Confirmation and the ravens
paradox 1
Seminar 3: Philosophy of the
Sciences
Wednesday, 21 September 2011
1
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)
2
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)
3
Tutorials
Tutorials will start on this Friday 23 September
Class 1: 1 PM - 2 PM seminar room 305
Class 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
4
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
5
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?
6
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); and
ii) 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)
8
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
9
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)); and
ii) Fa.~Ga disconfirms (xFx).(x(Fx  Gx)).
10
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 because
a) 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.
12
Two kinds of confirmation relation
3-place: E confirms H relative to background
knowledge K
2-place: E confirms H absolutely
Connection 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
13
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 ~H
iii) Otherwise E neither confirms or disconfirms
H
In favour of HDT: HDT fits many episodes in the
history of science well.
14
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.
15
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!
16
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’.
17
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)
18
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)
19
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 either
i) There are 100 black ravens, no non-black
ravens and 1 million other birds
ii) 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
either
i) There are 100 black ravens, no non-black
ravens and 1 million other birds
ii) 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).
21
Good on absolute confirmation
Good also claimed that it might be that Ra.Ba
fails to confirm x(Rx  Bx) absolutely.
Discuss unicorn case.
22
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.
23
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).
24