I do not understand the statement of your current understanding, in particular point 1. Could you please state in different words and/or state it formally and/or give an example?
By scanning the graphical proof, I don’t see any issue on the following generalization of the Mediator Determines Redund Theorem: Let X1,…,Xn,Λ and Λ′ be random variables and let X1,…,Xm be any not-empty subset of X1,…,Xn that satisfy the following conditions: - Λ Mediation: X1,…,Xm are independent given Λ - Λ′ Redundancy: ∀j∈{1,…,m}Λ′←Xj→Λ′ Then Λ′←Λ→Λ′.
In the above, I’ve weaken the Λ′ Redundancy hypothesis, requiring that the redundancy of any subset of random variables is enough to conclude the thesis. Does the above generalization work (if don’t, why?).
If the above stands true, then just one observational random variable (with agreement) is enough to satisfy the Redundancy condition (Mediation is trivially true with one variable), an therefore ΛA is determined by ΛB. Moreover, in the general approximation case, if we have various sets of random variables that meet the naturality condition, we can choose the one that will minimize the errors (there’s some kind of trade-off between ϵmed and ϵred errors).
Ah, yes, that is almost correct. You need redundancy over TWO distinct observables (i.e. the subset must be at least size two), not just one, but otherwise yes. With just one observable, you don’t have two branches to dangle a Λ′ off of in the graphical proof, so we can’t get Λ between two copies of Λ′.
I do not understand the statement of your current understanding, in particular point 1. Could you please state in different words and/or state it formally and/or give an example?
By scanning the graphical proof, I don’t see any issue on the following generalization of the Mediator Determines Redund Theorem:
Let X1,…,Xn,Λ and Λ′ be random variables and let X1,…,Xm be any not-empty subset of X1,…,Xn that satisfy the following conditions:
- Λ Mediation: X1,…,Xm are independent given Λ
- Λ′ Redundancy: ∀j∈{1,…,m}Λ′←Xj→Λ′
Then Λ′←Λ→Λ′.
In the above, I’ve weaken the Λ′ Redundancy hypothesis, requiring that the redundancy of any subset of random variables is enough to conclude the thesis.
Does the above generalization work (if don’t, why?).
If the above stands true, then just one observational random variable (with agreement) is enough to satisfy the Redundancy condition (Mediation is trivially true with one variable), an therefore ΛA is determined by ΛB. Moreover, in the general approximation case, if we have various sets of random variables that meet the naturality condition, we can choose the one that will minimize the errors (there’s some kind of trade-off between ϵmed and ϵred errors).
Ah, yes, that is almost correct. You need redundancy over TWO distinct observables (i.e. the subset must be at least size two), not just one, but otherwise yes. With just one observable, you don’t have two branches to dangle a Λ′ off of in the graphical proof, so we can’t get Λ between two copies of Λ′.