Interventions and identification: the do-operator, backdoor and frontdoor adjustment, do-calculus

Before this lesson: lesson 03 (d-separation, backdoor paths, colliders) and lesson 02 (potential outcomes, exchangeability). From probability: the law of total probability with conditioning (probability 04). Time: about 90 minutes, plus the notebook.

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Why this matters #

• Decisions need "doing". A churn model predicts who leaves; a retention team wants to know who leaves less when given a discount. The first question is $P(y \mid x)$, the second is $P(y \mid \mathrm{do}(t), x)$. Most ML data are observational, so we need the conditions under which the second can be computed from the first kind of data.
• Choosing features for adjustment. Lessons 05 and 08 estimate effects with regression, matching, weighting and double ML. All of them adjust for a set of covariates, and all of them fail if the set is wrong. This lesson tells you which set is right.
• Bad controls are common. Adjusting for a post-treatment variable (a mediator), for a variable caused by the outcome, or for a pre-treatment collider gives biased answers even with infinite data.
• Identification before estimation. Libraries such as DoWhy and the ID algorithm in causal software first check identification with the rules of this lesson, then estimate.

1. Predict first #

2. The picture #

2.1 Seeing versus doing #

Take three binary variables: severity $X$, drug $T$, recovery $Y$, with the graph $X \to T$, $X \to Y$, $T \to Y$. Severe patients ($X = 1$) get the drug more often, and severity lowers recovery.

• Seeing $T = 1$: we look at the patients who took the drug. Many of them are severe, because severity caused the treatment. Their recovery rate mixes the effect of the drug with the effect of severity.
• Doing $T = 1$: we give the drug to everybody. Now severity no longer decides who is treated. In the graph, the arrow $X \to T$ is cut. The share of severe patients is the same as in the whole population.

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2.2 From surgery to a formula #

The SCM of the figure factorizes as $p(x, t, y) = p(x)\,p(t \mid x)\,p(y \mid t, x)$. The surgery replaces the factor of $T$ by "$T = t$ with probability 1" and keeps the other factors:

Sum over $x$ and compare with ordinary conditioning (law of total probability):

The only difference is the weight of each severity group: the population share $P(x)$ for doing, the share among the treated $P(x \mid t)$ for seeing. If $T$ did not depend on $X$, the two would be equal.

Numbers. In the figure $P(Y = 1 \mid T, X) = 0.6 + 0.2T - 0.4X$, $P(X = 1) = 0.5$, and with strength $0.6$, $P(X = 1 \mid T = 1) = 0.8$ and $P(X = 1 \mid T = 0) = 0.2$.

Doing uses the weights $P(x) = (0.5, 0.5)$:

Seeing uses the weights $P(x \mid t)$:

In each sum, the first term is the mild group ($X = 0$) and the second the severe group. Doing gives the effect $0.60 - 0.40 = 0.20$; seeing gives $0.48 - 0.52 = -0.04$.

2.3 Which variables to adjust for #

The formula of section 2.2 adjusts for $X$. In a bigger graph, which set $Z$ should we use? Lesson 03 gives the tool. Association between $T$ and $Y$ flows along two kinds of paths:

• causal paths $T \to \dots \to Y$: they carry the effect; we must keep them open;
• backdoor paths $T \leftarrow \dots\, Y$: they carry association that is not the effect; we must block them.

A good set $Z$ blocks every backdoor path, does not block causal paths, and opens no new path. The backdoor criterion of section 3.4 says exactly this.

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3. Definitions and main results #

3.1 Interventions #

Definition (intervention). Let $\mathcal M$ be an SCM with graph $G$. The intervention $\mathrm{do}(T = t)$ gives a new SCM $\mathcal M_t$: the assignment $T := f_T(\mathrm{pa}_T, U_T)$ is replaced by $T := t$; all other assignments stay the same. Its graph is $G_{\overline T}$, the graph $G$ with all arrows into $T$ removed. The distribution of $\mathcal M_t$ is the interventional distribution $P(\cdot \mid \mathrm{do}(T = t))$, short $P(\cdot \mid \mathrm{do}(t))$.

Two assumptions hide in this definition. Modularity: changing one assignment does not change the others (each mechanism is autonomous). Well-defined intervention: "set $T$ to $t$" is something that could be done (lesson 02 called this consistency).

Theorem (truncated factorization). If $p(v) = \prod_j p(v_j \mid \mathrm{pa}_j)$, then for every $v$ with $T = t$,

and $p(v \mid \mathrm{do}(t)) = 0$ for $v$ with $T \ne t$.

