SLIDE 1
Solomonoff Induction Violates Nicods Criterion Jan Leike and Marcus - - PowerPoint PPT Presentation
Solomonoff Induction Violates Nicods Criterion Jan Leike and Marcus - - PowerPoint PPT Presentation
Solomonoff Induction Violates Nicods Criterion Jan Leike and Marcus Hutter http://jan.leike.name/ Australian National University 23 June 2015 Outline The Paradox of Confirmation Solomonoff Induction Results Resolving the Paradox of
SLIDE 2
SLIDE 3
Motivation
What does this green apple tell you about black ravens?
SLIDE 4
The Paradox of Confirmation
Proposed by Hempel [Hem45]. H = all ravens are black
SLIDE 5
The Paradox of Confirmation
Proposed by Hempel [Hem45]. H = all ravens are black H′ = all nonblack objects are nonravens
SLIDE 6
The Paradox of Confirmation
Proposed by Hempel [Hem45]. H = all ravens are black H′ = all nonblack objects are nonravens
◮ Nicod’s criterion:
Something that is F and G confirms “all Fs are Gs” = ⇒ A nonblack nonraven confirms H′
SLIDE 7
The Paradox of Confirmation
Proposed by Hempel [Hem45]. H = all ravens are black H′ = all nonblack objects are nonravens
◮ Nicod’s criterion:
Something that is F and G confirms “all Fs are Gs” = ⇒ A nonblack nonraven confirms H′
◮ Equivalence condition:
Logically equivalent hypotheses are confirmed by the same evidence = ⇒ A nonblack nonraven confirms H
SLIDE 8
The Paradox of Confirmation
Proposed by Hempel [Hem45]. H = all ravens are black H′ = all nonblack objects are nonravens
◮ Nicod’s criterion:
Something that is F and G confirms “all Fs are Gs” = ⇒ A nonblack nonraven confirms H′
◮ Equivalence condition:
Logically equivalent hypotheses are confirmed by the same evidence = ⇒ A nonblack nonraven confirms H Paradox?
SLIDE 9
Outline
The Paradox of Confirmation Solomonoff Induction Results Resolving the Paradox of Confirmation References
SLIDE 10
Solomonoff Induction
Let U be a universal monotone Turing machine. Solomonoff’s universal prior [Sol64]: M(x) :=
- p: U(p)=x...
2−|p| M is a semimeasure (probability distribution on X ∞ ∪ X ∗).
SLIDE 11
Solomonoff Induction
Let U be a universal monotone Turing machine. Solomonoff’s universal prior [Sol64]: M(x) :=
- p: U(p)=x...
2−|p| M is a semimeasure (probability distribution on X ∞ ∪ X ∗). Solomonoff normalization: Mnorm(ǫ) := 1 and Mnorm(xa) := Mnorm(x) M(xa)
- b∈X M(xb)
Mnorm is a measure (probability distribution on X ∞).
SLIDE 12
Properties of Solomonoff Induction
◮ M is lower semicomputable, but M(xy | x) is incomputable
SLIDE 13
Properties of Solomonoff Induction
◮ M is lower semicomputable, but M(xy | x) is incomputable ◮ M ×
≥ µ for every computable measure µ
SLIDE 14
Properties of Solomonoff Induction
◮ M is lower semicomputable, but M(xy | x) is incomputable ◮ M ×
≥ µ for every computable measure µ
◮ At most E + O(
√ E) errors when predicting computable measure µ (E = errors of the predictor that knows µ) [Hut01]
SLIDE 15
Properties of Solomonoff Induction
◮ M is lower semicomputable, but M(xy | x) is incomputable ◮ M ×
≥ µ for every computable measure µ
◮ At most E + O(
√ E) errors when predicting computable measure µ (E = errors of the predictor that knows µ) [Hut01]
◮ M merges with any computable measure µ [BD62]:
sup
H measurable
M(H | x<t) − µ(H | x<t) → 0 µ-a.s. as t → ∞ where x<t := x1x2 . . . xt−1
SLIDE 16
Properties of Solomonoff Induction
◮ M is lower semicomputable, but M(xy | x) is incomputable ◮ M ×
≥ µ for every computable measure µ
◮ At most E + O(
√ E) errors when predicting computable measure µ (E = errors of the predictor that knows µ) [Hut01]
◮ M merges with any computable measure µ [BD62]:
sup
H measurable
M(H | x<t) − µ(H | x<t) → 0 µ-a.s. as t → ∞ where x<t := x1x2 . . . xt−1 = ⇒ M is really good at induction!
SLIDE 17
Outline
The Paradox of Confirmation Solomonoff Induction Results Resolving the Paradox of Confirmation References
SLIDE 18
Setup
◮ Alphabet = observations:
X := {BR, BR, BR, BR}
SLIDE 19
Setup
◮ Alphabet = observations:
X := {BR, BR, BR, BR}
◮ Hypothesis H = “all ravens are black”:
H := {x ∈ X ∞ ∪ X ∗ | x does not contain BR}
SLIDE 20
Setup
◮ Alphabet = observations:
X := {BR, BR, BR, BR}
◮ Hypothesis H = “all ravens are black”:
H := {x ∈ X ∞ ∪ X ∗ | x does not contain BR}
◮ Data x<t drawn from a computable measure µ for t = 1, 2, . . .
