I think that we agree that neither of the definitions offered in the post are correct.
Can you see any problem with “e is evidence of h iff P(h|e) > P(h)”, other than cases where evidence interacts in some complex manner such that P(h|e1)>P(h); P(h|e2)>P(h); but P(h|e1&e1)<P(h) (I’m not sure that is even possible, but I think it can be done with three mutually exclusive hypotheses).
Yes, we agree on that. There is an example that copes with the structure you just mentioned. Suppose that
h: I will get rid of the flu
e1: I took Fluminex
e2: I took Fluminalva
b: Fluminex and Fluminalva cancel each other’s effect against flu
Now suppose that both, Fluminex and Fluminalva, are effective against flu. Given this setting, P(h|b&e1)>P(h|b) and P(h|b&e2)>P(h|b), but P(h|b&e1&e2)<P(h|b). If the use of background b is bothering you, just embed the information about the canceling of effects in each of the pieces of evidence e1 and e2.
I see further problems with the Positive Relevance account, like the one that lies in saying that the fact that a swimmer is swimming is evidence that she will drown—just because swimming increases the probability of drowning. I see more hope for a combination of these two accounts, but one in which quantification over the background b is very important. We shouldn’t require that in order for e to be evidence that h it has to increase the probability of h conditional in any background b.
I don’t understand what it would mean to divorce a hypothesis h from the background b.
Suppose you have the flu (background b); there is zero chance that you don’t have the flu, so P(~b)=0 and P(x&~b)=0, therefore P(x|~b)=0 (or undefined, but can be treated as zero for these purposes).
Since P(x)=P(x|b)+P(x|~b), P(x)=P(x|b) EDIT: As pointed out below, P(x)=P(x|b)P(b)+P(x|~b)P(~b). This changes nothing else . If we change the background information, we change b and are dealing with a new hypothetical universe (for example, one in which taking both Fluminex and Fluminalva increases the duration of a flu.)
In that universe, you need prior beliefs about whether you are taking Fluminex and Fluminalva, (and both, if they aren’t independent) as well as their effectiveness separately and together, in order to come to a conclusion.
P, h, and e are all dependent on the universe b existing, and a different universe (even one that only varies in a tiny bit of information) means a different h, even if the same words are used to describe it. Evidence exists only in the (possibly hypothetical) universe that it actually exists in.
Given a deck of cards shuffled and arranged in a circle, the odds of the northernmost card being the Ace of Spades should be 1⁄52. h=the northernmost card is the Ace of Spades (AoS)
Turning over a card at random which is neither the AoS nor the northernmost card is evidence for h.
Omega providing the true statement “The AoS is between the KoD and 5oC” is not evidence for or against, unless the card we turned over is either adjacent to the northernmost card or one of the referenced cards.
If we select another card at random, we can update again- either to 2%, 50%, 1, or 0. (2% if none of the referenced cards are shown, 50% if an adjacent card is picked and it is either KoD or 5oC, 1 if the northernmost card is picked and it is the AoS, and 0 if one of the referenced cards turns up where it shouldn’t be.)
That seems enough proof that evidence can alter the evidential value of other evidence.
I think that we agree that neither of the definitions offered in the post are correct.
Can you see any problem with “e is evidence of h iff P(h|e) > P(h)”, other than cases where evidence interacts in some complex manner such that P(h|e1)>P(h); P(h|e2)>P(h); but P(h|e1&e1)<P(h) (I’m not sure that is even possible, but I think it can be done with three mutually exclusive hypotheses).
Yes, we agree on that. There is an example that copes with the structure you just mentioned. Suppose that
h: I will get rid of the flu
e1: I took Fluminex
e2: I took Fluminalva
b: Fluminex and Fluminalva cancel each other’s effect against flu
Now suppose that both, Fluminex and Fluminalva, are effective against flu. Given this setting, P(h|b&e1)>P(h|b) and P(h|b&e2)>P(h|b), but P(h|b&e1&e2)<P(h|b). If the use of background b is bothering you, just embed the information about the canceling of effects in each of the pieces of evidence e1 and e2.
I see further problems with the Positive Relevance account, like the one that lies in saying that the fact that a swimmer is swimming is evidence that she will drown—just because swimming increases the probability of drowning. I see more hope for a combination of these two accounts, but one in which quantification over the background b is very important. We shouldn’t require that in order for e to be evidence that h it has to increase the probability of h conditional in any background b.
I don’t understand what it would mean to divorce a hypothesis h from the background b.
Suppose you have the flu (background b); there is zero chance that you don’t have the flu, so P(~b)=0 and P(x&~b)=0, therefore P(x|~b)=0 (or undefined, but can be treated as zero for these purposes).
Since P(x)=P(x|b)+P(x|~b), P(x)=P(x|b) EDIT: As pointed out below, P(x)=P(x|b)P(b)+P(x|~b)P(~b). This changes nothing else . If we change the background information, we change b and are dealing with a new hypothetical universe (for example, one in which taking both Fluminex and Fluminalva increases the duration of a flu.)
In that universe, you need prior beliefs about whether you are taking Fluminex and Fluminalva, (and both, if they aren’t independent) as well as their effectiveness separately and together, in order to come to a conclusion.
P, h, and e are all dependent on the universe b existing, and a different universe (even one that only varies in a tiny bit of information) means a different h, even if the same words are used to describe it. Evidence exists only in the (possibly hypothetical) universe that it actually exists in.
Me neither—but I am not thinking that it is a good idea to divorce h from b.
Just a technical point: P(x) = P(x|b)P(b) + P(x|~b)P(~b)
Given a deck of cards shuffled and arranged in a circle, the odds of the northernmost card being the Ace of Spades should be 1⁄52. h=the northernmost card is the Ace of Spades (AoS)
Turning over a card at random which is neither the AoS nor the northernmost card is evidence for h.
Omega providing the true statement “The AoS is between the KoD and 5oC” is not evidence for or against, unless the card we turned over is either adjacent to the northernmost card or one of the referenced cards.
If we select another card at random, we can update again- either to 2%, 50%, 1, or 0. (2% if none of the referenced cards are shown, 50% if an adjacent card is picked and it is either KoD or 5oC, 1 if the northernmost card is picked and it is the AoS, and 0 if one of the referenced cards turns up where it shouldn’t be.)
That seems enough proof that evidence can alter the evidential value of other evidence.