Day 55: The Rules of do-Calculus · Math for Causal Inference

Theory

Modified graphs

For sets of nodes $\mathbf X, \mathbf Z$ in $G$ :

$G_{\overline{\mathbf X}}$ : delete all arrows into $\mathbf X$ (the graph after intervening on $\mathbf X$ );
$G_{\underline{\mathbf Z}}$ : delete all arrows out of $\mathbf Z$ (only back-door paths from $\mathbf Z$ remain);
$G_{\overline{\mathbf X}\underline{\mathbf Z}}$ : both.

The three rules

For disjoint sets $\mathbf X, \mathbf Y, \mathbf Z, \mathbf W$ :

Rule 1 (insert or delete observations).

P(\mathbf y \mid \operatorname{do}(\mathbf x), \mathbf z, \mathbf w) = P(\mathbf y \mid \operatorname{do}(\mathbf x), \mathbf w) \quad \text{if } (\mathbf Y \perp\!\!\!\perp \mathbf Z \mid \mathbf X, \mathbf W)_{G_{\overline{\mathbf X}}}.

Rule 2 (exchange action and observation).

P(\mathbf y \mid \operatorname{do}(\mathbf x), \operatorname{do}(\mathbf z), \mathbf w) = P(\mathbf y \mid \operatorname{do}(\mathbf x), \mathbf z, \mathbf w) \quad \text{if } (\mathbf Y \perp\!\!\!\perp \mathbf Z \mid \mathbf X, \mathbf W)_{G_{\overline{\mathbf X}\underline{\mathbf Z}}}.

Rule 3 (insert or delete actions).

P(\mathbf y \mid \operatorname{do}(\mathbf x), \operatorname{do}(\mathbf z), \mathbf w) = P(\mathbf y \mid \operatorname{do}(\mathbf x), \mathbf w) \quad \text{if } (\mathbf Y \perp\!\!\!\perp \mathbf Z \mid \mathbf X, \mathbf W)_{G_{\overline{\mathbf X}\overline{\mathbf Z(\mathbf W)}}},

where $\mathbf Z(\mathbf W)$ is the set of $\mathbf Z$ -nodes that are not ancestors of any $\mathbf W$ -node in $G_{\overline{\mathbf X}}$ .

Reading the rules

Rule 1 is d-separation applied in the post-intervention world.
Rule 2 says: if $\mathbf Z$ has no open back-door paths to $\mathbf Y$ (given $\mathbf X, \mathbf W$ ), seeing $\mathbf z$ is as good as setting it. It generalises the back-door criterion.
Rule 3 says: if $\mathbf Z$ cannot reach $\mathbf Y$ through any causal path (given the rest), setting it changes nothing. The $\mathbf Z(\mathbf W)$ subtlety handles conditioning on descendants of $\mathbf Z$ .

Identification as derivation

A causal query $P(\mathbf y \mid \operatorname{do}(\mathbf x))$ is identified if the rules, plus ordinary probability, transform it into an expression with no $\operatorname{do}$ . The rules are complete: if no derivation exists, the effect is not identified from observational data in that graph (Shpitser & Pearl, 2006). Day 56 gives an algorithm that finds the derivation automatically.

Standard moves

Condition and sum: $P(y \mid \operatorname{do}(x)) = \sum_z P(y \mid \operatorname{do}(x), z)P(z \mid \operatorname{do}(x))$ .
Remove the $\operatorname{do}$ from $P(z \mid \operatorname{do}(x))$ with Rule 3 (if $Z$ is not a descendant) or Rule 2.
Turn $\operatorname{do}(x)$ into $x$ inside $P(y \mid \operatorname{do}(x), z)$ with Rule 2.

Part 1 · Warm-up

Exercises 1 and 4 use $G$ : $Z \to X \to Y$ , $Z \to Y$ , $X \to W$ .

Exercise 1 · Modified graphs

Write the edges of $G_{\overline X}$ , $G_{\underline X}$ and $G_{\overline Z}$ .