SLIDE 21
Setup
◮ Alphabet = observations:
X := {BR, BR, BR, BR}
◮ Hypothesis H = “all ravens are black”:
H := {x ∈ X ∞ ∪ X ∗ | x does not contain BR}
◮ Data x<t drawn from a computable measure µ for t = 1, 2, . . . ◮ M(H | x<t) is subjective belief in H at time step t
SLIDE 22
Setup
◮ Alphabet = observations:
X := {BR, BR, BR, BR}
◮ Hypothesis H = “all ravens are black”:
H := {x ∈ X ∞ ∪ X ∗ | x does not contain BR}
◮ Data x<t drawn from a computable measure µ for t = 1, 2, . . . ◮ M(H | x<t) is subjective belief in H at time step t
Confirmation and disconfirmation: µ(H) = 0 = ⇒ ∃t. M(H | x<t) = 0 µ-a.s. µ(H) = 1 = ⇒ M(H | x<t) → 1 µ-a.s.
SLIDE 23
Setup
◮ Alphabet = observations:
X := {BR, BR, BR, BR}
◮ Hypothesis H = “all ravens are black”:
H := {x ∈ X ∞ ∪ X ∗ | x does not contain BR}
◮ Data x<t drawn from a computable measure µ for t = 1, 2, . . . ◮ M(H | x<t) is subjective belief in H at time step t
Confirmation and disconfirmation: µ(H) = 0 = ⇒ ∃t. M(H | x<t) = 0 µ-a.s. µ(H) = 1 = ⇒ M(H | x<t) → 1 µ-a.s. Question: Does a black raven confirm H: M(H | x<t) < M(H | x<tBR)?
SLIDE 24
Solomonoff Induction and Nicod’s Criterion
Theorem (Counterfactual Black Raven Disconfirms H)
Let x1:∞ ∈ H ⊂ X ∞ be computable and xt = BR infinitely often. = ⇒ ∃t ∈ N (with xt = BR) s.t. M(H | x<tBR) < M(H | x<t)
SLIDE 25
Solomonoff Induction and Nicod’s Criterion
Theorem (Counterfactual Black Raven Disconfirms H)
Let x1:∞ ∈ H ⊂ X ∞ be computable and xt = BR infinitely often. = ⇒ ∃t ∈ N (with xt = BR) s.t. M(H | x<tBR) < M(H | x<t)
Theorem (Disconfirmation Infinitely Often for M)
Let x1:∞ ∈ H be computable. = ⇒ M(H | x1:t) < M(H | x<t) infinitely often.
SLIDE 26
Solomonoff Induction and Nicod’s Criterion
Theorem (Counterfactual Black Raven Disconfirms H)
Let x1:∞ ∈ H ⊂ X ∞ be computable and xt = BR infinitely often. = ⇒ ∃t ∈ N (with xt = BR) s.t. M(H | x<tBR) < M(H | x<t)
Theorem (Disconfirmation Infinitely Often for M)
Let x1:∞ ∈ H be computable. = ⇒ M(H | x1:t) < M(H | x<t) infinitely often.
Theorem (Disconfirmation Finitely Often for Mnorm)
Let x1:∞ ∈ H be computable. = ⇒ ∃t0∀t > t0. Mnorm(H | x1:t) > Mnorm(H | x<t).
SLIDE 27
Solomonoff Induction and Nicod’s Criterion
Theorem (Counterfactual Black Raven Disconfirms H)
Let x1:∞ ∈ H ⊂ X ∞ be computable and xt = BR infinitely often. = ⇒ ∃t ∈ N (with xt = BR) s.t. M(H | x<tBR) < M(H | x<t)
Theorem (Disconfirmation Infinitely Often for M)
Let x1:∞ ∈ H be computable. = ⇒ M(H | x1:t) < M(H | x<t) infinitely often.
Theorem (Disconfirmation Finitely Often for Mnorm)
Let x1:∞ ∈ H be computable. = ⇒ ∃t0∀t > t0. Mnorm(H | x1:t) > Mnorm(H | x<t).
Theorem (Disconfirmation Infinitely Often for Mnorm)
There is an (incomputable) x1:∞ ∈ H s.t. Mnorm(H | x1:t) < Mnorm(H | x<t) infinitely often.
SLIDE 28
Proof
M( · ∩ · ) H Hc
- a=xt Γx<ta
A B Γx1:t C D {x<t} E
Lemma
(i) 0 < A, B, C, D, E < 1 (ii) A + B
×
≤ 2−K(t) (iii) A, B
×
≥ 2−K(t) (iv) C
×
≥ 1 (v) D
×
≥ 2−m(t) (vi) D → 0 as t → ∞ (vii) E → 0 as t → ∞ M(H | x1:t) > M(H | x<t) if and only if AD + DE < BC
SLIDE 29
Outline
The Paradox of Confirmation Solomonoff Induction Results Resolving the Paradox of Confirmation References
SLIDE 30
Resolving the Paradox of Confirmation I
Solution: Reject Nicod’s criterion! Not all black ravens confirm H.
SLIDE 31
Resolving the Paradox of Confirmation II
- E. T. Jaynes [Jay03, p. 144]:
In the literature there are perhaps 100 ‘paradoxes’ and controversies which are like this, in that they arise from faulty intuition rather than faulty mathematics. Someone asserts a general principle that seems to him intuitively
- right. Then, when probability analysis reveals the error,
instead of taking this opportunity to educate his intuition, he reacts by rejecting the probability analysis.
SLIDE 32
Outline
The Paradox of Confirmation Solomonoff Induction Results Resolving the Paradox of Confirmation References
SLIDE 33