probabilities and indicative conditionals
TRANSCRIPT
Probabilities and Indicative Conditionals
Robert van Rooij, Katrin Schulz
Conditionals and Causality courseBeijing, June, 2021
Robert van Rooij, Probabilities and Indicative Conditionals 1 / 39
Meaning of Indicative Conditionals
To account for the meaning of indicative conditionals, we will
1 Introduce a new connective ⇒ not expressible in standard logic(se extend language with a new connective)
2 Formulate a new inference relation involving ‘⇒’
for 1, we need to introduce probabilities
for 2, we need to relate logic with probabilities
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Probabilities and Logic
1 Probabilities and Logic
2 Indicative conditionals and material implication
3 Conditionals and Conditional Probability
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Probabilities and Logic
Probabilities and Logic
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Probabilities and Logic
Probabilities and Logic
Normally assumed
Logic for1 consistency2 deductive inference
Probability for1 representation of uncertainty2 learning and induction
But inference with uncertain premisses is closely related with logic
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Probabilities and Logic
Axioms Probability Theory
Any function p from L (prop logic) to [0, 1] is a probability function iff
1 ∀S ∈ L : 0 ≤ p(S) ≤ 1
2 ∀S, T : If S |=cl T , then p(S) ≤ p(T )
3 If |=cl S, then p(S) = 1
4 If S and T mutual exclusive, then p(S ∨ T ) = p(S) + p(T )
Prove: p(¬S) = 1− p(S) (for any p)
Prove: p(S ∧ T ) ≥ p(S) (for any p)
Robert van Rooij, Probabilities and Indicative Conditionals 6 / 39
Probabilities and Logic
Some Exercises + facts, I
Prove: p(¬φ) = 1− p(φ) (for any p)1 =by 3 p(φ ∨ ¬φ) =by 4 p(φ) + p(¬φ) thus 1− p(φ) = p(¬φ)Prove: p(φ ∧ ψ) ≥ p(φ) (for any p)p(φ) =logic,2 p((φ ∧ ψ) ∨ (φ ∧ ¬ψ)) =by 4 p(φ ∧ ψ) + p(φ ∧ ¬ψ)thus p(φ ∧ ψ) = p(φ)− p(φ ∧ ¬ψ)p(φ ∨ ψ) = p(φ) + p(ψ)− p(φ ∧ ψ)thus if p(φ ∧ ψ) = 0, then p(φ ∨ ψ) = p(φ) + p(ψ)
p(E ∨ L) =
p(E) + p(L)− p(E ∧ L) = (p1 + p2) + (p1 + p3)− p1 = p1 + p2 + p3
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Probabilities and Logic
Probability and worlds
World as state description. Suppose only 2 sentences, E and L, then:1 = E ∧ L, 2 = E ∧ ¬L, 3 = ¬E ∧ L, 4 = ¬E ∧ ¬L:
p a function from W to [0, 1] s.t.∑
w∈W p(w) = 1
Assume every sentence φ denotes a proposition [[φ]]
p(φ) = p([[φ]]) =∑
w∈[[φ]] p(w)
e.g. p(E) = p1 + p2
Thus: p(⊥) = 0, p(>) = 1 and if [[φ]] ⊆ [[ψ]], then p(φ) ≤ p(ψ)
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Probabilities and Logic
Some facts, II
p(φ ∧ ψ) = p(φ) + p(ψ)− p(φ ∨ ψ)
Probability of φ ∧ ψ is not determined by probabilities of φ and of ψ⇒ p is not compositional
But probability of complex sentence depends on probabilities of statedescriptions: p(φ ∨ ψ) = p(φ ∧ ψ) + p(φ ∧ ¬ψ) + p(¬φ ∧ ψ)
A valuation function for propositional logic is a (very opinionated)probability function
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Probabilities and Logic
Some more facts relating to logic
if φ1, φ2 |=cl ψ and if p(φ1) = p(φ2) = 1, then p(ψ) = 1 (for all p)
This holds in general: also for Γ |=cl φ, if Γ has many premisses
If φ1, φ2 6|=cl ψ, then ∃p such that p(φ1) = p(φ2) = 1 and p(ψ) = 0
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Probabilities and Logic
Inference: probability can decrease
What if premisses don’t have probability 1?
A ∨B,A→ B |=cl B
Can prove: p(B) = p(A ∨B) + p(A→ B)− 1.For example, if p(A ∨B) = 6
7 and p(A→ B) = 57 , then p(B) = 4
7 .