Show solution

Step 1. $G_{\overline X}$ : remove $Z \to X$ . Left: $X \to Y$ , $Z \to Y$ , $X \to W$ .

Step 2. $G_{\underline X}$ : remove $X \to Y$ , $X \to W$ . Left: $Z \to X$ , $Z \to Y$ .

Step 3. $G_{\overline Z}$ : $Z$ has no incoming arrows, so $G_{\overline Z} = G$ .

Exercise 2 · Randomised treatment

In $X \to Y$ with $X$ randomised (no parents), use Rule 2 to show $P(y \mid \operatorname{do}(x)) = P(y \mid x)$ .

Show solution

Step 1. Rule 2 with empty outer $\mathbf X$ and $\mathbf W$ , and $\mathbf Z = \{X\}$ : condition $(Y \perp\!\!\!\perp X)_{G_{\underline X}}$ .

Step 2. $G_{\underline X}$ removes $X \to Y$ : no edges remain, so $X$ and $Y$ are d-separated.

Answer: Rule 2 applies: $P(y \mid \operatorname{do}(x)) = P(y \mid x)$ .

Exercise 3 · Actions that cannot matter

If $X$ is not an ancestor of $Y$ , show $P(y \mid \operatorname{do}(x)) = P(y)$ with Rule 3.

Show solution

Step 1. Rule 3 with empty $\mathbf W$ : condition $(Y \perp\!\!\!\perp X)_{G_{\overline X}}$ .

Step 2. In $G_{\overline X}$ , $X$ has no incoming arrows; any path from $X$ starts $X \to \cdots$ .

Step 3. Since $X$ is not an ancestor of $Y$ , such a path must contain a collider (it cannot be directed all the way to $Y$ ); unconditioned, it is blocked.

Answer: d-separated, so $P(y \mid \operatorname{do}(x)) = P(y)$ .

Exercise 4 · Rule 1 in action

In $G$ , is $P(y \mid \operatorname{do}(x), w) = P(y \mid \operatorname{do}(x))$ ?

Show solution

Step 1. Rule 1 condition: $(Y \perp\!\!\!\perp W \mid X)_{G_{\overline X}}$ .

Step 2. In $G_{\overline X}$ , paths $Y$ – $W$ : $Y \leftarrow X \to W$ (fork at $X$ , conditioned: blocked).

Step 3. $Y \leftarrow Z \to Y$ is not a path to $W$ ; and $Z \to X$ is gone, so no other path exists.

Answer: yes; once $X$ is set, $W$ (a side effect of $X$ ) carries no information about $Y$ .

Exercise 5 · Name the rule

Which rule justifies each step? (a) $P(z \mid \operatorname{do}(x)) = P(z)$ for a pre-treatment $Z$ . (b) $P(y \mid \operatorname{do}(x), z) = P(y \mid x, z)$ when $Z$ blocks the back-door paths. (c) Dropping a measured side effect of treatment from the conditioning set.

Show solution

Step 1. (a) Rule 3 (deleting an action that cannot affect $Z$ ).

Step 2. (b) Rule 2 (exchanging the action $\operatorname{do}(x)$ for the observation $x$ ).

Step 3. (c) Rule 1 (deleting an observation).

Part 2 · Proofs

Exercise 6 · The back-door formula by do-calculus

Let $\mathbf Z$ satisfy the back-door criterion for $(X, Y)$ . Derive $P(y \mid \operatorname{do}(x)) = \sum_{\mathbf z} P(y \mid x, \mathbf z)P(\mathbf z)$ .

Show solution

Step 1. Condition and sum: $P(y \mid \operatorname{do}(x)) = \sum_{\mathbf z} P(y \mid \operatorname{do}(x), \mathbf z)P(\mathbf z \mid \operatorname{do}(x))$ .

