is any hypothesis at all.

• Hence, according to HDT, E confirms H.S.• Moreover, by SPC, E confirms S. But S can be

anything at all!

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The probability raising theory of confirmation

PRT for absolute confirmation:i) E confirms H iff P(H|E) > P(H)ii) E disconfirms H iff P(H|E) < P(H)

where ‘P(H)’ means ‘the probability of H’, and ‘P(H|E)’ means ‘the probability of H given E’.

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The probability raising theory of confirmation (cont)

PRT for relative confirmation:i) E confirms H relative to background

knowledge K iff P(H|E.B) > P(H|K)ii) E disconfirms H relative to background

knowledge K iff P(H|E.B) < P(H|K)

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Quantitative probability raising theories of confirmation

Def: c(H,E,K) = the degree to which E confirms H relative to background knowledge K

A popular account of c among PRT theorists:D) c(H,E,K) = P(H|E.K) - P(H|K)

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Good’s response to the raven paradox

Good’s claim: Whether E=Ra.Ba confirms H= x(Rx Bx) relative to K depends on what the background knowledge K is.

Example: E won’t confirm H if K is the knowledge that eitheri) There are 100 black ravens, no non-black

ravens and 1 million other birdsii) There are 1000 black ravens, 1 white raven,

and 1 million other birds 20

Good’s example

E won’t confirm H if K is the knowledge that eitheri) There are 100 black ravens, no non-black

ravens and 1 million other birdsii) There are 1000 black ravens, 1 white raven,

and 1 million other birds In this case P(E|H.K) < P(E|~H.K), from which it can be proved that P(H|E.K) < P(H|K).

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Good on absolute confirmation

Good also claimed that it might be that Ra.Ba fails to confirm x(Rx Bx) absolutely.

Discuss unicorn case.

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The standard Bayesian strategy to solve the ravens paradox

Show that given plausible assumptions about our background knowledge, Ra.Ba confirms x(Rx Bx) relative to K more than ~Ra.~Ba.

The result if established can then be used to explain why (PC) seems false.

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Hawthorne and Fitelson’s attempt

Given the assumptions about K given by (K-ass), H+F show that the following theorem holds.K-ass: i) P(H|Ba.Ra.K), P(H|~Ba.~Ra.K), and P(~Ba.Ra|K) aren’t 0 or 1; and ii) P(~Ba|K) > P(Ra|K).

Theorem: If P(H|Ra.K) ≥ P(H|~Ba.K), then P(H|Ba.Ra.K) > P(H|~Ba.~Ra.K).

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