Day 54: The Front-Door Criterion · Math for Causal Inference

Theory

When back-door fails

Suppose $X \leftarrow U \to Y$ with $U$ unobserved. No back-door set exists. But if $X$ affects $Y$ only through a mediator $M$ that is not itself confounded with $X$ , we can still identify the effect:

U \to X, \quad U \to Y, \quad X \to M \to Y.

Classic example: smoking $X$ , tar deposits $M$ , lung cancer $Y$ , unobserved genotype $U$ .

The criterion

A set $\mathbf M$ satisfies the front-door criterion relative to $(X, Y)$ if

$\mathbf M$ intercepts all directed paths from $X$ to $Y$ ;
there is no unblocked back-door path from $X$ to $\mathbf M$ ;
all back-door paths from $\mathbf M$ to $Y$ are blocked by $X$ .

The formula

If $\mathbf M$ satisfies the criterion and $P(x, \mathbf m) > 0$ ,

P(y \mid \operatorname{do}(x)) = \sum_{\mathbf m} P(\mathbf m \mid x)\sum_{x'} P(y \mid \mathbf m, x')\,P(x').

The logic in two steps

Effect of $X$ on $M$ . By condition 2, $P(m \mid \operatorname{do}(x)) = P(m \mid x)$ .
Effect of $M$ on $Y$ . By condition 3, $X$ is a back-door set for $M \to Y$ : $P(y \mid \operatorname{do}(m)) = \sum_{x'} P(y \mid m, x')P(x')$ .
Chain them. By condition 1, $X$ affects $Y$ only via $M$ : $P(y \mid \operatorname{do}(x)) = \sum_m P(m \mid \operatorname{do}(x))\,P(y \mid \operatorname{do}(m))$ .

The linear case

With $M = aX + e_M$ and $Y = bM + cU + e_Y$ , $X$ correlated with $U$ : estimate $a$ by regressing $M$ on $X$ , and $b$ by regressing $Y$ on $M$ and $X$ . The effect is $\hat a\hat b$ .

Plausibility

The conditions are strong: no direct effect of $X$ on $Y$ outside $M$ , and no unobserved common cause of $M$ and $Y$ . They hold most credibly when $M$ is a mechanical step: signing up versus attending, being offered versus receiving, prescribed versus dispensed.

Part 1 · Warm-up

Exercise 1 · Check the conditions

For each graph, does $\{M\}$ satisfy the front-door criterion for $X \to Y$ ? (a) $U \to X$ , $U \to Y$ , $X \to M \to Y$ . (b) Add $X \to Y$ . (c) Instead add $U \to M$ . (d) Instead add $V \to M$ , $V \to Y$ with $V$ unobserved.

Show solution

Step 1. (a) $M$ intercepts $X \to M \to Y$ ; back-door paths from $X$ to $M$ : $X \leftarrow U \to Y \leftarrow M$ , blocked at the collider $Y$ ; back-door from $M$ to $Y$ : $M \leftarrow X \leftarrow U \to Y$ , blocked by $X$ . Satisfied.

Step 2. (b) $X \to Y$ is a directed path not through $M$ : condition 1 fails.

Step 3. (c) $X \leftarrow U \to M$ is an open back-door path from $X$ to $M$ : condition 2 fails.

Step 4. (d) $M \leftarrow V \to Y$ is a back-door path from $M$ to $Y$ not blocked by $X$ : condition 3 fails.

Exercise 2 · Linear front door

Regressing $M$ on $X$ gives 0.5; regressing $Y$ on $M$ and $X$ gives 2.0 on $M$ . The front-door conditions hold. What is the effect of a one-unit increase in $X$ ?

Show solution

Step 1. Effect of $X$ on $M$ : 0.5. Effect of $M$ on $Y$ : 2.0.

Answer: $0.5 \times 2.0 = 1.0$ .

Exercise 3 · The first step

Why is $P(m \mid \operatorname{do}(x)) = P(m \mid x)$ in graph (a) of Exercise 1?

Show solution

Step 1. The only back-door path from $X$ to $M$ is $X \leftarrow U \to Y \leftarrow M$ .

Step 2. It has a collider at $Y$ , so it is blocked by the empty set.

Answer: $\varnothing$ satisfies the back-door criterion for $X \to M$ , so conditioning equals intervening.

Exercise 4 · A small front-door computation

$P(X = 1) = 0.4$ . $P(M = 1 \mid X = 1) = 0.8$ , $P(M = 1 \mid X = 0) = 0.3$ . $P(Y = 1 \mid M, X)$ : $(M, X) = (1, 1)$ : 0.6; $(0, 1)$ : 0.4; $(1, 0)$ : 0.5; $(0, 0)$ : 0.3. Compute $P(Y = 1 \mid \operatorname{do}(X = 1))$ by the front-door formula.