Step 2. Rule 3 for $P(\mathbf z \mid \operatorname{do}(x))$ : need $(\mathbf Z \perp\!\!\!\perp X)_{G_{\overline X}}$ . In $G_{\overline X}$ , paths from $X$ leave through its children; since no $\mathbf Z$ node is a descendant of $X$ , each such path meets a collider and is blocked. So $P(\mathbf z \mid \operatorname{do}(x)) = P(\mathbf z)$ .

Step 3. Rule 2 for $P(y \mid \operatorname{do}(x), \mathbf z)$ : need $(Y \perp\!\!\!\perp X \mid \mathbf Z)_{G_{\underline X}}$ . In $G_{\underline X}$ only back-door paths from $X$ remain, and $\mathbf Z$ blocks them all. So $P(y \mid \operatorname{do}(x), \mathbf z) = P(y \mid x, \mathbf z)$ .

Step 4. Substitute. $\blacksquare$ (This also proves fact (b) assumed on Day 52.)

Exercise 7 · Front door, first piece

In the front-door graph $U \to X$ , $U \to Y$ , $X \to M \to Y$ ( $U$ unobserved), show $P(m \mid \operatorname{do}(x)) = P(m \mid x)$ .

Show solution

Step 1. Rule 2 condition: $(M \perp\!\!\!\perp X)_{G_{\underline X}}$ .

Step 2. $G_{\underline X}$ removes $X \to M$ . Remaining paths $X$ – $M$ : $X \leftarrow U \to Y \leftarrow M$ , with a collider at $Y$ : blocked.

Answer: Rule 2 applies. $\blacksquare$

Exercise 8 · Front door, second piece

In the same graph, show $P(y \mid \operatorname{do}(m)) = \sum_x P(y \mid m, x)P(x)$ .

Show solution

Step 1. Condition and sum: $P(y \mid \operatorname{do}(m)) = \sum_x P(y \mid \operatorname{do}(m), x)P(x \mid \operatorname{do}(m))$ .

Step 2. Rule 2 on the first factor: $(Y \perp\!\!\!\perp M \mid X)_{G_{\underline M}}$ . Removing $M \to Y$ leaves the path $M \leftarrow X \leftarrow U \to Y$ , blocked by $X$ (chain). So $P(y \mid \operatorname{do}(m), x) = P(y \mid m, x)$ .

Step 3. Rule 3 on the second factor: $(X \perp\!\!\!\perp M)_{G_{\overline M}}$ . Removing $X \to M$ leaves $M \to Y \leftarrow U \to X$ with a collider at $Y$ : blocked. So $P(x \mid \operatorname{do}(m)) = P(x)$ .

Step 4. Substitute. $\blacksquare$

Exercise 9 · Front door, assembled

Complete the derivation of $P(y \mid \operatorname{do}(x)) = \sum_m P(m \mid x)\sum_{x'} P(y \mid m, x')P(x')$ .

Show solution

Step 1. Condition and sum: $P(y \mid \operatorname{do}(x)) = \sum_m P(y \mid \operatorname{do}(x), m)P(m \mid \operatorname{do}(x))$ .

Step 2. The second factor is $P(m \mid x)$ (Exercise 7).

Step 3. Rule 2 on the first: $(Y \perp\!\!\!\perp M \mid X)_{G_{\overline X}\underline M}$ . Removing arrows into $X$ and out of $M$ leaves no path $M$ – $Y$ except through $X$ , which is conditioned. So $P(y \mid \operatorname{do}(x), m) = P(y \mid \operatorname{do}(x), \operatorname{do}(m))$ .

Step 4. Rule 3: $(Y \perp\!\!\!\perp X \mid M)_{G_{\overline X\,\overline M}}$ . With arrows into $X$ and $M$ removed, $X$ is isolated. So $P(y \mid \operatorname{do}(x), \operatorname{do}(m)) = P(y \mid \operatorname{do}(m))$ .