Proof. $\mathcal M_t$ has the same assignments, and so the same conditional distributions $p(v_j \mid \mathrm{pa}_j)$, for all $V_j \ne T$; its graph is $G_{\overline T}$, in which $T$ has no parents and the constant value $t$. Apply the Markov factorization (lesson 03) to $\mathcal M_t$. The factor of $T$ is 1 at $T = t$ and 0 elsewhere. $\blacksquare$

3.2 The link to potential outcomes #

The potential outcome $Y(t)$ of lesson 02 is the value of $Y$ in the modified model $\mathcal M_t$, with the same noises $U$. So

The two languages describe the same objects. Graphs make the assumptions visible; potential outcomes make the estimands precise. A useful bridge (Pearl 2009, ch. 11): if a set $Z$ satisfies the backdoor criterion of section 3.4, then conditional exchangeability holds, $Y(t) \perp\!\!\!\perp T \mid Z$. Intuition: $Y(t)$ is a function of the noises of $Y$ and of the non-descendants of $T$ that matter; once $Z$ blocks every backdoor path, nothing links those noises to the assignment of $T$.

3.3 Identification #

Definition (identified). A causal quantity, such as $P(y \mid \mathrm{do}(t))$, is identified from the graph $G$ if it is the same for every SCM with graph $G$ that gives the same observational distribution $P(V)$ (with positive probabilities). Then it can be written as a formula in $P(V)$ only.

Identification is about infinite data. It asks "is the answer in the distribution at all?", not "how do I estimate it well from $n$ points?". Section 3.8 returns to estimation.

Positivity. The formulas below divide by, or condition on, probabilities like $P(t \mid z)$. They need $P(T = t \mid Z = z) > 0$ for every $z$ with $P(z) > 0$: in every group, some units get each treatment value. Positivity can be checked with data (lesson 05 shows how).

3.4 The backdoor criterion and the adjustment formula #

Definition (backdoor criterion; Pearl 1993). A set $Z$ of observed variables satisfies the backdoor criterion relative to $(T, Y)$ if

1. no node in $Z$ is a descendant of $T$; and
2. $Z$ blocks every path between $T$ and $Y$ that starts with an arrow into $T$ (every backdoor path).

Theorem (backdoor adjustment). If $Z$ satisfies the backdoor criterion and positivity holds, then

Why each condition is needed:

• Condition 2 removes confounding: after conditioning on $Z$, the only open paths between $T$ and $Y$ are causal.
• Condition 1 stops two errors: a mediator in $Z$ blocks part of the effect, and a descendant of $T$ that is also a collider (or a descendant of $Y$) opens a new path.

Proof for $Z = \mathrm{pa}_T$, the parents of $T$.

1. Truncated factorization: $p(v \mid \mathrm{do}(t)) = \prod_{j:\, V_j \ne T} p(v_j \mid \mathrm{pa}_j)$ at $T = t$.
2. Multiply and divide by the missing factor $p(t \mid \mathrm{pa}_T)$ (positive by positivity): $p(v \mid \mathrm{do}(t)) = p(v) / p(t \mid \mathrm{pa}_T)$, where $v$ includes $T = t$.
3. Sum over all variables except $Y$, $T$ (fixed at $t$) and $\mathrm{pa}_T$. The denominator depends only on $\mathrm{pa}_T$, so it stays outside the sum, and the sum of $p(v)$ gives $p(y, t, \mathrm{pa}_T)$. Then sum over $\mathrm{pa}_T$: $P(y \mid \mathrm{do}(t)) = \sum_{\mathrm{pa}_T} p(y, t, \mathrm{pa}_T) / p(t \mid \mathrm{pa}_T)$.
4. Write $p(y, t, \mathrm{pa}_T) = p(y \mid t, \mathrm{pa}_T)\, p(t \mid \mathrm{pa}_T)\, p(\mathrm{pa}_T)$ and cancel: $P(y \mid \mathrm{do}(t)) = \sum_{\mathrm{pa}_T} p(y \mid t, \mathrm{pa}_T)\, p(\mathrm{pa}_T)$. $\blacksquare$

The parents of $T$ always satisfy the backdoor criterion (every backdoor path starts $T \leftarrow P$ with a parent $P$ as a non-collider). But parents are often unobserved, and other sets can work.

Proof for a general backdoor set $Z$

This proof uses rules 2 and 3 of do-calculus (section 3.7).

