Probabilities and Indicative Conditionals

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


Probabilities and Logic

1 Probabilities and Logic

2 Indicative conditionals and material implication

3 Conditionals and Conditional Probability




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



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)



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



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)



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



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



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



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



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.



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



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.



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



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



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)



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!



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.



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.



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!



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



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


Conditionals and Conditional Probability

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)



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√



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.



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



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



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



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



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)



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



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.


Probabilities and Indicative Conditionals

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