Fantastic heuristic! It’s like x=y·(z/y)+(1-y)·(x-z)/(1-y) for the rationalist’s soul :)
It’s worth noting, though, that you can rationally expect your credence in a certain belief “to increase”, in the following sense: If I roll a die, and I’m about to show you the result, your credence that it didn’t land 6 is now 5⁄6, and you’re 5⁄6 sure that this credence it about to increase to 1.
I think this is what makes people feel like they can have a non-trivial expected value for their new beliefs: you can expect an increase or expect a decrease, but quantitatively the two possibilities exactly cancel each out in the expected value of your belief.
It’s worth noting, though, that you can rationally expect your credence in a certain belief “to increase”, in the following sense: If I roll a die, and I’m about to show you the result, your credence that it didn’t land 6 is now 5⁄6, and you’re 5⁄6 sure that this credence it about to increase to 1.
No, you can’t, because you also expect with 1⁄6 probability that your credence will go down to zero: 5⁄6 + (5/6 1⁄6) + (1/6 −5/6) = 5⁄6.
In order to fully understand this concept, it helped me to think about it this way: any evidence shifting your expectated change in confidence will necessarily cause a corresponding shift in your actual confidence. Suppose you hold some belief B with confidence C. Now some new experiment is being performed that will produce more data about B. If you had some prior evidence that the new data is expected to shift your confidence to C’, that same evidence would already have shifted C to C’, thus maintaining the conservation of expected evidence.
Consider the following example: initially, if someone were to ask you to bet on the veracity of B, you would choose odds C:(1-C). Suppose an oracle reveals to you that there is a 1⁄3 chance of the new data shifting your confidence to C+ and a 2⁄3 chance of it shifting to C-, giving C’=(C + (C+)/3 − 2C(-)/3). What would you then consider to be fair odds on B’s correctness?
Fantastic heuristic! It’s like x=y·(z/y)+(1-y)·(x-z)/(1-y) for the rationalist’s soul :)
It’s worth noting, though, that you can rationally expect your credence in a certain belief “to increase”, in the following sense: If I roll a die, and I’m about to show you the result, your credence that it didn’t land 6 is now 5⁄6, and you’re 5⁄6 sure that this credence it about to increase to 1.
I think this is what makes people feel like they can have a non-trivial expected value for their new beliefs: you can expect an increase or expect a decrease, but quantitatively the two possibilities exactly cancel each out in the expected value of your belief.
No, you can’t, because you also expect with 1⁄6 probability that your credence will go down to zero: 5⁄6 + (5/6 1⁄6) + (1/6 −5/6) = 5⁄6.
In order to fully understand this concept, it helped me to think about it this way: any evidence shifting your expectated change in confidence will necessarily cause a corresponding shift in your actual confidence. Suppose you hold some belief B with confidence C. Now some new experiment is being performed that will produce more data about B. If you had some prior evidence that the new data is expected to shift your confidence to C’, that same evidence would already have shifted C to C’, thus maintaining the conservation of expected evidence.
Consider the following example: initially, if someone were to ask you to bet on the veracity of B, you would choose odds C:(1-C). Suppose an oracle reveals to you that there is a 1⁄3 chance of the new data shifting your confidence to C+ and a 2⁄3 chance of it shifting to C-, giving C’=(C + (C+)/3 − 2C(-)/3). What would you then consider to be fair odds on B’s correctness?