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Induction

(Work In Progress: output at 07/08/2020 20:28:10)

(For earlier versions of this Note, see the table at the end)


Background

“The lesson of Hempel’s Paradox is that whether observations confirm a hypothesis is never independent of background information.” Discuss.

Introduction
What is Hempel’s Paradox?
Is the argument for the Paradox sound?
Nelson Goodman’s Explanation of Hempel’s Paradox
Bayesian Explanation of Hempel’s Paradox
  1. Quick Proof of Bayes’ Theorem
    • We start off with the basic truth of conditional probability, irrespective of the interpretation of the events E and H, that p(E&H) = p(E) x p(H|E). Similarly, p(H&E) = p(H) x p(E|H). So, given that p(E&H) = p(H&E) we have the simple form of Bayes’ theorem, that:
        p(H|E) / p(H) = p(E|H) / p(E)
    • The interpretation of this is as follows. The probability of an hypothesis, H, given a piece of evidence E, divided by the probability of that hypothesis before the evidence, equals the probability of the evidence, E, on the assumption the hypothesis is true, divided by the probability of the evidence irrespective of the truth of the hypothesis.
    • What we want is evidence that increases the probability of our hypothesis, ie. we want p(H|E) > p(H). This obviously occurs if p(H|E) / p(H) > 1, and, consequently (from our equation), if p(E|H) / p(E) >1. So, our evidence increases the probability of our hypothesis if p(E|H) / p(E) > 1. This is also evident from the standard form of Bayes’ Theorem:
        p(H|E) = p(H) * p(E|H) / p(E) …………. (*)
  2. Application of Bayes’ Theorem to Coin-Tossing
    • It is useful to see Bayes’ Theorem in action with a simple example, to highlight issues that will arise in the ensuing discussion, so we’ll imagine we’re tossing a coin, and our hypothesis H is that it’s double-headed. This raises the issue of sample spaces. We restrict the outcome of the experiment of coin tossing to heads or tails. We ignore the possibility of outcomes where coins roll away, land on their edge or never come down.
    • We’ll suppose that we’re initially highly sceptical, and give H only a 1% chance. This points out the first limitation of Bayesian methods – we have to “seed” the formula with some initial probability, which is simply our initial degree of confidence in the hypothesis. As we will see, it’s easy to have inconsistent beliefs where degrees of confidence are concerned.
    • If the first toss lands tails, our scepticism is rewarded, for (using (*) and assuming that p(E) = 0.5 since most coins are of the head/tails variety) p(H|E) = 0.01*0/0.5 = 0. This is because p(E|H), the probability of tails on the hypothesis that the coin is double-headed is zero. Our evidence has falsified the hypothesis. Since p(H|E) = 0 becomes the p(H) in the next round of trials, p(H|E) stays at zero throughout subsequent trials, no matter how many heads we get. This is in accord with intuition. If we’ve seen a tail, the coin can’t be double-headed.
    • Say the first toss lands heads. Then, p(H|E) = 0.01*1/0.5 = 0.02, since p(E|H), the probability of heads on the hypothesis that the coin is double-headed, is 1. What about the second round of trials? We set p(H) to 0.02 and p(E) is again 1. What about p(E), presupposing that E is another head? There seems no good reason to budge from our original presupposition of an unbiased coin, so p(E) is again 0.5, and p(H|E) = 0.02*1/0.5 = 0.04. It’s easy to see that after 7 successive heads, we’d end up with p(H|E) = 1.28, which is invalid! There are two lessons to be learnt from this:
      1. Firstly, since we can consider 7 successive heads as one piece of evidence, our initial confidence of 1% was inconsistent with this piece of potential evidence and our belief that we had a normal coin, demonstrated by assigning p(E) = 1/128.
      2. Secondly, and relatedly, as the number of consecutive heads increases, our confidence in a fair coin must decrease if we are to maintain consistent beliefs.
  3. Application of Bayes’ Theorem to the Raven Paradox
    • Now let’s interpret Bayes’ Theorem using H1 (“all ravens are black”) for H, and two separate, alternative pieces of evidence, E1 and E2, for E.
      • E1 is the occurrence of a black raven, and
      • E2 is the occurrence of a non-black non-raven.
    • Clearly, we cannot estimate p(Ei|H1) or p(Ei) without background information – that is, without some knowledge of the world. To come to a conclusion, we also need to choose our sample spaces carefully.
      1. For E1, I’ll assume our sample space is of ravens, to simplify calculating the probabilities. Then, if H1 (all ravens are black) is true, E1 (a black raven) would be certain (since our sample space is restricted to ravens, we’re guaranteed a raven). So, p(E1|H1) = 1. However, p(E1) is to be determined in the absence of the hypothesis. What’s the probability of a sample raven being black, if we don’t know what colour they’re supposed to be? This depends on where we sampled the raven, so let’s assume it was selected from an English garden2 where birds are one of four colours (for the sake of argument; black, white, brown and “other”, in equal measure). Hence p(E1) = 0.25. So, p(E1|H1) / p(E1) = 4, and our confidence that all ravens are black has increased fourfold as a result of finding a black raven. Of course, setting p(E1) = 0.25 would be inconsistent with holding p(H1) > 0.25, as then we’d end up with the impossible p(H|E) > 1. For subsequent trials, we have the same dilemma as in the coin-tossing case. Our expectation p(E) of a black raven has to rise in line with the credence we give to H1 (all ravens are black), otherwise we have inconsistent beliefs.
      2. For E2, I’ll assume our sample space is “non-black things likely to be found in a garden”. The thing we’ve actually found, E2, is a non-black non-raven. We want to know the probability that it’s not a raven, given that it’s non-black. If H1 is true, ie. all ravens are black, then anything that isn’t black can’t be a raven. So, p(E2|H1) = 1. What about p(E2)? This is the probability of selecting a non-raven from all the nonblack things in my garden. Well, there are probably millions of non-black things in my garden (depending on what counts as a thing), and, even if there are non-black ravens, there aren’t likely to be many of them in my garden, since there are rarely any ravens there at all, at least compared with the number of other things. So, p(E2)≈1. Hence, our evidence of a green leaf has hardly shifted our confidence that all ravens are black at all.
    • This shows the importance of the background evidence. However, the killer blow is supplied by considering possible worlds in which the evidence of a green leaf would be strong support for the hypothesis that all ravens are black. Such a world is one in which most things are black and most things are ravens, (though it is still theoretically possible for some of the few non-black things to be ravens). We return to our two cases:
      1. For E1, we again assume our sample space is ravens. Again, if H1 (all ravens are black) is true, E1 (a black raven) would be certain. So, p(E1|H1) = 1. However, p(E1) is to be determined in the absence of the hypothesis. What’s the probability of a sample raven being black, in our new black- and raven-dominated world? Let’s suppose 95% of things are black, then it seems reasonable to suppose p(E1) = 0.95. So, p(E1|H1) / p(E1) = 1.00/0.95, and our confidence that all ravens are black has only increased by 5% as a result of finding a black raven.
      2. For E2, I’ll assume our sample space is “non-black things”. The thing we’ve actually found, E2, is a non-black non-raven. If H1 is true, ie. all ravens are black, then again anything that isn’t black can’t be a raven. So, p(E2|H1) = 1. What about p(E2) this time? Let’s suppose 90% of things are ravens. What we want is the probability of selecting a non-raven from all the non-black things. Well, in this strange world, the colour of the average raven could be anything, so we’ve no reason to think our nonblack thing to be any less likely to be a raven than the frequency of ravens in the general population. So, we have a 90% chance of our non-black thing being a raven, ie. only a 10% chance of it being a non-raven. So, p(E2) 0.1, and p(E2|H1) / p(E2) = 1.0/0.1 = 10. Hence, our evidence of a green leaf has increased our confidence that all ravens are black by a factor of 10.
Conclusion



