ID returns incorrect estimand for napkin example

[djinnome]

In the README.md,

from y0.dsl import P, X, Y
from y0.examples import napkin
from y0.algorithm.identify import identify_outcomes

estimand = identify_outcomes(napkin, treatments=X, outcomes=Y)
assert estimand == P(Y | X)

However, the correct estimand for the napkin problem should be:

$$\frac{\sum_{Z_2} P(X, Y\mid Z_1, Z_2)P(Z_2)}{\sum_{Z_2} P(X\mid Z_1, Z_2)P(Z_2)}$$

[cthoyt]

I checked - this isn't just an issue in the README, but this is actually what the code returns

[djinnome]

@callahanr @ShantMahserejian @emilydogbatsePNNL2025 and I developed unit tests for the napkin problem in test_id_napkin and test_counterfactual_transportability.py. It turns out that transport_unconditional_counterfactual_query produces the correct estimand for the napkin problem. Emily will update the cyclic Id algorithm to generalize identify_districts instead of calling identify_outcomes, and I will figure out why identify_outcomes produces the wrong answer.

[djinnome]

@emilydogbatsePNNL2025 and @djinnome looked into this today, and we have determined that the Shpitser ID algorithm is incapable of producing the correct estimand for the napkin problem.

Proof: The ID algorithm lacks a step for generating fractions in the estimand. Only the IDC algorithm generates fractions in the estimand. Therefore, the correct estimand for the napkin problem:

$$\frac{\sum_{Z_2} P(X, Y\mid Z_1, Z_2)P(Z_2)}{\sum_{Z_2} P(X\mid Z_1,Z_2)P(Z_2)}$$

is not possible to generate from the ID algorithm.

Question: Is $P(Y|X)$ the same as $\frac{\sum_{Z_2} P(X, Y\mid Z_1, Z_2)P(Z_2)}{\sum_{Z_2} P(X\mid Z_1,Z_2)P(Z_2)}$?

Answer: No.
Proof:

$$P(Y|X) = \frac{P(X, Y)}{P(X)} = \frac{\sum_{Z_1, Z_2} P(X, Y, Z_1, Z_2)}{\sum_{Z_1, Z_2} P(X, Z_1, Z_2)} = \frac{\sum_{Z_1, Z_2} P(X, Y\mid Z_1, Z_2)P(Z_1\mid Z_2)P(Z_2)}{\sum_{Z_1,Z_2} P(X\mid Z_1, Z_2)P(Z_1 \mid Z_2)P(Z_2)} = \frac{\sum_{Z_1, Z_2} P(X, Y\mid Z_1, Z_2)P(Z_2)}{\sum_{Z_1,Z_2} P(X\mid Z_1, Z_2)P(Z_2)}$$

$$ \neq \frac{\sum_{Z_2} P(X, Y\mid Z_1, Z_2)P(Z_2)}{\sum_{Z_2} P(X\mid Z_1,Z_2)P(Z_2)}$$

Note, however, that the Tian and Pearl algorithm is capable of generating the correct estimand for the napkin problem.

Furthermore, Bareinboim solves the napkin problem and provides a derivation in terms of the do calculus rules:

Figure 7.5 - Napkin derivation steps.pdf

[djinnome]

Well, apparently $P(Y|do(X)) = P(Y|X)$ for all values of $Z_1$, so it doesn't matter that we didn't marginalize out $Z_1$ above.

https://github.com/rrtucci/napkin-do-calc?tab=readme-ov-file

http://stats.stackexchange.com/questions/514615/do-calculus-for-causal-diagram-7-5-from-the-book-of-why-napkin-problem

[djinnome]

@ShantMahserejian @njmmerrill and @djinnome proved that both solutions to the napkin problem are correct.