But cannot always calculate the exact probability of the conclusion.Lower (and higher) bounds is in general all we can prove.
FACT: If p(A→ B) ≥ a, p(A) ≥ b, then p(B) ≥ a+ b− 1.
More specifically: if p(A→ B) ≥ 1− ε and p(A) ≥ 1− ε (i.e. if bothpremisses are almost true), then p(B) ≥ 1− 2ε
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Probabilities and Logic
Probabilistic Entailment
Γ |=1 φ iff ∀p : if ∀γ ∈ Γ : p(γ) = 1, then p(φ) = 1 Nothing new
Γ |=2 φ iff ∀p : if ∀γ ∈ Γ : p(γ) ≥ n, then p(φ) ≥ nMP not valid, Lottery paradox
FACT: If φ1, · · · , φn |=cl ψ andfor each φi : p(φi) ≥ 1− ε, then P (ψ) ≥ 1− nε.(i.e., for each φi : p(¬φi) < ε, then p(¬ψ) < nε)
Up(φ) =df p(¬φ) = 1− p(φ) Up(φ) stands for uncertainty
FACT: If Γ |=cl φ, then for all p : Up(φ) ≤∑
γ∈Γ Up(γ)
Proposal: Γ |=p φ iffdf for all p : Up(φ) ≤∑
γ∈Γ Up(γ)φ1, · · · , φn |=n ψ iff ∀p : ∀φi : p(φi) ≥ 1− ε, then p(ψ) ≥ 1− nε
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Probabilities and Logic
Suppes, Adams and |=cl
Γ |=s φ iff for all p : p(¬φ) ≤∑
γ∈Γ p(¬γ)
Γ |=a φ iff for all ε > 0 there is a δ > 0 such that for all p: ifp(¬γ) < δ for all γ ∈ Γ, then p(¬φ) < ε
Suppose language is classical (with → material implication)
Fact: (a) Γ |=cl φ iff (b) Γ |=s φ iff (c) Γ |=a φ
Proof: For (a) ⇒ (b), suppose Γ 6|=s φ, so there is a P , such that1− p(φ) >
∑γ∈Γ p(¬γ). From this it follows that
1 >∑
γ∈Γ p(¬γ) + p(φ). Thus (p(¬a ∨ b) = p(¬a) + p(b)− p(¬a ∧ b)),1 >
∑γ∈Γ p(¬γ) + p(φ) ≥ p(γ1 ∧ · · · ∧ γn → φ) for any subset
{γ1, · · · , γn} ⊆ Γ. Since all tautologies must have probability 1, nosuch implication γ1 ∧ · · · ∧ γn → φ is a tautology.
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Probabilities and Logic
From Suppes to Adams
(b) Γ |=s φ iff for all P : p(¬φ) ≤∑
γ∈Γ p(¬γ)
(c) Γ |=a φ iff for all ε > 0 there is a δ > 0 such that for P : ifp(¬γ) < δ for all γ ∈ Γ, then p(¬φ) < ε
For (b) ⇒ (c), suppose Γ |=s φ and ε > 0. If Γ is finite, chooseδ = ε
|Γ| . Then we have p(¬φ) ≤∑
γ∈Γ p(¬γ) < |Γ| × δ = ε. Hence
p(¬φ) < ε, and Γ |=a φ.
The case of Γ infinite is similar
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Probabilities and Logic
From Adams to Classical
(c) Γ |=a φ iff for all ε > 0 there is a δ > 0 such that for P : ifp(¬γ) < δ for all γ ∈ Γ, then p(¬φ) < ε
For (c) to (a), suppose Γ 6|=cl φ. For any ε > 0 we can simply take avaluation witnessing Γ 6|=cl φ, which is itself a probability function Pmaking p(φ) = 0 and p(γ) = 1, for all γ ∈ Γ. Hence Γ 6|=a φ.1
1Notice that essentially the same proof works to show (b) to (a).Robert van Rooij, Probabilities and Indicative Conditionals 15 / 39
Indicative conditionals and material implication
Indicative Conditionals
and Material Implication
Robert van Rooij, Probabilities and Indicative Conditionals 16 / 39
Indicative conditionals and material implication
Indicative versus Subjunctive conditionals
(i) If Oswald didn’t shoot Kennedy then someone else did.
(ii) If Oswald hadn’t shot Kennedy, someone else would have.