1. Law of total probability in the intervened model: $P(y \mid \mathrm{do}(t)) = \sum_z P(y \mid \mathrm{do}(t), z)\, P(z \mid \mathrm{do}(t))$.
2. $P(z \mid \mathrm{do}(t)) = P(z)$ by rule 3: in $G_{\overline T}$, $T$ has no parents, so every path from $T$ starts $T \to \dots$. Such a path reaches a non-descendant of $T$ only after it turns against an arrow, at a collider. With an empty conditioning set the collider blocks. $Z$ has no descendant of $T$ (condition 1), so $(Z \perp\!\!\!\perp T)_{G_{\overline T}}$.
3. $P(y \mid \mathrm{do}(t), z) = P(y \mid t, z)$ by rule 2: in $G_{\underline T}$ (arrows out of $T$ removed), the paths between $T$ and $Y$ are exactly the backdoor paths of $G$, and $Z$ blocks them (condition 2). (A collider on such a path has the same descendants in $Z$ in both graphs, because a node of $Z$ reached only through $T$ would be a descendant of $T$.) So $(Y \perp\!\!\!\perp T \mid Z)_{G_{\underline T}}$.
4. Put steps 2 and 3 into step 1. $\blacksquare$

Kidney stones again (lesson 01). Stone size $S$ affects the choice of treatment and the success: $S \to T$, $S \to Y$, $T \to Y$. $\{S\}$ satisfies the backdoor criterion. With $P(\text{small}) = 357/700 = 0.51$:

Treatment A is better by about 5 percentage points, although the pooled rates (78.0% vs 82.6%) say the opposite. The adjustment formula is the "standardized rate" of lesson 01, now with a reason.

Several valid sets. Different sets can satisfy the criterion. They all give the same $P(y \mid \mathrm{do}(t))$ in the population, but different variances in a sample. A set that also contains causes of $Y$ gives a more precise estimate; a set that contains strong causes of $T$ gives a less precise one.

3.5 Good and bad controls #

Cinelli, Forney and Pearl (2022) sort candidate controls by their position in the graph. The next figure puts six of them in one linear SCM with a true total effect of 1.

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The table gives the population value of the coefficient of $T$, and its standard error for $n = 500$, for the SCM of the figure:

Adjustment set	Coefficient of $T$ ($n \to \infty$)	SE ($n = 500$)	Verdict
$\{\}$ (naive)	$1.24$	$0.050$	bias: the backdoor path through $X$ is open
$\{X\}$ (confounder)	$1.00$	$0.050$	good: closes the backdoor path
$\{X, P\}$ (cause of $Y$ only)	$1.00$	$0.041$	good: no bias, smaller SE
$\{X, Z\}$ (instrument)	$1.00$	$0.063$	bad for precision: no bias, larger SE
$\{Z\}$	$1.35$	$0.059$	bad: bias grows from $0.24$ to $0.35$ (bias amplification)
$\{X, M\}$ (mediator)	$0.50$	$0.091$	bad: only the direct effect
$\{X, C\}$ (collider of $T$ and $Y$)	$-0.53$	$0.045$	bad: strong bias, even the sign flips
$\{X, Mb\}$ (M-bias node)	$0.89$	$0.049$	bad: opens $T \leftarrow U_1 \to Mb \leftarrow U_2 \to Y$

How to read each row:

• Confounder $X$ ($T \leftarrow X \to Y$): adjust.
• Precision variable $P$ (a cause of $Y$ only): no backdoor path goes through it, so it does not change the bias. It explains part of the noise of $Y$, so the standard error goes down.
• Instrument $Z$ (a cause of $T$ only): it does not change the bias either, but it removes the part of the variation of $T$ that is "as good as random", so the standard error goes up. If a backdoor path stays open (here without $X$), the remaining variation of $T$ is more confounded and the bias grows. Lesson 06 uses $Z$ in a different, good way.
• Mediator $M$ ($T \to M \to Y$): a descendant of $T$. Adjusting for it blocks part of the effect. You get the direct effect, a different estimand (lesson 09).
• Collider $C$ ($T \to C \leftarrow Y$; it is also a descendant of $Y$): adjusting for it opens a path and can destroy the answer. For the SCM $Y = T + 2X + \varepsilon_Y$, $C = T + Y + \varepsilon_C$ with independent Gaussian noises of equal variance, the coefficient of $T$ given $(X, C)$ is exactly 0, because $\mathbb{E}[Y \mid T, X, C] = X + C/2$: given $(T, X, C)$ we know $\varepsilon_Y + \varepsilon_C = C - 2T - 2X$, and we expect half of this sum to be $\varepsilon_Y$.
• M-bias node $Mb$: pre-treatment, but a collider on a backdoor path (lesson 03).
• Proxy of a hidden confounder (not in the figure): a noisy measurement of $U$. It closes the path only partly; adjusting usually reduces the bias but does not remove it.

