The overall order is reversed after looking at each group. If the question is that, if a child is to be sent to one of the states to receive education, which one would be a better choice, then the overall index cannot be trusted.
The reason why this paradox happens is that the distribution of the covariate X, which is the unit's race, is different between the two groups. This is equivalent to the difference in assignment mechanism W across each race group. The following example explains:
group1
group2
Treat
a
b
Control
c
d
ba=dc⟺ca=db
This is math from elementary school.
So we need to know the assignment mechanism before conducting correct inferences.
Note that even in CREs, Simpson's paradox can still happen. If it happens that the distribution of the covariate differs between the treated group and the control group, though it doesn't affect the validity of the estimator.
Definitions for Assignment Mechanism
Three Concepts
Definition (Assignment Mechanism) Given a population of N units, the assignment mechanism is a row-exchangeable function P(W∣X,Y(0),Y(1)), taking on values in [0,1], satisfying
W∈{0,1}N∑P(W∣X,Y(0),Y(1))=1,∀X,Y(0),Y(1)
Definition (Unit Assignment Probability)
pi(X,Y(0),Y(1))=P(Wi=1∣X,Y(0),Y(1))
Definition (Finite Population Propensity Score) The propensity score at x is the average unit assignment probability for units with Xi=x,
e(x)=N(x)1i:Xi=x∑pi(X,Y(1),Y(0))
where N(x)=#{i∈{1,2,…,N}:Xi=x} is the number of units with Xi=x. If N(x)=0,e(x) is defined to be 0.
Examples
Some notes on example III:
Because there is no covariate, the propensity score degrades to
e(x)=N1i=1∑Npi(Y(1),Y(0))
Whatever Y(1),Y(0) are, e(x)=21. It's easy to verify.
Below is a more complex example.
A sequential tree should be used to analyze this example.
Let A=Y1(1)−Y2(0),B=Y1(0)−Y2(1). Then the value space of Y(1),Y(0) can be divided into four mutually exclusive parts:
(1)(2)(3)(4)A>0,B>0A>0,B<0A<0,B>0A<0,B<0
For the first part, W can only be (1,0,1) and (0,1,0) with equal probs 0.5. Similarly we can get the full assignment mechanism.
As for the unit assignment prob., we can easily verify that
p1(X,Y(0),Y(1))=p2(X,Y(0),Y(1))=21
(Note that it doesn't matter if we add X or not because the assignment mechanism, in this setting, has nothing to do with X.)
The four examples above are somewhat complex already though the number of the units is rather small. Cannot imagine how hard it would be if the number of units becomes large. So we need some assumptions to simplify our analysis.
Assumptions
Definition (Individualistic Assignment) An assignment mechanism P(W∣X,Y(0),Y(1)) is individualistic if, for some function q(⋅)∈[0,1],
for (W,X,Y(0),Y(1))∈A, for some set A, and 0 elsewhere (c is the constant that ensures that the probabilities sum to unity).
The line ⋯∏… is weaker than saying that Wi's are independent given X,Y(0),Y(1). For example, complete randomization satisfies this.
Example IV above doesn't satisfy this.
Given individualistic assignment, the propensity score simplifies to
e(x)=Nx1i:Xi=x∑q(Xi,Yi(0),Yi(1))
3N-variate function → 3-variate function.
Definition (Probabilistic Assignment) An assignment mechanism P(W∣X,Y(0),Y(1)) is probabilistic if the probability of assignment to treatment for unit i is strictly between zero and one:
pi(X,Y(0),Y(1))∈(0,1),for each possibleX,Y(0),Y(1)
for all i=1,…,N.
Don't know the purpose of this assumption.
Definition (Unconfounded Assignment) An assignment mechanism is unconfounded if it does not depend on the potential outcomes:
P(W∣X,Y(0),Y(1))=P(W∣X,Y′(0),Y′(1))
for all W,X,Y(0),Y(1),Y′(0),Y′(1).
If an assignment mechanism is unconfounded, we can drop the two potential outcomes and write the assignment mechanism as P(W∣X).
This kind of assignment can avoid Simpson's paradox. Why?
Application
Here is an example of application, which will be used here. Suppose Xi∈{f,m}. Then
