This is not a case where we have two definitions talking about two sorts of things (like sound waves versus perception of sound waves). This is a case where we have two rival mathematical definitions to account for the relation of evidential support.
What is this "relation of evidential support", that is a given thing in front of us? From your paraphrase of Achinstein, and the blurb of his book, it is clear that there is no such thing, any more than "sound" means something distinct from either "vibrations" or "aural perceptions". "Sound" is a word that covers both of these, and since both are generally present when we ordinarily talk of sound, the unheard falling tree appears paradoxical, leading us to grasp around for something else that "sound" must mean. "Evidence" is a word that covers both of the two definitions offered, and several others, but the fact that our use of the word does not seem to match any one of them does not mean that there must be something else in the world that is the true meaning of "evidence".
The analogy with unheard falling trees is exact.
What would you expect to accomplish by discovering whether some particular e really is "evidence" for some h, that would not be accomplished by discovering whether each of the concrete definitions is satisfied? If you know whether e is "fortitudinence" for h (increases its probability), and you know whether e is "veritescence" for h (gives a posterior probability above 1/2), what else do you want to know?
BTW, around here "fortitudinence" is generally called "Bayesian evidence" for reasons connected with Bayes theorem, but again, that's just a definition. There are reasons why that is an especially useful concept, but however strong those reasons, one is not discovering what the word "evidence" "really means".
Thanks. I would say that what we have in front of us are clear cases where someone have evidence for something else. In the example given, we have in front of us that both, e1 and e2 (together with the assumption that the NYT and WP are reliable) are evidence for g. So, presumably, there is an agreement between people offering the truth conditions for 'e is evidence that h' about the range of cases where there is evidence - while the is no agreement between people answering the question about the sound of the three, because the don't agree on the range of ...
I would like to share a doubt with you. Peter Achinstein, in his The Book of Evidence considers two probabilistic views about the conditions that must be satisfied in order for e to be evidence that h. The first one says that e is evidence that h when e increases the probability of h when added to some background information b:
The second one says that e is evidence that h when the probability of h conditional on e is higher than some threshold k:
A plausible way of interpreting the second definition is by saying that k = 1/2. When one takes k to have such fixed value, it turns out that P(h|e) > k has the same truth-conditions as P(h|e) > P(~h|e) - at least if we are assuming that P is a function obeying Kolmogorov's axioms of the probability calculus. Now, Achinstein takes P(h|e) > k to be a necessary but insufficient condition for e to be evidence that h - while he claims that P(h|e&b) > P(h|b) is neither necessary nor sufficient for e to be evidence that h. That may seem shocking for those that take the condition fleshed out in (Increase in Probability) at least as a necessary condition for evidential support (I take it that the claim that it is necessary and sufficient is far from accepted - presumably one also wants to qualify e as true, or as known, or as justifiably believed, etc). So I would like to check one of Achinstein's counter-examples to the claim that increase in probability is a necessary condition for evidential support.
The relevant example is as follows:
The point now is that, although it seems right to regard e2 as being evidence in favor of h, it fails to increase h's probability conditional on (b&e1) - at least so says Achinstein. According to his example, the following is true:
Well, I have my doubts about this counterexample. The problem with it seems to me to be this: that e1 and e2 are taken to be the same piece of evidence. Let me explain. If e1 and e2 increase the probability of h, that is because they increase the probability of a further proposition:
and, as it happens, g increases the probability of h. That The New York Times reports g, assuming that the New York Times is reliable, increases the probability of g - and the same can be said about The Washington Post reporting g. But the counterexample seems to assume that both e1 and e2 are equivalent with g, and they're not. Now, it is clear that P(h|b&g) = P(h|b&g&g), but this does not show that e2 fails to increase h's probability on (b&e1). So, if it is true that e2 increases the probability of g conditional on e1, that is, if P(g|e1&e2) > P(g|e1), and if it is true that g increases the probability of h, then it is also true that e2 increases the probability of h. I may be missing something, but this reasoning sounds right to me - the example wouldn't be a counterexample. What do you think?