3.6 The frontdoor criterion #

Sometimes the confounder is hidden and no backdoor set exists. Pearl's example: smoking $T$ affects cancer $Y$ only through tar deposits $M$, and a hidden genotype $U$ affects both smoking and cancer.

Definition (frontdoor criterion; Pearl 1995). A set $M$ satisfies the frontdoor criterion relative to $(T, Y)$ if

1. $M$ intercepts every directed path from $T$ to $Y$ (all of the effect goes through $M$);
2. there is no open backdoor path from $T$ to $M$;
3. every backdoor path from $M$ to $Y$ is blocked by $T$.

Theorem (frontdoor adjustment). If $M$ satisfies the frontdoor criterion and positivity holds,

Derivation: two backdoor adjustments in a chain.

1. Effect of $T$ on $M$. By condition 2 there is no open backdoor path from $T$ to $M$, so (backdoor with the empty set) $P(m \mid \mathrm{do}(t)) = P(m \mid t)$.
2. Effect of $M$ on $Y$. The backdoor path $M \leftarrow T \leftarrow U \to Y$ is blocked by $T$ (condition 3), so $\{T\}$ is a backdoor set: $P(y \mid \mathrm{do}(m)) = \sum_{t'} P(y \mid m, t')\, P(t')$. The sum is over all values $t'$ of $T$, weighted by the population share $P(t')$.
3. Chain them. All of the effect of $T$ goes through $M$ (condition 1), and $M$ is not confounded with $Y$ except through $T$. So setting $T = t$ changes $Y$ only by changing the distribution of $M$: $P(y \mid \mathrm{do}(t)) = \sum_m P(m \mid \mathrm{do}(t))\, P(y \mid \mathrm{do}(m))$. Insert steps 1 and 2.

The formal version of step 3 uses do-calculus; the folded block in section 3.7 goes through it.

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How often does the frontdoor work in practice? Rarely in a clean form. It needs a mediator that carries all of the effect and that shares no hidden cause with the treatment or with the outcome. Real tar deposits are probably affected by the genotype too. The frontdoor idea is still useful as a check, and in settings where a mechanism is fully measured (for example an ad that can only act through a click, or a policy that acts only through an eligibility status).

3.7 Do-calculus #

The backdoor and frontdoor formulas are special cases. Pearl's do-calculus (1995) has three rules that change one expression with $\mathrm{do}$ into another. Each rule has a condition, a d-separation in a modified graph. Notation: $X, Y, Z, W$ are disjoint sets of nodes; $G_{\overline X}$ removes the arrows into $X$; $G_{\underline Z}$ removes the arrows out of $Z$.

Rule 1 (insert or delete an observation). $P(y \mid \mathrm{do}(x), z, w) = P(y \mid \mathrm{do}(x), w)$ if $(Y \perp\!\!\!\perp Z \mid X, W)_{G_{\overline X}}$. In words: after the intervention, observing $Z$ tells nothing more about $Y$. Example: in $T \to M \to Y$, $P(y \mid \mathrm{do}(m), t) = P(y \mid \mathrm{do}(m))$, because in $G_{\overline M}$ the node $T$ is cut off from $Y$.

Rule 2 (exchange an action and an observation). $P(y \mid \mathrm{do}(x), \mathrm{do}(z), w) = P(y \mid \mathrm{do}(x), z, w)$ if $(Y \perp\!\!\!\perp Z \mid X, W)_{G_{\overline X \underline Z}}$. In words: if, apart from its own effect, nothing links $Z$ and $Y$ (no open backdoor path), then setting $Z$ and seeing $Z$ are the same. Example: with no confounding ($T \to Y$ only), $P(y \mid \mathrm{do}(t)) = P(y \mid t)$.

Rule 3 (insert or delete an action). $P(y \mid \mathrm{do}(x), \mathrm{do}(z), w) = P(y \mid \mathrm{do}(x), w)$ if $(Y \perp\!\!\!\perp Z \mid X, W)_{G_{\overline X\, \overline{Z(W)}}}$, where $Z(W)$ is the set of nodes of $Z$ that are not ancestors of any node of $W$ in $G_{\overline X}$. In words: setting $Z$ does not change $Y$ when $Z$ has no causal path to $Y$. Example: if $Y \to T$ (the outcome causes the "treatment"), $P(y \mid \mathrm{do}(t)) = P(y)$.

Completeness (Huang and Valtorta 2006; Shpitser and Pearl 2006). If $P(y \mid \mathrm{do}(x))$ is identified in a graph with hidden variables, then a finite sequence of the three rules (and the rules of probability) derives a formula without $\mathrm{do}$. The ID algorithm finds such a formula or proves that none exists. So "the rules cannot do it" means "no method can do it" with this graph and no further assumptions.