Syntactic distinction: Indicative vs. counterfactualsimple past vs. past perfect
We accept (i) but not (ii). A ∨B |= ¬A⇒ B (OR-to-IF) valid forindicatives (if A ∨B certain) but not for subjunctive conditionals.
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Indicative conditionals and material implication
Material Implication
Suppose we represent indicative conditional ‘If A, B’ by A⇒ B.
Should we interpret A⇒ B as material implication?
A→ B ≡ ¬A ∨B ≡ ¬(A ∧ ¬B)
Argument in favour 1: good of ad absurdum conditionals
1 If Trump becomes the US-president again, I’ll be the new emperor ofChina.
2 If you can keep the ball high 20 times, I’ll eat my hat
Claim ‘If A, then B’ to make the point that ¬A. because B is absurd
Argument in favour 2: Gibbard: If A⇒ B is stronger than A→ B,either Conditional Proof or Import-Export has to be given up
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Indicative conditionals and material implication
Proof Gibbard
(P1) A⇒ B |= A→ B assumption
(P2) if A |= B, then |= A⇒ B Conditional Proof
(P3) A⇒ (B ⇒ C) ≡ (A ∧B)⇒ C Import-Export
(C1) |= (A→ B)⇒ (A⇒ B) ≡ ((A→ B) ∧A)⇒ B instance of (P3)
(C2) |= ((A→ B) ∧A)⇒ B by (A→ B) ∧A |= B and (P2)
(C3) |= (A→ B)⇒ (A⇒ B) by (C1) and (C2)
(C4) (A→ B)⇒ (A⇒ B) |= (A→ B)→ (A⇒ B) by (P1)
(C5) |= (A→ B)→ (A⇒ B) by (C3) and (C4)
(C6) A→ B |= A⇒ B by Deduction Theorem
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Indicative conditionals and material implication
Problems Material Implication
Paradoxes of Material Implication
B |= A→ B
¬A |= A→ B
“If the butler didn’t do it, the gardener did.” ¬(¬B → G)can be denied without affirming butler’s innocence.But ¬(¬B → G) ≡ ¬(B ∨G) ≡ ¬B ∧ ¬G
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Indicative conditionals and material implication
Strict Implication
Perhaps we should analyse A⇒ B as strict implication?
A⇒ B = 2(A→ B) = ∀w ∈W : if w |= A, then w |= B
No paradoxes of Material Implication
But paradoxes of Strict Implication
2B |= 2(A→ B)
2¬A |= 2(A→ B)
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Indicative conditionals and material implication
Problems for Material & Strict implication
Adams: There are clear counterexamples to the following
A⇒ C |= (A ∧B)⇒ C Strengthening Antecedent
A⇒ B |= ¬B ⇒ ¬A Contraposition
A⇒ B,B ⇒ C |= A⇒ C Hypothetical Syllogism
A ∨B |= ¬A⇒ B OR-to-IF
None of them are valid on similarity-based analysis of counterfactuals!
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Indicative conditionals and material implication
Counterexamples I
(i) If there is sugar in the coffee, then it will taste good. therefore(ii) If there is sugar in the coffee and diesel-oil as well, then it willtaste good.
(i) If it is after 3 o’clock, it is not much after 3 o’clock. therefore(ii) If it is much after 3 o’clock, it is not after 3 o’clock.
(i) If Jones wins the election, Smith will retire to private life.(ii) If Smith dies before the election, Jones will win it. therefore(iii) If Smith dies before the election, he will retire to private life.
(i) Hitler started the second world war or some Martians did. therefore(ii) If Hitler didn’t start the second world war, some Martians did.
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Indicative conditionals and material implication
Counterexamples II
If Tweety is a bird, she can fly. thereforeIf Tweety is a penguin (and thus a bird), she can fly.
If Lucy does well for her job interview, then she will be hired. thereforeIf Lucy does not get hired, she did not do well in her job interview.
If Lucy does well in her job interview, then she will be hired.If Lucy is hired, then she will move to live closer. thereforeIf Lucy does well in her job interview, then she will move to live closer.
Either it rained or it snowed in Hangzhou in the winter of 2015.. thereforeIf it didn’t rain in Hangzhou in the winter of 2015, it snowed.
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Indicative conditionals and material implication
Pragmatics to the rescue?