Step 5. By Exercise 8, $P(y \mid \operatorname{do}(m)) = \sum_{x'} P(y \mid m, x')P(x')$ . Substitute. $\blacksquare$

Exercise 10 · Two models, one distribution

In the "bow" graph $X \to Y$ with $X \leftrightarrow Y$ , show that $P(y \mid \operatorname{do}(x))$ is not identified by exhibiting two SCMs with the same $P(x, y)$ and different interventional distributions.

Show solution

Step 1. Model A: $U \sim \text{Bernoulli}(0.5)$ , $X := U$ , $Y := U$ . $X$ has no effect on $Y$ .

Step 2. Model B: $X := U \sim \text{Bernoulli}(0.5)$ , $Y := X$ . Effect of 1.

Step 3. Both give $P(X = Y = 1) = P(X = Y = 0) = 0.5$ : identical observational distributions.

Step 4. $P(Y = 1 \mid \operatorname{do}(X = 1))$ is 0.5 in Model A and 1 in Model B.

Answer: data cannot distinguish the models, so no formula can recover the effect: it is not identified. $\blacksquare$

Part 3 · Applied

Exercise 11 · A tax credit and labour supply

An earned-income tax credit $X$ (eligibility depends on number of children $C$ ) affects hours worked $Y$ ; children also affect hours directly. Graph: $C \to X$ , $C \to Y$ , $X \to Y$ . Derive $P(y \mid \operatorname{do}(x))$ with the rules, naming each.

Show solution

Step 1. Condition and sum: $P(y \mid \operatorname{do}(x)) = \sum_c P(y \mid \operatorname{do}(x), c)P(c \mid \operatorname{do}(x))$ .

Step 2. Rule 3: in $G_{\overline X}$ the arrow $C \to X$ is gone; paths $X$ – $C$ : $X \to Y \leftarrow C$ , collider: blocked. So $P(c \mid \operatorname{do}(x)) = P(c)$ .

Step 3. Rule 2: in $G_{\underline X}$ , paths $X$ – $Y$ : $X \leftarrow C \to Y$ , blocked by $C$ . So $P(y \mid \operatorname{do}(x), c) = P(y \mid x, c)$ .

Answer: $P(y \mid \operatorname{do}(x)) = \sum_c P(y \mid x, c)P(c)$ : compare hours of eligible and ineligible families with the same number of children, and average over the distribution of family size.

Exercise 12 · Effects within subgroups

In Exercise 11's graph plus a pre-treatment region indicator $R$ ( $R \to C$ , $R \to Y$ ), derive the region-specific effect $P(y \mid \operatorname{do}(x), r)$ .

Show solution

Step 1. Condition and sum within region: $P(y \mid \operatorname{do}(x), r) = \sum_c P(y \mid \operatorname{do}(x), c, r)P(c \mid \operatorname{do}(x), r)$ .

Step 2. Rule 3 (with $\mathbf W = \{R\}$ ): in $G_{\overline X}$ , paths $X$ – $C$ go through the collider $Y$ ( $X \to Y \leftarrow C$ or $X \to Y \leftarrow R \to C$ ), unconditioned: blocked. So $P(c \mid \operatorname{do}(x), r) = P(c \mid r)$ .

Step 3. Rule 2: in $G_{\underline X}$ , back-door paths $X \leftarrow C \to Y$ and $X \leftarrow C \leftarrow R \to Y$ are blocked by $C$ . So $P(y \mid \operatorname{do}(x), c, r) = P(y \mid x, c, r)$ .

Answer: $P(y \mid \operatorname{do}(x), r) = \sum_c P(y \mid x, c, r)P(c \mid r)$ , a conditional (CATE-type) version of the adjustment formula.

Exercise 13 · Two rounds of HIV treatment

A clinic decides treatment $X_1$ at the first visit; CD4 count $L$ is measured; treatment $X_2$ is decided at the second visit based on $L$ ; viral suppression $Y$ is measured. Graph: $X_1 \to L$ , $X_1 \to Y$ , $L \to X_2$ , $X_2 \to Y$ , $U \to L$ , $U \to Y$ ( $U$ = unobserved health status); $X_1$ is randomised. Derive $P(y \mid \operatorname{do}(x_1), \operatorname{do}(x_2))$ .