An effect that is not identified: the bow-tie. Graph: $T \leftarrow U \to Y$, $T \to Y$, with $U$ hidden. Two SCMs show that the data cannot decide. Let $U \sim \mathrm{Bernoulli}(0.5)$ and $T := U$. In model 1, $Y := U$ (the drug does nothing). In model 2, $Y := T$ (the drug decides everything). Both give the same data: $T = Y$ always, each value with probability $0.5$. But $P(Y{=}1 \mid \mathrm{do}(T{=}1))$ is $0.5$ in model 1 and $1$ in model 2. No formula in the observational distribution can give both answers. To make progress you need more: a measured confounder, an instrument (lesson 06), a frontdoor mediator, or bounds and sensitivity analysis (lesson 05).

The frontdoor formula, derived with the three rules

The figure after this block steps through the same derivation. In symbols:

1. $P(y \mid \mathrm{do}(t)) = \sum_m P(y \mid \mathrm{do}(t), m)\,P(m \mid \mathrm{do}(t))$ (probability).
2. $P(m \mid \mathrm{do}(t)) = P(m \mid t)$: rule 2, since $(M \perp\!\!\!\perp T)_{G_{\underline T}}$.
3. $P(y \mid \mathrm{do}(t), m) = P(y \mid \mathrm{do}(t), \mathrm{do}(m))$: rule 2, since $(Y \perp\!\!\!\perp M \mid T)_{G_{\overline T \underline M}}$.
4. $P(y \mid \mathrm{do}(t), \mathrm{do}(m)) = P(y \mid \mathrm{do}(m))$: rule 3, since $(Y \perp\!\!\!\perp T \mid M)_{G_{\overline M\, \overline T}}$.
5. $P(y \mid \mathrm{do}(m)) = \sum_{t'} P(y \mid \mathrm{do}(m), t')\,P(t' \mid \mathrm{do}(m))$ (probability).
6. $P(t' \mid \mathrm{do}(m)) = P(t')$: rule 3, since $(T \perp\!\!\!\perp M)_{G_{\overline M}}$.
7. $P(y \mid \mathrm{do}(m), t') = P(y \mid m, t')$: rule 2, since $(Y \perp\!\!\!\perp M \mid T)_{G_{\underline M}}$.
8. Insert 2–7 into 1: the frontdoor formula.

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3.8 From identification to estimation #

An identification formula such as $\sum_z P(y \mid t, z) P(z)$ is a statistical estimand: a property of the observational distribution. To use it with data we need an estimator. With few discrete covariates, plug in the sample frequencies (this lesson's notebook). With many or continuous covariates, use regression, matching, inverse probability weighting or doubly robust methods (lesson 05) or ML-based estimators (lesson 08). The graph decides what to estimate; statistics decides how well.

Answers to section 1. (1) Yes. With strength 0.6 in section 2.1, the drug group recovers less often ($0.48$ vs $0.52$), although the drug raises recovery by $0.2$: the treated are more often severe. (2) Worse. $Z$ is an instrument: it removes no bias, increases the variance, and amplifies any remaining bias. (3) Yes, if tar carries the whole effect and shares no hidden cause with smoking or with cancer: the frontdoor formula.

4. Worked example #

(a) All valid adjustment sets in the butterfly graph. Edges: $X_1 \to X_3$, $X_2 \to X_3$, $X_1 \to T$, $X_3 \to T$, $X_2 \to Y$, $X_3 \to Y$, $T \to Y$. No $X_i$ is a descendant of $T$, so only condition 2 matters.

Step 1: backdoor paths.

• $p_1$: $T \leftarrow X_3 \to Y$ ($X_3$ is a fork)
• $p_2$: $T \leftarrow X_1 \to X_3 \to Y$ ($X_1$ fork, $X_3$ chain)
• $p_3$: $T \leftarrow X_3 \leftarrow X_2 \to Y$ ($X_3$ chain, $X_2$ fork)
• $p_4$: $T \leftarrow X_1 \to X_3 \leftarrow X_2 \to Y$ ($X_1$ fork, $X_3$ collider, $X_2$ fork)

Step 2: check the 8 subsets of $\{X_1, X_2, X_3\}$.

• $X_3$ is a fork on $p_1$ and nothing else is on $p_1$, so every valid set contains $X_3$.
• With $X_3$ in the set, $p_1, p_2, p_3$ are blocked at $X_3$, but $p_4$ opens at the collider $X_3$. It must be closed at $X_1$ or $X_2$.
• Valid sets: $\{X_1, X_3\}$, $\{X_2, X_3\}$, $\{X_1, X_2, X_3\}$. Not valid: $\{\}$, $\{X_1\}$, $\{X_2\}$, $\{X_1, X_2\}$ ($p_1$ open), $\{X_3\}$ ($p_4$ open).