Show solution

Step 1. Inner sums: $m = 1$ : $0.4(0.6) + 0.6(0.5) = 0.24 + 0.30 = 0.54$ ; $m = 0$ : $0.4(0.4) + 0.6(0.3) = 0.16 + 0.18 = 0.34$ .

Step 2. Outer sum with $P(m \mid X = 1)$ : $0.8(0.54) + 0.2(0.34) = 0.432 + 0.068 = 0.50$ .

Answer: 0.50.

Exercise 5 · Why average over $x'$ ?

In the second step, why do we average $P(y \mid m, x')$ over $P(x')$ instead of using $P(y \mid m, x)$ at the treatment value $x$ ?

Show solution

Step 1. We need the effect of $M$ on $Y$ in the whole population, $P(y \mid \operatorname{do}(m))$ .

Step 2. $X$ is used only as a back-door adjustment variable for $M \to Y$ (it blocks $M \leftarrow X \leftarrow U \to Y$ ), so it must be averaged over its marginal distribution, as in any back-door formula.

Answer: using $P(y \mid m, x)$ alone would describe the $U$ -mix of people with $X = x$ , which is confounded.

Part 2 · Proofs

Exercise 6 · Step 1 in general

Prove $P(m \mid \operatorname{do}(x)) = P(m \mid x)$ under condition 2.

Show solution

Step 1. Condition 2 says $\varnothing$ blocks every back-door path from $X$ to $M$ , and $\varnothing$ contains no descendants of $X$ .

Step 2. So $\varnothing$ satisfies the back-door criterion for $(X, M)$ , and the adjustment formula reduces to $P(m \mid \operatorname{do}(x)) = P(m \mid x)$ . $\blacksquare$

Exercise 7 · Step 2 in general

Prove $P(y \mid \operatorname{do}(m)) = \sum_{x} P(y \mid m, x)P(x)$ under the front-door conditions.

Show solution

Step 1. Condition 3: $\{X\}$ blocks all back-door paths from $M$ to $Y$ .

Step 2. $X$ is not a descendant of $M$ : $M$ lies on the directed paths from $X$ , and acyclicity forbids $M \to \cdots \to X$ .

Step 3. So $\{X\}$ satisfies the back-door criterion for $(M, Y)$ and the adjustment formula gives the result. $\blacksquare$

Exercise 8 · Chaining the steps

Using the SCM $U \to X$ , $U \to Y$ , $X \to M \to Y$ , prove $P(y \mid \operatorname{do}(x)) = \sum_m P(m \mid \operatorname{do}(x))P(y \mid \operatorname{do}(m))$ .

Show solution

Step 1. Truncated factorisation for $\operatorname{do}(x)$ : $P(u, m, y \mid \operatorname{do}(x)) = P(u)P(m \mid x)P(y \mid m, u)$ .

Step 2. Sum over $u$ and $m$ : $P(y \mid \operatorname{do}(x)) = \sum_m P(m \mid x)\sum_u P(y \mid m, u)P(u)$ .

Step 3. Truncated factorisation for $\operatorname{do}(m)$ : $P(u, x, y \mid \operatorname{do}(m)) = P(u)P(x \mid u)P(y \mid m, u)$ ; summing over $u, x$ gives $P(y \mid \operatorname{do}(m)) = \sum_u P(y \mid m, u)P(u)$ .

Step 4. By Exercise 6, $P(m \mid x) = P(m \mid \operatorname{do}(x))$ . Substitute both into Step 2. $\blacksquare$ Combined with Exercise 7, this proves the front-door formula.

Exercise 9 · The linear product

With $X = gU + e_X$ , $M = aX + e_M$ , $Y = bM + cU + e_Y$ (all disturbances independent), prove the population coefficient on $M$ in the regression of $Y$ on $(M, X)$ is $b$ , so $\hat a\hat b \to ab$ .

Show solution

Step 1. $E[Y \mid M, X] = bM + cE[U \mid M, X]$ .

Step 2. $M = aX + e_M$ with $e_M$ independent of $(U, X)$ ; so given $X$ , $M$ carries no information about $U$ : $E[U \mid M, X] = E[U \mid X]$ , a linear function of $X$ only.

Step 3. So $E[Y \mid M, X] = bM + (\text{function of } X)$ : the coefficient on $M$ is $b$ .

Step 4. The regression of $M$ on $X$ recovers $a$ because $e_M$ is independent of $X$ . $\blacksquare$

Exercise 10 · A direct effect breaks it

Add a direct effect, $Y = bM + dX + cU + e_Y$ . What does $\hat a\hat b$ converge to, and what is the true total effect?

Show solution

Step 1. The argument of Exercise 9 still gives coefficient $b$ on $M$ (the $dX$ term is a function of $X$ ).

Step 2. So $\hat a\hat b \to ab$ .