Paradoxes of Strict Implication
2B |= 2(A→ B)2¬A |= 2(A→ B)
Pragmatics: when (in which contexts) is a sentence appropriate?Grice: Don’t say a weaker proposition if you can say (believe) astronger one
2B |= 2(A→ B), thus if say conditional, you indicate that ¬2BIf claim 2B then cancell this implicature, though conditional still true
Counterexample Or-to-If: A ∨B suggests ¬2A and ¬2B. Butcounterexample is one where 2A. Thus, counterexample is one wherethe disjunction could not be uttered appropriately.
Similarly with 2¬A for A→ B? Not really, claiming 2¬A suggeststhat you have no good evidence for conditional at all!
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Indicative conditionals and material implication
Other problems: Conditionals and Probabilities
To what degree would one believe the following sentences, given thata card has been picked at random from a standard 52 card deck?
1 The selected card is a king, if it’s red.2 It’s clubs, if it’s black.3 It’s spades, if it is a nine.
The obvious answers are 113 ,
12 , and 1
4 , respectively.
But p(R→ K) = p(¬R ∨K) = 2652 + 4
52 >113
Similar for other examples.
Thus, Material/strict implication don’t seem correct
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Indicative conditionals and material implication
Other problems with probabilities
Yog and Zog play chess, Yog has black 9 out of 10 times, and drawsare not allowed. We don’t know who won what game, but we doknow that of the hundred games they played up to now, Yog won 80times when he had black and lost all 10 times that he had white.Under these circumstances, the following sentences seem true:
(i) If Yog had black, there is a probability of 8/9 that he won.
(ii) If Yog didn’t win, there’s probability of 1/2 that he didn’t have black.
How to represent the conditionals, if using material implication?
A→ (p(C) = n)? No: wrong truth conditions, because?the probability that Yog won seems 0.8, and the consequent would thusbe false for both examples.p(A→ C) = n? No: because? exerciseproblem with contraposition
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Conditionals and Conditional Probability
Conditionals and
Conditional Probability
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Conditionals and Conditional Probability
Conditional probability and relevance
p(ψ/φ) = p(φ∧ψ)p(φ) , if p(φ) 6= 0
Note: ‘/’ is not a connective: p(χ/(ψ/φ)), p(¬(ψ/φ)) andp((ψ/φ) ∨ (χ/α)) are not interpretable
It is possible that p(ψ/φ ∧ χ) ≤ p(ψ/φ)⇒ p is not monotone
Note, p(φ ∧ ψ) = p(ψ/φ)× p(φ)
Chain rule: p(φ1 ∧ φ2 ∧ φ3) = p(φ3/φ1 ∧ φ2)× p(φ2/φ1)× p(φ1)
φ positively relevant to ψ, if p(ψ/φ) > p(ψ) iff. p(φ ∧ ψ) > p(φ)× p(ψ)
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Conditionals and Conditional Probability
Conditionals and conditional probability
Suggestion (Adams): analyse indicative conditionals A⇒ B suchthat p(A⇒ B) = p(B/A) or perhaps Ass(A⇒ B) = p(B/A)
The selected card is a king, if it is red. p(K/R) = 113
√
If Yog had black, there is a probability of 8/9 that he won.p(Won/Black) = 8/9
√
If Yog didn’t win, there’s probability of 1/2 that he didn’t have black.p(¬Black/¬Won) = 1/2
√
Contraposition is not valid, so the claims are mutually consistent√
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Conditionals and Conditional Probability
Conditionals as conditional probability
Recall: Γ |=p φ iffdf for all p : Up(φ) ≤∑
γ∈Γ Up(γ)
|=p can also be used if we extend language with ‘⇒’ !
FACT: If p(φ⇒ ψ) ≥ 1− ε and p(φ) ≥ 1− ε, then p(ψ) ≥ 1− 2ε.
Proof: If p(ψ/φ) ≥ 1− ε and p(φ) ≥ 1− ε, then p(ψ) ≥ (1− ε)2,and if ε > 0, (1− ε)2 = 1− 2ε+ ε2 > 1− 2ε.
Thus Modus Ponens, φ⇒ ψ, φ |=p ψ, is predicted to be valid.