(b) The adjustment formula from a table. A study records sex $X$, treatment $T$ and recovery $Y$, with graph $X \to T$, $X \to Y$, $T \to Y$:

	$P(X)$	$P(Y{=}1 \mid T{=}1, X)$	$P(Y{=}1 \mid T{=}0, X)$
$X = 0$	$0.6$	$0.7$	$0.5$
$X = 1$	$0.4$	$0.4$	$0.3$

The average treatment effect is $0.58 - 0.42 = 0.16$. Note that we never needed $P(T \mid X)$: it matters for seeing, not for doing (it matters for estimation, through positivity and variance).

(c) The frontdoor formula with numbers. Use the SCM of the frontdoor figure: $P(U{=}1) = 0.5$, $P(T{=}1 \mid U) = 0.2$ or $0.8$, $P(M{=}1 \mid T) = 0.1$ or $0.9$, $P(Y{=}1 \mid M, U) = 0.1 + 0.5M + 0.3U$. The observational distribution (what infinite data show) gives $P(T{=}1) = 0.5$ and

	$t' = 0$	$t' = 1$
$P(Y{=}1 \mid M{=}0, t')$	$0.16$	$0.34$
$P(Y{=}1 \mid M{=}1, t')$	$0.66$	$0.84$

(For example, $P(U{=}1 \mid T{=}0) = 0.5 \cdot 0.2 / (0.5 \cdot 0.2 + 0.5 \cdot 0.8) = 0.2$, and $M$ does not depend on $U$ given $T$, so $P(Y{=}1 \mid M{=}0, T{=}0) = 0.1 + 0.3 \cdot 0.2 = 0.16$.)

Step 1. $P(M{=}1 \mid T{=}1) = 0.9$, $P(M{=}1 \mid T{=}0) = 0.1$.

Step 2. $\sum_{t'} P(Y{=}1 \mid m, t')P(t')$: for $m = 0$, $0.5 \cdot 0.16 + 0.5 \cdot 0.34 = 0.25$; for $m = 1$, $0.5 \cdot 0.66 + 0.5 \cdot 0.84 = 0.75$.

Step 3. $P(Y{=}1 \mid \mathrm{do}(T{=}1)) = 0.1 \cdot 0.25 + 0.9 \cdot 0.75 = 0.70$ and $P(Y{=}1 \mid \mathrm{do}(T{=}0)) = 0.9 \cdot 0.25 + 0.1 \cdot 0.75 = 0.30$.

The effect is $0.40$. The truth from the hidden SCM is the same: $P(Y{=}1 \mid \mathrm{do}(T{=}1)) = 0.1 + 0.5 \cdot 0.9 + 0.3 \cdot 0.5 = 0.70$. The naive difference is $P(Y{=}1 \mid T{=}1) - P(Y{=}1 \mid T{=}0) = 0.79 - 0.21 = 0.58$.

5. In code #

The code checks the kidney-stone adjustment and the frontdoor example. The truth is computed by simulating the intervention: run the SCM again with $T$ set by hand.

import numpy as np

# (1) kidney stones: backdoor adjustment over stone size (Charig et al. 1986)
success = {"A": {"small": (81, 87), "large": (192, 263)},
           "B": {"small": (234, 270), "large": (55, 80)}}
p_small = (87 + 270) / 700
for t, d in success.items():
    pooled = sum(s for s, n in d.values()) / sum(n for s, n in d.values())
    adjusted = (p_small * d["small"][0] / d["small"][1]
                + (1 - p_small) * d["large"][0] / d["large"][1])
    print(t, round(pooled, 3), round(adjusted, 3))
# A 0.78 0.833
# B 0.826 0.779

# (2) frontdoor: hidden genotype U, smoking T, tar M, cancer Y
def simulate(n, seed, do_t=None):
    rng = np.random.default_rng(seed)
    U = rng.random(n) < 0.5
    T = rng.random(n) < np.where(U, 0.8, 0.2) if do_t is None else np.full(n, bool(do_t))
    M = rng.random(n) < np.where(T, 0.9, 0.1)
    Y = rng.random(n) < 0.1 + 0.5 * M + 0.3 * U
    return T.astype(int), M.astype(int), Y.astype(int)