Show solution

Step 1. Condition on $L$ and sum: $P(y \mid \operatorname{do}(x_1, x_2)) = \sum_l P(y \mid \operatorname{do}(x_1, x_2), l)\,P(l \mid \operatorname{do}(x_1, x_2))$ .

Step 2. Rule 3 removes $\operatorname{do}(x_2)$ from the second factor: $X_2$ is not an ancestor of $L$ . Then Rule 2 turns $\operatorname{do}(x_1)$ into $x_1$ : $X_1$ has no back-door paths. So $P(l \mid \operatorname{do}(x_1, x_2)) = P(l \mid x_1)$ .

Step 3. Rule 2 for $X_2$ in the first factor: in $G_{\overline{X_1}\underline{X_2}}$ , back-door paths from $X_2$ ( $X_2 \leftarrow L \leftarrow U \to Y$ , $X_2 \leftarrow L \leftarrow X_1 \to Y$ ) are blocked by $L$ (and $X_1$ ). So it equals $P(y \mid \operatorname{do}(x_1), x_2, l)$ .

Step 4. Rule 2 for $X_1$ : in $G_{\underline{X_1}}$ , $X_1$ has no remaining edges. So it equals $P(y \mid x_1, x_2, l)$ .

Answer: the g-formula $P(y \mid \operatorname{do}(x_1, x_2)) = \sum_l P(y \mid x_1, l, x_2)\,P(l \mid x_1)$ . Note that regressing $Y$ on $X_1, X_2, L$ would be wrong: $L$ is a mediator for $X_1$ and a collider ( $X_1 \to L \leftarrow U$ ). Day 96 develops this.

Exercise 14 · Ice cream and drowning

Hot weather $H$ raises ice-cream sales $I$ and swimming, hence drownings $D$ : $H \to I$ , $H \to D$ . A mayor proposes banning ice cream to cut drownings. Use Rule 3 to evaluate the proposal, and contrast with the observational association.

Show solution

Step 1. Rule 3: $(D \perp\!\!\!\perp I)_{G_{\overline I}}$ ? Removing $H \to I$ isolates $I$ : d-separated.

Step 2. So $P(d \mid \operatorname{do}(i)) = P(d)$ : setting ice-cream sales to zero changes nothing.

Step 3. Observationally, $P(d \mid i)$ rises with $i$ through the fork $I \leftarrow H \to D$ .

Answer: the ban would not reduce drownings; the association is pure confounding by weather.

Exercise 15 · A non-identified effect, numerically

A health economist has data only on insurance status $X$ and medical spending $Y$ (both binary), with unobserved health $U$ affecting both. Build two SCMs matching $P(X = 1, Y = 1) = 0.3$ , $P(X = 1, Y = 0) = 0.2$ , $P(X = 0, Y = 1) = 0.1$ , $P(X = 0, Y = 0) = 0.4$ , one where insurance has no effect and one where it does.

Show solution

Step 1. Model A (no effect): let $U$ take four values that encode $(X, Y)$ directly, with $P(U = (x, y))$ equal to the table; set $X := U_1$ , $Y := U_2$ . The table is reproduced and $Y$ does not depend on $X$ : $P(Y = 1 \mid \operatorname{do}(x)) = P(Y = 1) = 0.4$ for both $x$ .

Step 2. Model B (no confounding): $X \sim \text{Bernoulli}(0.5)$ and $Y := \mathbf 1[V < p_X]$ with $V$ uniform, $p_1 = 0.3/0.5 = 0.6$ , $p_0 = 0.1/0.5 = 0.2$ . Same table; causal effect $0.6 - 0.2 = 0.4$ .

Answer: both models fit the data perfectly; the effect could be 0 or 0.4. Without an admissible set, a front door or an instrument, the data are silent.

