Imagine that I had a bit, and it’s either a 0 or a 1, and it’s either blue or green. And these two facts are primitive and independently generated. And I also have this other concept that’s like, “Is it grue or bleen?”, which is the XOR of blue/green and 0⁄1.
There’s a sense in which we’re inferring X is before Y, and in that case, we can infer that blueness is before grueness. And that’s pointing at the fact that blueness is more primitive, and grueness is a derived property.
I found it helpful to work through this example. Let’s say A is “is the bit 0 or 1” (resp. probabilities 1⁄4 and 3⁄4), B is “is the bit blue or green” (independent of A, 1⁄3 and 2⁄3), and C is “is the bit bleen (blue/0 or green/1) or grue (blue/1 or green/0)” (7/12 and 5⁄12).
We have A and B independent of each other, but A and C aren’t independent, and B and C aren’t independent.
We have
A is orthogonal to B
A can be computed from B and C
So by the theorem you proved, A is before C.
(And applying the theorem a second time, we have B is before C.)
And you can’t do this in reverse. Someone might try to say: “no, the bit is bleen or grue with probabilities 7⁄12 and 5⁄12, and then it’s blue if it’s bleen/0 or grue/1 and which has probability 1/3”. But they couldn’t tell you that bleen/grue and 0⁄1 are independent; you can reframe all you like, but that won’t change. And so the theorem won’t apply, we’ll still have B before C and not C before B.
(But per a discussion elsethread, this wouldn’t work if the probabilities were all 1⁄2, because then A and C are independent and so are B and C.)
I found it helpful to work through this example. Let’s say A is “is the bit 0 or 1” (resp. probabilities 1⁄4 and 3⁄4), B is “is the bit blue or green” (independent of A, 1⁄3 and 2⁄3), and C is “is the bit bleen (blue/0 or green/1) or grue (blue/1 or green/0)” (7/12 and 5⁄12).
We have A and B independent of each other, but A and C aren’t independent, and B and C aren’t independent.
We have
A is orthogonal to B
A can be computed from B and C
So by the theorem you proved, A is before C.
(And applying the theorem a second time, we have B is before C.)
And you can’t do this in reverse. Someone might try to say: “no, the bit is bleen or grue with probabilities 7⁄12 and 5⁄12, and then it’s blue if it’s bleen/0 or grue/1 and which has probability 1/3”. But they couldn’t tell you that bleen/grue and 0⁄1 are independent; you can reframe all you like, but that won’t change. And so the theorem won’t apply, we’ll still have B before C and not C before B.
(But per a discussion elsethread, this wouldn’t work if the probabilities were all 1⁄2, because then A and C are independent and so are B and C.)