T, M, Y = simulate(1_000_000, seed=0)      # observational data; U is not returned

def frontdoor(T, M, Y, t):
    pt = {tp: (T == tp).mean() for tp in (0, 1)}
    total = 0.0
    for m in (0, 1):
        p_m_t = (M[T == t] == m).mean()                                        # step 1
        inner = sum(Y[(M == m) & (T == tp)].mean() * pt[tp] for tp in (0, 1))  # step 2
        total += p_m_t * inner                                                 # step 3
    return total

naive = Y[T == 1].mean() - Y[T == 0].mean()
fd = frontdoor(T, M, Y, 1) - frontdoor(T, M, Y, 0)
truth = simulate(1_000_000, 1, do_t=1)[2].mean() - simulate(1_000_000, 2, do_t=0)[2].mean()
print(f"naive {naive:.3f}  frontdoor {fd:.3f}  truth by intervention {truth:.3f}")
# naive 0.580  frontdoor 0.400  truth by intervention 0.400

6. Where it shows up #

• Observational studies in medicine and economics. Analysts draw the DAG, find a backdoor set, check positivity, and then estimate (lessons 05, 08). Target-trial emulation (Hernán and Robins) is this workflow with careful time ordering.
• Product and marketing analytics. "Did the discount increase retention?" when the discount was given by a rule or by sales staff. The variables that the rule used are the backdoor set; variables measured after the discount (sessions, support tickets) are mediators or colliders and must stay out.
• Recommender systems and ads. The effect of showing an item is $P(\text{click} \mid \mathrm{do}(\text{show}))$; logged data give $P(\text{click} \mid \text{show})$, confounded by the old policy's features. Off-policy evaluation (lesson 11) is backdoor adjustment over those features.
• Fairness. Path-specific and counterfactual fairness definitions are statements about interventions on a causal graph (lesson 09).
• ML for science. In Earth-system and biomedical data, interventions are often impossible (climate) or expensive (gene knockouts). Identification from graphs tells which observational comparisons answer an interventional question, and which need an experiment.

7. Common mistakes #

• "Adjust for every variable you have." Mediators, colliders, descendants of $Y$ and M-bias nodes create bias. Check: draw the graph and test the backdoor criterion for the set.
• Adjusting for a variable measured after treatment. It is often a descendant of $T$ (condition 1 fails). Check: was this variable fixed before the treatment was decided?
• "Predicts the treatment strongly" = "confounder". A cause of $T$ only is an instrument; adjusting for it costs precision and can amplify bias. A confounder must also be linked to $Y$ by a path that does not go through $T$.
• Writing $P(y \mid t)$ when you mean $P(y \mid \mathrm{do}(t))$. They are equal only when there is no open backdoor path. Check: list the backdoor paths.
• Mixing identification and estimation. "The effect is identified" says nothing about the sample size needed; "the estimate is significant" says nothing about whether it estimates the right quantity.
• Using the frontdoor formula without checking its conditions. If the mediator shares a hidden cause with $T$ or $Y$, or if part of the effect bypasses it, the formula is biased with any amount of data.

8. Check yourself #

1. What does $\mathrm{do}(T = t)$ change in an SCM, and what does it change in the graph?
   Answer

   It replaces the assignment of $T$ by the constant $t$ and keeps all other assignments. In the graph it removes every arrow into $T$ (graph $G_{\overline T}$).
2. Graph: $X \to T$, $X \to Y$, $T \to M \to Y$. Is $\{X, M\}$ a valid adjustment set for the total effect of $T$ on $Y$? Why?
   Answer

   No. $M$ is a descendant of $T$ (condition 1 fails). Adjusting for $M$ blocks the causal path $T \to M \to Y$. $\{X\}$ alone is valid.
3. Write the truncated factorization for the graph $X \to T$, $X \to Y$, $T \to Y$, $Y \to S$ under $\mathrm{do}(t)$.
   Answer

   $p(x, y, s \mid \mathrm{do}(t)) = p(x)\,p(y \mid t, x)\,p(s \mid y)$: drop the factor $p(t \mid x)$ and put $T = t$ in the others.
4. Why is the effect of $T$ on $Y$ not identified in the graph $T \leftarrow U \to Y$, $T \to Y$ with $U$ hidden?
   Answer

   Two SCMs with this graph give the same observational distribution but different $P(y \mid \mathrm{do}(t))$, for example $T := U$ with $Y := U$ (no effect) or $Y := T$ (full effect). A quantity that differs between models with the same data cannot be a function of the data.
5. State the three conditions of the frontdoor criterion for a mediator $M$.
   Answer

   (1) $M$ intercepts all directed paths from $T$ to $Y$; (2) no open backdoor path from $T$ to $M$; (3) every backdoor path from $M$ to $Y$ is blocked by $T$.
6. Two sets $Z_1$ and $Z_2$ both satisfy the backdoor criterion. Do the two adjusted estimates agree?
   Answer

   In the population, yes: both give $P(y \mid \mathrm{do}(t))$. In a finite sample they differ by noise, and their variances differ: sets with causes of $Y$ are more precise, sets with strong causes of $T$ less precise.