Step 3. True total effect: $ab + d$ .

Answer: the front-door estimate misses the direct effect entirely; condition 1 is essential. $\blacksquare$

Part 3 · Applied

Exercise 11 · Smoking, tar and cancer

Observational data: $P(X = 1) = 0.5$ smokers; $P(M = 1 \mid X = 1) = 0.9$ and $P(M = 1 \mid X = 0) = 0.1$ heavy tar deposits; cancer rates $P(Y = 1 \mid M, X)$ : $(1, 1)$ : 0.44; $(0, 1)$ : 0.34; $(1, 0)$ : 0.26; $(0, 0)$ : 0.16. Compute the naive risk difference and the front-door effect.

Show solution

Step 1. Naive: $P(Y = 1 \mid X = 1) = 0.9(0.44) + 0.1(0.34) = 0.43$ and $P(Y = 1 \mid X = 0) = 0.1(0.26) + 0.9(0.16) = 0.17$ ; difference 0.26.

Step 2. Effect of tar, averaging over smoking: $P(Y \mid \operatorname{do}(M = 1)) = 0.5(0.44) + 0.5(0.26) = 0.35$ ; $P(Y \mid \operatorname{do}(M = 0)) = 0.5(0.34) + 0.5(0.16) = 0.25$ .

Step 3. $P(Y \mid \operatorname{do}(X = 1)) = 0.9(0.35) + 0.1(0.25) = 0.34$ .

Step 4. $P(Y \mid \operatorname{do}(X = 0)) = 0.1(0.35) + 0.9(0.25) = 0.26$ .

Answer: causal risk difference 0.08, versus a naive 0.26. (These numbers come from an SCM with a genotype that raises both smoking and cancer; the code checks that 0.08 is the true effect.)

Exercise 12 · Signing up versus showing up

A job-training programme records who signed up ( $X$ ), who attended ( $M$ ) and later earnings ( $Y$ ). Motivation is unobserved. Discuss each front-door condition for the effect of signing up.

Show solution

Step 1. Condition 1: signing up should affect earnings only through attending. Plausible, unless signing up itself changes behaviour (e.g. it triggers other services or job-search requirements).

Step 2. Condition 2: no unblocked confounding between signing up and attending. Questionable: motivated people both sign up and attend.

Step 3. Condition 3: given sign-up status, attendance is unconfounded with earnings. Questionable for the same reason.

Answer: among sign-ups, attendance is nearly mechanical (most attend), which makes conditions 2–3 more credible; studies of this design find estimates close to experimental benchmarks for some groups and not others. Plausibility must be argued, not assumed.

Exercise 13 · Mobile money and savings

A mobile-money app ( $X$ , adopted or not) can raise household savings $Y$ only through account usage $M$ (deposits per month). Adoption and savings share unobserved financial literacy. Regressions: usage on adoption gives 0.6 deposits; savings on usage and adoption gives $50 per deposit on usage. Estimate the effect of adoption on savings.

Show solution

Step 1. Effect of adoption on usage: $\hat a = 0.6$ .

Step 2. Effect of usage on savings, adjusting for adoption: $\hat b = 50$ .

Answer: $0.6 \times 50 = 30$ dollars per household. The naive regression of savings on adoption would also pick up literacy.

Exercise 14 · What if the app has a direct effect?

Suppose the app also raises savings by $10 directly (through reminders that work even without deposits). What does the front-door estimate from Exercise 13 converge to, and what is the true effect?

Show solution

Step 1. By Exercise 10, $\hat a\hat b \to ab = 30$ .

Step 2. True total effect: $30 + 10 = 40$ .

Answer: the front-door estimate understates the effect by $10 (25%). In the simulation below, the front-door estimate is about 30 and the truth 40.

Exercise 15 · Nurse staffing and mortality

A hospital network asks whether higher nurse staffing ( $X$ ) lowers in-hospital mortality ( $Y$ ). Wards with sicker patients get more nurses, and severity is poorly measured. The analyst proposes using monitoring frequency ( $M$ , vital-sign checks per day) as a front-door variable. List what must be true and one likely failure for each condition.

Show solution

Step 1. Condition 1: staffing affects mortality only through monitoring. Likely failure: nurses also deliver medication on time and prevent falls; those channels bypass $M$ .

Step 2. Condition 2: nothing unmeasured affects both staffing and monitoring. Likely failure: severity raises staffing and monitoring protocols directly.

Step 3. Condition 3: given staffing, monitoring is unconfounded with mortality. Likely failure: sicker patients are monitored more and die more.

Answer: monitoring fails all three; the front door is not credible here. A better design might use staffing shocks (nurse absences, Phase 5 instruments).

