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 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.