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Conditionals and Conditional Probability
Quasi-Truth Values and probabilities
state descriptions A B A⇒ B
1 T T T
2 T F F
3 F T · · ·4 F F · · ·
with probabilities
state descriptions p A B A⇒ B
1 0.2 T T T
2 0.3 T F F
3 0.4 F T · · ·4 0.1 F F · · ·p 1 0.5 0.6 0.4
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Conditionals and Conditional Probability
Paradox of material implication
φ1, · · · , φn |=p ψ iff for all p : Up(φ1) + · · ·+ Up(φn) ≥ Up(ψ)
If only one premisse: φ |=p ψ iff for all p: Up(φ) ≥ Up(ψ)iff for all p: p(φ) ≤ p(ψ)
B 6|=p A⇒ B
state descriptions p A B A⇒ B
1 0.01 T T T
2 0.09 T F F
3 0.89 F T · · ·4 0.01 F F · · ·p 1 0.1 0.9 0.01
0.1 = 0.1
p(B) = 0.9, but p(A⇒ B) = 0.1
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Conditionals and Conditional Probability
A general theorem
Can show in general for all p
Up(φ1) + · · ·+ Up(φn) < Up(ψ) thus φ1, · · · , φn 6|=p ψ
iff
p(φ1) ≥ 0.9 and · · · and p(φn) ≥ 0.9, but p(ψ) ≤ 0.1
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Conditionals and Conditional Probability
Some more inferences
Inference 1 A→ B |=? A⇒ B
Inference 2 A ∨B |=? ¬A⇒ B Or-to-If
Inference 3 A⇒ B |=? ¬B ⇒ ¬A Contraposition
Inference 4 A⇒ B |=? A→ B
Inference 5 A,A⇒ B |=? B Modus Ponens
Inference 6 B ⇒ C |=? (A ∧B)⇒ C Strengthening Antecedent
Inference 7 A⇒ B,B ⇒ C |=? A⇒ C Transitivity
Inference 8 A⇒ B, (A ∧B)⇒ C |=? A⇒ C (cautious) Cut
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Conditionals and Conditional Probability
Inferences 1-3
We know already that A→ B 6|=p A⇒ B (cardgame example)A ∨B 6|=p ¬A⇒ BA⇒ B 6|=p ¬B ⇒ ¬A
regions p A B A ∨B ¬A⇒ B A⇒ B ¬B ⇒ ¬A1 0.900 T T T · · · T · · ·2 0.090 T F T · · · F F3 0.009 F F F F · · · T4 0.001 F T T T · · · · · ·
p 0.99 0.901 0.991 0.1 0.909 0.091
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Conditionals and Conditional Probability
Inferences 4 and 5 are valid
A⇒ B |=p A→ B
p(A→ B) = x+ z + w ≥ z/(y + z) = p(A⇒ B), because1− (x+ z + w) = y ≤ y/(y + z) (because y + z ≤ 1), andy/(y + z) = 1− z/(y + z)
A,A⇒ B |=p B Modus Ponens
Up(B) = x+ y, Up(A) = x+ w, and Up(A⇒ B) = y/(y + z), andtherefore up(B) ≤ Up(A) + Up(A⇒ B) becausex+ y ≤ x+ w + [y/(y + z)], due to y ≤ y/(y + z)
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Conditionals and Conditional Probability
Inferences 6 and 7 are invalid
B ⇒ C 6|=p (A ∧B)⇒ C Strengthening Antecedent
A⇒ B,B ⇒ C 6|=p A⇒ C Transitivity/Hypothetical Syllogism
region p A B C A⇒ B B ⇒ C (A ∧ B)⇒ C A⇒ C— — —T— —T— —T— —T— —T— —T— —T—3 0.09 T T F T F F F— — —T— —F— —T— —F— —· · ·— —· · ·— —T—2 0.00001 T F F F · · · · · · F5 0.9 F T T · · · T · · · · · ·6 0.0009 F T F · · · F · · · · · ·4 0.00009 F F T · · · · · · · · · · · ·1 0.009 F F F · · · · · · · · · · · ·
p 0.0901 0.9909 0.900091 0.9999 0.9083 0 0
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Conditionals and Conditional Probability
Arguments in favour of probabilistic analysis
Many doubtfull inferences not valid on a probabilistic analysis
Paradox of Material Implication1 B 6|=p A ⇒ B
Strengthening the Antecedent A⇒ C 6|=p (A ∧B)⇒ C
Contraposition A⇒ B 6|=p ¬B ⇒ ¬A
Hypothetical reasoning A⇒ B,B ⇒ C 6|=p A⇒ C
Or-to-If A ∨B 6|=p ¬A⇒ B.
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