Verify in Python

import itertools
import numpy as np

# Exercise 11: the SCM behind the table (U = genotype, unobserved)
pU = .5; pX = {1: .8, 0: .2}; pM = {1: .9, 0: .1}; pY = {(0, 0): .1, (1, 0): .2, (0, 1): .4, (1, 1): .5}   # key (m, u)
truth = {x: sum((pM[x] if m else 1 - pM[x]) * sum((pU if u else 1 - pU) * pY[m, u] for u in (0, 1)) for m in (0, 1)) for x in (0, 1)}
J = {(u, x, m): (pU if u else 1 - pU) * (pX[u] if x else 1 - pX[u]) * (pM[x] if m else 1 - pM[x])
     for u, x, m in itertools.product((0, 1), repeat=3)}
Px = {x: sum(v for k, v in J.items() if k[1] == x) for x in (0, 1)}
Pm_x = {(m, x): sum(v for k, v in J.items() if k[1] == x and k[2] == m) / Px[x] for m in (0, 1) for x in (0, 1)}
Py_mx = {(m, x): sum(v * pY[m, k[0]] for k, v in J.items() if k[1] == x and k[2] == m) / (Px[x] * Pm_x[m, x]) for m in (0, 1) for x in (0, 1)}
print("observed P(Y=1|M,X):", {k: round(v, 3) for k, v in Py_mx.items()})
naive = {x: sum(Pm_x[m, x] * Py_mx[m, x] for m in (0, 1)) for x in (0, 1)}
front = {x: sum(Pm_x[m, x] * sum(Py_mx[m, xp] * Px[xp] for xp in (0, 1)) for m in (0, 1)) for x in (0, 1)}
print(f"naive {naive[1] - naive[0]:.3f}  front-door {front[1] - front[0]:.3f}  truth {truth[1] - truth[0]:.3f}")

# Exercises 13-14: linear front door with and without a direct effect
rng = np.random.default_rng(54); n = 500_000
L = rng.normal(size=n)                                 # financial literacy (unobserved)
X = (L + rng.normal(size=n) > 0).astype(float)
M = 0.6 * X + rng.normal(size=n)
for d in (0, 10):
    Y = 50 * M + d * X + 40 * L + rng.normal(0, 20, n)
    a = np.polyfit(X, M, 1)[0]
    b = np.linalg.lstsq(np.column_stack([np.ones(n), M, X]), Y, rcond=None)[0][1]
    naive = Y[X == 1].mean() - Y[X == 0].mean()
    print(f"direct effect {d}: naive {naive:.1f}, front-door {a * b:.1f}, truth {30 + d}")

Key takeaways

The front door identifies an effect despite unobserved treatment–outcome confounding, if a mediator captures the whole effect, is unconfounded with the treatment, and is unconfounded with the outcome given the treatment.
Formula: $P(y \mid \operatorname{do}(x)) = \sum_m P(m \mid x)\sum_{x'} P(y \mid m, x')P(x')$ ; in linear models, the product of two regression coefficients.
Its assumptions are strong; they are most credible when the mediator is a mechanical step of the treatment.

Day 54. The Front-Door Criterion

Theory

When back-door fails

The criterion

The formula

The logic in two steps

The linear case

Plausibility

Part 1 · Warm-up

Exercise 1 · Check the conditions

Exercise 2 · Linear front door

Exercise 3 · The first step

Exercise 4 · A small front-door computation

Exercise 5 · Why average over $x'$ ?

Part 2 · Proofs

Exercise 6 · Step 1 in general

Exercise 7 · Step 2 in general

Exercise 8 · Chaining the steps

Exercise 9 · The linear product

Exercise 10 · A direct effect breaks it

Part 3 · Applied

Exercise 11 · Smoking, tar and cancer

Exercise 12 · Signing up versus showing up

Exercise 13 · Mobile money and savings

Exercise 14 · What if the app has a direct effect?

Exercise 15 · Nurse staffing and mortality

Verify in Python

Key takeaways

Simulation Lab: The Front Door

Day 54. The Front-Door Criterion

Theory

When back-door fails

The criterion

The formula

The logic in two steps

The linear case

Plausibility

Part 1 · Warm-up

Exercise 1 · Check the conditions

Exercise 2 · Linear front door

Exercise 3 · The first step

Exercise 4 · A small front-door computation

Exercise 5 · Why average over x′x'x′?

Part 2 · Proofs

Exercise 6 · Step 1 in general

Exercise 7 · Step 2 in general

Exercise 8 · Chaining the steps

Exercise 9 · The linear product

Exercise 10 · A direct effect breaks it

Part 3 · Applied

Exercise 11 · Smoking, tar and cancer

Exercise 12 · Signing up versus showing up

Exercise 13 · Mobile money and savings

Exercise 14 · What if the app has a direct effect?

Exercise 15 · Nurse staffing and mortality

Verify in Python

Key takeaways

Simulation Lab: The Front Door

Exercise 5 · Why average over $x'$ ?