In-Page Footnotes:

Footnote 1: Footnote 2:


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Previous Version of this Note:

Date Length Title
01/08/2017 00:11:31 300 Induction



Note last updated Reading List for this Topic Parent Topic
07/08/2020 20:28:15 None available None



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Authors, Books & Papers Citing this Note

Author Title Medium Extra Links Read?
Hains (Brigid) & Hains (Paul) Aeon: 2019+ Paper High Quality Abstract   Yes



References & Reading List

Author Title Medium Source Read?
Earman (John), Ed. Inference, Explanation and Other Philosophical Frustrations Book - By Subtopic (via Paper By Subtopic) Low Quality Abstract Earman (John), Ed. - Inference, Explanation and Other Philosophical Frustrations 4%
Goodman (Nelson) Fact, Fiction and Forecast Book - Cited Medium Quality Abstract Goodman (Nelson) - Fact, Fiction and Forecast Yes
Goodman (Nelson) The New Riddle of Induction Paper - Cited High Quality Abstract Goodman - Fact, Fiction and Forecast, 4th Edition, 1983, Chapter 3 Yes
Hains (Brigid) & Hains (Paul) Aeon: 2019+ Paper - Referencing Hains (Brigid) & Hains (Paul) - Aeon Yes
Hains (Brigid) & Hains (Paul) Aeon: A-B (& General) Book - Referencing (via Paper Referencing) Low Quality Abstract Bibliographical details to be supplied 100%
Hintikka (Jaakko) The Concept of Induction in the Light of the Interrogative Approach to Inquiry Paper - By Subtopic Medium Quality Abstract Earman (John), Ed. - Inference, Explanation and Other Philosophical Frustrations No



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