In your theorem, I don’t see how you get that E1[f(x,Y,Z)∣Y≥c]=E2[f(x,Y,Z)∣Y≥c]. Just because the conditional expectation of Z is the same doesn’t mean the conditional expectation of f(x,Y,Z) is the same (e.g. you could have two different distributions over Z with the same expected value conditional on Y≥c but different shapes, and then have f depend non-linearly on Z, or something similar with Y). It seems like you’d need some stronger assumptions on f or whatever to get this to work. Or am I misunderstanding something?
Ack, I think you’re right. I think I need to replace the assumption that 𝔼₁[Z | Y≥c] = 𝔼₂[Z | Y≥c] with the assumption that ℙ₁[(Y,Z)|Y≥c] = ℙ₂[(Y,Z)|Y≥c] which will guarantee the equality you’re pointing out is true for all f. This seems totally fine since it’s just a construction, but it definitely looks like an error to me. I’ll fix that, thank you!
In your theorem, I don’t see how you get that E1[f(x,Y,Z)∣Y≥c]=E2[f(x,Y,Z)∣Y≥c]. Just because the conditional expectation of Z is the same doesn’t mean the conditional expectation of f(x,Y,Z) is the same (e.g. you could have two different distributions over Z with the same expected value conditional on Y≥c but different shapes, and then have f depend non-linearly on Z, or something similar with Y). It seems like you’d need some stronger assumptions on f or whatever to get this to work. Or am I misunderstanding something?
(Your overall point seems right, though)
Ack, I think you’re right. I think I need to replace the assumption that 𝔼₁[Z | Y≥c] = 𝔼₂[Z | Y≥c] with the assumption that ℙ₁[(Y,Z)|Y≥c] = ℙ₂[(Y,Z)|Y≥c] which will guarantee the equality you’re pointing out is true for all f. This seems totally fine since it’s just a construction, but it definitely looks like an error to me. I’ll fix that, thank you!