I am reading What If by Hernan and Robins. In Section 2, the definition of conditional exchangeability is given as $$Y^a \perp A | L, $$ where $\perp$ means conditional independent.
In other words, this means $P(Y^a| A = 1, L = 1) = P(Y^a| A = 0, L = 1)$, i.e. $Y^a \perp A | L=1$ for all $a$; and $P(Y^a| A = 1, L = 0) = P(Y^a| A = 0, L = 0)$, i.e. $Y^a \perp A | L=0$ for all $a$.
In plain English, it means that in the subset $L=1$, the treated and the untreated are exchangeable because the treated had they remained untreated, would have experienced the same average outcome as the untreated did, and vice versa.
However, what confused me is that in the same section, the author keep mentioning the conditional probability of receiving every value of treatment depends only on measured covariates $L$, which I think is also exchangeability.
For example, if the only outcome predictor that is unequally distributed between the treated and the untreated is $L$, then conditional exchangeability $Y^a \perp A | L$ holds and standardization or IP weighting can be used to identify the causal effect.
On the other hand, if, say, $S$ is another outcome predictor that is unequally distributed between the treated and the untreated - in the stratum $L=1$, there is lower proportion of treated ones with $S=1$ and higher proportion of untreated ones with $S=1$, then $Y^a \perp A | L=1$ does not hold.
To restate my question: let's denote
$Y^a \perp A | L$. In more detail, $P(Y^a| A = 1, L = 1) = P(Y^a| A = 0, L = 1)$, i.e. $Y^a \perp A | L=1$ for all $a$; and $P(Y^a| A = 1, L = 0) = P(Y^a| A = 0, L = 0)$, i.e. $Y^a \perp A | L=0$ for all $a$
$\forall T\ne L, P(T=1|A=1, L=1) = P(T=1|A=0, L=1)$. In plain English, this says for all risk factor (predictor of the outcome) other than $L$ such that within stratum $L=1$, the distribution of $T$ is the same between treated and untreated; similar statement holds within stratum $L=0$.
I want to ask is this correct - 1. $\iff$ 2. $\iff$ the concept of conditional exchangeability the author introduced.
PS. If you have suggestion on the title of this problem, please advice. Thanks!
Here is my answer: the treatment was randomly assigned to individuals in the group $L=0$ is equivalent with $Y^a \perp A | L=0$; similarly, the treatment was randomly assigned to individuals in the group $L=1$ is equivalent with $Y^a \perp A | L=1$. Therefore, 1. and 2. are if and only if relationship.
Suppose $S=1$ represents the person is a smoker, and $S=0$ is a nonsmoker. Within the group $L=0$, if we look at the treated people, and found there are 10% smokers, while there are 80% smoker among untreated people, then in this case 2. is violated.
Note $S$ is a risk factor, i.e. some predictor for the outcome. To that end, had the group of people who are $L=0$, $A=1$ not received any treatment, their risk of death would be differently from the one calculated from the group of people who actually did not receive any treatment; similarly, had the group of people who are $L=0$, $A=0$ received treatment, their risk of death would be differently from the one calculated from the group of people who actually did receive treatment. We can write mathematically that
$$P(Y^a | A=1, L=0) \ne P(Y^a | A=1, L=0),$$
which is exactly saying that $Y^a \not\perp A | L=0$.
One can argue the other direction using similar logic.