Verify in Python

import networkx as nx

dsep = getattr(nx, "is_d_separator", None) or nx.d_separated   # name changed in networkx 3.3

def cut(G, into=(), out_of=()):
    """Remove arrows into the nodes in `into` and out of the nodes in `out_of`."""
    H = G.copy()
    H.remove_edges_from([e for e in G.edges if e[1] in into or e[0] in out_of])
    return H

# Front-door graph (U explicit)
F = nx.DiGraph([("U", "X"), ("U", "Y"), ("X", "M"), ("M", "Y")])
print("Ex 7  rule 2:", dsep(cut(F, out_of={"X"}), {"M"}, {"X"}, set()))
print("Ex 8  rule 2:", dsep(cut(F, out_of={"M"}), {"Y"}, {"M"}, {"X"}), " rule 3:", dsep(cut(F, into={"M"}), {"X"}, {"M"}, set()))
print("Ex 9  rule 2:", dsep(cut(F, into={"X"}, out_of={"M"}), {"Y"}, {"M"}, {"X"}),
      " rule 3:", dsep(cut(F, into={"X", "M"}), {"Y"}, {"X"}, {"M"}))

# Exercise 13: two rounds of treatment
H = nx.DiGraph([("X1", "L"), ("X1", "Y"), ("L", "X2"), ("X2", "Y"), ("U", "L"), ("U", "Y")])
print("Ex 13 rule 3 (X2 on L):", dsep(cut(H, into={"X2"}), {"L"}, {"X2"}, set()),
      " rule 2 (X1 on L):", dsep(cut(H, out_of={"X1"}), {"L"}, {"X1"}, set()))
print("      rule 2 (X2 on Y):", dsep(cut(H, into={"X1"}, out_of={"X2"}), {"Y"}, {"X2"}, {"X1", "L"}),
      " rule 2 (X1 on Y):", dsep(cut(H, out_of={"X1"}), {"Y"}, {"X1"}, {"X2", "L"}))
print("      naive regression on L is invalid for X1: L is a descendant of X1:", "L" in nx.descendants(H, "X1"))

# Exercise 15: two SCMs, same data, different effects
table = {(1, 1): .3, (1, 0): .2, (0, 1): .1, (0, 0): .4}
modelA = {x: sum(v for (xx, y), v in table.items() if y == 1) for x in (0, 1)}            # Y ignores X
pX = {x: sum(v for (xx, y), v in table.items() if xx == x) for x in (0, 1)}
modelB = {x: table[x, 1] / pX[x] for x in (0, 1)}
print("P(Y=1|do(x)) model A:", modelA, " model B:", {k: round(v, 2) for k, v in modelB.items()})

Key takeaways

Rule 1 adds or removes observations, Rule 2 swaps actions and observations, Rule 3 adds or removes actions; each needs a d-separation in a modified graph.
The back-door and front-door formulas, and the g-formula for sequential treatments, are short derivations.
The rules are complete: if no derivation exists, two models with the same data but different effects can be built.

Day 55. The Rules of do-Calculus

Theory

Modified graphs

The three rules

Reading the rules

Identification as derivation

Standard moves

Part 1 · Warm-up

Exercise 1 · Modified graphs

Exercise 2 · Randomised treatment

Exercise 3 · Actions that cannot matter

Exercise 4 · Rule 1 in action

Exercise 5 · Name the rule

Part 2 · Proofs

Exercise 6 · The back-door formula by do-calculus

Exercise 7 · Front door, first piece

Exercise 8 · Front door, second piece

Exercise 9 · Front door, assembled

Exercise 10 · Two models, one distribution

Part 3 · Applied

Exercise 11 · A tax credit and labour supply

Exercise 12 · Effects within subgroups

Exercise 13 · Two rounds of HIV treatment

Exercise 14 · Ice cream and drowning

Exercise 15 · A non-identified effect, numerically

Verify in Python

Key takeaways

Simulation Lab: do-Calculus, Step by Step