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Probability and Theory Confirmation

Probability and Theory Confirmation. Why making a prediction that is borne out makes a theory more probably true. Probability can be thought of as:. Objective chance Physical propensity Degree of belief Long-run frequency Degree of confirmation. The probability function p(A).

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Probability and Theory Confirmation

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  1. Probability and Theory Confirmation Why making a prediction that is borne out makes a theory more probably true

  2. Probability can be thought of as: • Objective chance • Physical propensity • Degree of belief • Long-run frequency • Degree of confirmation

  3. The probability function p(A) • Applies to events or propositions A • Yields as output a real number in [0,1] • Linearizes what is otherwise a partial ordering of the propositions • … and must therefore satisfy certain constraints

  4. Constraints on p(A) • 0  p(A)  1 • If |– A , then p(A)=1 • If A |– , then p(A)=0 • If A |– B , then p(A)  p(B)

  5. Lemma 1A |– B B |– Ap(A)=p(B). Proof: By (4) we have p(A)p(B) and p(B)p(A). By anti-symmetry of ,we have p(A)=p(B).

  6. Definition AB is short for (AB)

  7. Definition If 0<p(B) then p(A|B)=p(AB)/p(B)

  8. Theorem 1 (Bayes)0<p(A) 0<p(B)p(A|B)=p(B|A).p(A)/p(B) Under the hypotheses we have, by definition, p(AB)=p(A|B).p(B) and p(BA)=p(B|A).p(A). By Lemma 1, p(AB)=p(BA). Hence p(A|B).p(B)=p(B|A).p(A). Dividing both sides by p(B), the result follows.

  9. Lemma 20<p(B) B |– A0<p(A). Proof: Suppose B|–A. By (4), p(B)  p(A). Now suppose 0<p(B). It follows that 0<p(A).

  10. Theorem 20<p(CB) 0<p(AB)p(A|CB).p(C|B)=p(C|AB).p(A|B) Proof: Suppose 0<p(CB). Note CB|–B. By Lemma 2, 0<p(B). By Lemma 1, p(ACB)=p(CAB). Hence p(ACB)/p(B) = p(CAB)/p(B). Hence p(ACB).p(CB)/p(CB).p(B) = p(CAB).p(AB)/p(AB).p(B). By definition, p(A|CB).p(C|B)=p(C|AB).p(A|B)

  11. Lemma 3B |– A 0<p(B)p(A|B)=1. Proof: Suppose B|–A. Then B is interdeducible with AB. So by Lemma 1 we have p(B)=p(AB). By definition we have p(A|B)=p(AB)/p(B). Substituting, p(A|B)=p(B)/p(B)=1.

  12. Lemma 4A,B |– C 0<p(AB)p(C|AB)=1. Proof: Suppose A,B|–C. Then AB|–CAB. Also, CAB|–AB. So by Lemma 1 we have p(CAB)=p(AB). By definition we have p(C|AB)=p(CAB)/p(AB). Substituting, p(C|AB)=p(AB)/p(AB)=1.

  13. Corollary 1A,B |– C 0<p(CB) 0<p(AB)p(A|CB).p(C|B)=p(A|B). Proof: The two inequalities mean Theorem 2 applies. So p(A|CB).p(C|B)=p(C|AB).p(A|B). But now the deducibility hypothesis means that the factor p(C|AB)=1. The result follows.

  14. Theorem 30<p(HK) H,K |– E p(E|K)<1p(H|EK)>p(H|K). Proof: P(E|HK) exists. Since H,K|–E, we have (by Lemma 4) p(E|HK)=1. We also have HK|–EK. By (4), p(HK)  p(EK). But 0<p(HK). So 0<p(EK). Now Theorem 2 applies: p(H|EK).p(E|K)=p(E|HK).p(H|K). Putting p(E|HK)=1, we get p(H|EK).p(E|K)=p(H|K). But p(E|K)<1. So p(H|EK)>p(H|K).

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