GuidesCheck whether an effect is identifiable

Check whether an effect is identifiable

Ask whether a causal graph lets you recover an effect from observational data, and get the formula to estimate if it does. You'll check five graphs, including two from discovery.

Run the program#

Identification needs no data. It reads only the graph. U marks a common cause you never measured: pass it in latent=.

from ergodic import identify_effect
from ergodic.graph import cpdag, dag, mixed

# 1. Four single graphs in which X may cause Y.
worlds = {
    "measured confounder": dag(["Z -> X", "Z -> Y", "X -> Y"]),
    "hidden confounder, mediator": dag(["U -> X", "U -> Y", "X -> M", "M -> Y"], latent=["U"]),
    "hidden confounder, instrument": dag(["Z -> X", "U -> X", "U -> Y", "X -> Y"], latent=["U"]),
    "hidden confounder only": dag(["U -> X", "U -> Y", "X -> Y"], latent=["U"]),
}
for name, graph in worlds.items():
    result = identify_effect(graph, "X", "Y")
    print(f"{name}: {result}")

# 2. The hedge explains why the last one fails.
print(result.hedge)

# 3. Two equivalence classes, as discovery returns them.
for graph in [cpdag(["Z --- X", "X --> Y", "W --> Y"]), cpdag(["Z --- X", "X --- Y"])]:
    result = identify_effect(graph, "X", "Y")
    print(bool(result), result.strategy, sorted(result.adjustment_set or []))

# 4. A graph without a verified face is refused, with advice.
try:
    identify_effect(mixed(["Z -> X", "Z -> Y", "X -> Y"]), "X", "Y")
except ValueError as error:
    print(error)

measured confounder: Identification(adjustment: P(Y | do(X)) = sum_z P(Y | X, z) P(z), with z = {Z})
hidden confounder, mediator: Identification(frontdoor: P(Y | do(X)) = sum_m P(m | X) sum_x' P(Y | m, x') P(x'), with m = {M})
hidden confounder, instrument: Identification(iv: identified through instrument(s) Z under a linear (parametric) model; needs an IV estimator)
hidden confounder only: Identification(not identifiable, ['X'] -> ['Y'])
not identifiable: a hedge formed by ['X', 'Y'] and the c-component ['Y']
True adjustment ['W', 'Z']
False none []
identification needs a verified graph face (dag, admg, mag, cpdag, or pag), not 'mixed'; every edge is directed and there is no directed cycle. If it came from a search that assumes no hidden confounding (PC or GES with domain knowledge orienting every edge, say) and you read it as one causal world, re-face it with graph.as_kind('dag'); if it came from FCI or RFCI (or otherwise means ancestry, where a directed edge can hide confounding), re-face it with graph.as_kind('mag') instead, which identification reads as an equivalence class

• Step 1. Each world gets a different strategy. With Z measured, adjust for it. With U hidden but a mediator M measured, use the front-door formula. With an instrument Z, the effect is identified under a linear model only. With nothing else, it isn't identifiable.
• Step 2. The hedge names the structure that blocks identification: here X and Y share a hidden cause, and X -> Y is the only path.
• Step 3. On a CPDAG, a success holds in every graph of the class. The first class orients X --> Y and every member agrees on {W, Z}. In the second, X --- Y may point either way, so the members disagree and the answer is no.
• Step 4. A mixed graph has no verified meaning. The error says which kind it fits and how to re-face it.

Read the result#

identify_effect(graph, treatment, outcome) returns an Identification. It is truthy when the effect is identifiable.

The strategies are tried in order. First, if no directed path runs from treatment to outcome, the effect is zero: "null". Then adjustment, the front-door and instruments. On a single graph, the general do-calculus algorithm (Shpitser and Pearl, 2006) runs last. It returns a formula or a hedge. When it recognizes the formula as adjustment or the front door, the strategy says so; otherwise it is "id".

What each graph kind allows#

A* and HillClimb also return a dag, but it is one member of an equivalence class, picked by tie-breaking among DAGs with the same score. Identify on cpdag_of_dag(graph) instead, so the answer doesn't rest on an arbitrary direction.

A graph that marks selection nodes (selection=) raises on every kind. The criteria assume the data were not selected on any node, and an effect generally isn't identifiable from a selected sample: with X -> S <- Y and S selected, X and Y are dependent in the sample even though X has no effect on Y.

On a class, the front-door and instruments need a single graph. To see each member's answer, use ida, described in Find adjustment sets, front-door sets and instruments.

Next steps#