-1

Prove that when an agent conditionalizes on new evidence, her credence in a proposition that entails the evidence cannot decrease. That is, when H entails E, it must be the case that cr2(H) is greater than or equal to cr1(H) when E is learned between t1 and t2.

4
  • question continued: is greater than or equal to cr1(H) when E is learned between t1 and t2. – Fiona717 Nov 21 '19 at 4:58
  • I'm pretty lost on this problem, and I do not know where to start – Fiona717 Nov 21 '19 at 4:59
  • Welcome to SE Philosophy! Please take a quick moment to take the tour (philosophy.stackexchange.com/tour) or see general help (philosophy.stackexchange.com/help). Please be aware that questions are subject to editing and closure, and that reflects the site's policies on acceptable questions and NOT a personal attack. (philosophy.stackexchange.com/help/dont-ask). Questions, including those that are closed, can be edited to bring them within guidelines. Additional clarification at(philosophy.meta.stackexchange.com). – J D Nov 21 '19 at 15:42
  • Please visit our Help Center to see what questions we answer and how to ask. Is this a HW question? We tend not to answer those unless more context is given, your own thinking on the issue presented, and specific difficulty identified. (In this case, go beyond just giving us the problem to do for you!) For a logic example, see: philosophy.stackexchange.com/questions/63249/… – J D Nov 21 '19 at 15:46
1

(I'm guessing you're learning about this in the context of Bayesian epistemology, right? "Credence" plays the roll that a probability assignment does in regular probability theory. I'll answer the question using the common probability theory notation. It should be an easy but useful exercise for you to then translate the concepts into your "credence" notation.)

First, let's symbolize the problem:

"H entails E" is symbolized as P(E|H) = 1; the probability of E being true given the truth of H is 1, or certain.

Next, we need to symbolize the probability of H before and after we learn about new evidence. Before we learn about E, we assign some prior probability to H, P(H). After learning about E, we conditionalize on the new evidence and get P(H|E), the probability of H given the new evidence E.

Putting it all together, we want to prove that if H entails E, then the probability of H given that E, cannot be less than the probability of H before learning that E. So in symbols we say...

Theorem: If P(E|H) = 1, then P(H|E) ≥ P(H)


Proof: P(H|E) = [P(E|H) * P(H)] / P(E)        Bayes' theorem
       P(H|E) = P(H) / P(E)                   Since we know that P(E|H) = 1, we can remove it
       P(H) = P(H|E) * P(E)                   Multiply both sides of he equality by P(E)

       Since P(E) is greater than 0 and at most 1 (axiom of probability), P(H|E) must
       be greater than or equal to P(H|E) * P(E). But since P(H|E) * P(E) is equal to
       P(H) it must be that P(H|E) is also greater than or equal to P(H). Which is what
       we wanted to prove.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.