Recall cardsAnswer in your head first, then reveal. Add the cards to your review deck to see them again just before you would forget.

1. What does $\mathrm{do}(T = t)$ do to the causal graph?

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2. Write the truncated factorization.

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3. State the two conditions of the backdoor criterion for a set $Z$.

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4. Write the backdoor adjustment formula.

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5. What is positivity, and why does the adjustment formula need it?

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6. Kidney stones: $P(\text{small}) = 0.51$; A succeeds in 93.1% (small) and 73.0% (large). What is $P(\text{success} \mid \mathrm{do}(A))$?

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7. Why is adjusting for a mediator a bad control for the total effect?

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8. Why is an instrument a bad control?

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9. Write the frontdoor formula.

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10. What does completeness of do-calculus mean?

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9. Interview angle #

• "Which variables would you control for to estimate the effect of a discount on retention?" Draw the graph first. Control for the variables that decided who got the discount and that also affect retention (tenure, past spend, plan type), measured before the discount. Do not control for post-discount behavior (sessions, purchases), which are mediators, or for anything caused by retention.
• "What is the difference between $P(y \mid x)$ and $P(y \mid \mathrm{do}(x))$?" The first describes units that happen to have $X = x$; the second describes what happens if we set $X = x$ for everybody. They differ when a backdoor path is open, as with sicker patients getting a stronger drug.
• "Is more control always better in a regression for causal effects?" No. Give the mediator, the collider and the instrument examples, each with its graph.
• "The confounder is not measured. Is all hope lost?" Not always: a frontdoor mediator, an instrument (lesson 06), a design such as regression discontinuity or difference-in-differences (lesson 07), or a sensitivity analysis (lesson 05). In the bow-tie graph with nothing else, the effect is not identified.

10. Practice #

The notebook for this lesson has five exercises, each with a check cell and a folded solution:

1. intervene(n, seed, do_t): simulate the SCM of section 2.1 with and without the intervention, and compare $P(Y \mid T)$ with $P(Y \mid \mathrm{do}(T))$.
2. backdoor_adjust(df, t, y, Z): the plug-in adjustment formula for discrete data; check it on the kidney-stone counts and on a simulated SCM.
3. is_backdoor_set(parents, T, Y, Z) with the d-separation function of lesson 03 on $G_{\underline T}$; list all valid sets in the butterfly graph.
4. Good and bad controls by simulation: the estimate and its standard deviation over 200 samples for each adjustment set of section 3.5.
5. A frontdoor estimator on simulated discrete data with a hidden $U$; compare with the truth $0.40$ and the naive $0.58$, then add the arrow $U \to M$ and watch it break.

11. Watch and read #

4.6 - The Backdoor AdjustmentBrady Neal · Introduction to Causal Inference · 10 min

The backdoor criterion and the adjustment formula, with a proof from the truncated factorization.

5.3 - The Frontdoor AdjustmentBrady Neal · Introduction to Causal Inference · 6 min

The frontdoor formula as two chained backdoor adjustments.

5.4 - Pearl’s do-calculusBrady Neal · Introduction to Causal Inference · 13 min

The three rules and their graph conditions.

Statistical Rethinking 2023 - 06 - Good & Bad ControlsRichard McElreath · 86 min

Simulations of good and bad controls, including colliders and the "table 2 fallacy".

Keynote: Judea Pearl - The New Science of Cause and EffectPyData · 66 min

The ladder of causation and the do-operator, from the author of do-calculus.

• Judea Pearl, Madelyn Glymour and Nicholas Jewell, Causal Inference in Statistics: A Primer (Wiley 2016), ch. 3 (interventions, backdoor, frontdoor); companion site.
• Carlos Cinelli, Andrew Forney and Judea Pearl, "A Crash Course in Good and Bad Controls", Sociological Methods & Research (2022).
• Brady Neal, Introduction to Causal Inference, free course notes, ch. 4 (do-operator, backdoor) and ch. 6 (frontdoor, do-calculus).
• Judea Pearl, "Causal diagrams for empirical research", Biometrika 82 (1995): the original do-calculus and frontdoor paper.
• Ilya Shpitser and Judea Pearl, "Identification of joint interventional distributions in recursive semi-Markovian causal models", AAAI 2006: the ID algorithm and completeness.