Day 51: Interventions & the do-Operator · Math for Causal Inference

Theory

Intervening on a model

An intervention $\operatorname{do}(X = x)$ replaces the structural equation $X := f_X(\operatorname{pa}(X), U_X)$ by the constant $X := x$ , leaving every other equation unchanged. The modified model is $\mathcal M_x$ ; its graph $G_{\overline X}$ is the mutilated graph, with all arrows into $X$ removed.

The distribution of $Y$ in $\mathcal M_x$ is written $P(y \mid \operatorname{do}(x))$ . It answers "what would happen to $Y$ if we set $X$ to $x$ for everyone?"

Seeing versus doing

$P(y \mid x)$ describes $Y$ among units that happen to have $X = x$ . Those units may differ in other ways (they have different parents of $X$ ). $P(y \mid \operatorname{do}(x))$ describes the whole population when $X$ is set to $x$ . They coincide only when nothing that affects $Y$ also affects $X$ .

The truncated factorisation

For a Markovian model, delete the factor for $X$ and plug in $X = x$ everywhere else:

P(\mathbf v \mid \operatorname{do}(x)) = \prod_{i: V_i \ne X} P\big(v_i \mid \operatorname{pa}(v_i)\big)\Big|_{X = x} \quad \text{for } \mathbf v \text{ consistent with } x.

Adjusting for the parents

Summing the truncated factorisation over everything except $Y$ and the parents $\mathbf{PA} = \operatorname{pa}(X)$ gives the adjustment formula

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

Compare with conditioning, which weights by $P(\mathbf{pa} \mid x)$ instead of $P(\mathbf{pa})$ :

P(y \mid x) = \sum_{\mathbf{pa}} P(y \mid x, \mathbf{pa})\,P(\mathbf{pa} \mid x).

The difference is exactly confounding. (When the parents are unobserved, Days 52–56 look for other sets.)

Causal effects

The average causal effect of a binary $X$ is $E[Y \mid \operatorname{do}(X = 1)] - E[Y \mid \operatorname{do}(X = 0)]$ . It is the ATE of Phase 3: in an SCM, the potential outcome is $Y_i(x) = Y$ computed in $\mathcal M_x$ with unit $i$ 's disturbances (Day 59).

Stochastic (policy) interventions

Real policies rarely force $X$ on everyone; they shift its distribution, for example a subsidy that raises vaccination rates. Replacing $P(x \mid \operatorname{pa})$ by a new $q(x \mid \operatorname{pa})$ gives

P_q(y) = \sum_{\mathbf{pa}, x} P(y \mid x, \mathbf{pa})\,q(x \mid \mathbf{pa})\,P(\mathbf{pa}).

Part 1 · Warm-up

Exercise 1 · Mutilated graphs

For $Z \to X$ , $Z \to Y$ , $X \to M \to Y$ , $W \to X$ , draw $G_{\overline X}$ and $G_{\overline M}$ .

Show solution

Step 1. $G_{\overline X}$ removes arrows into $X$ ( $Z \to X$ , $W \to X$ ): left with $Z \to Y$ , $X \to M \to Y$ , and $W$ isolated.

Step 2. $G_{\overline M}$ removes $X \to M$ : left with $Z \to X$ , $W \to X$ , $Z \to Y$ , $M \to Y$ .

Exercise 2 · Truncated factorisation

Write $P(z, m, y \mid \operatorname{do}(x))$ for the graph in Exercise 1 (ignore $W$ ).

Show solution

Step 1. Full factorisation: $P(z)P(x \mid z)P(m \mid x)P(y \mid z, m)$ .

Step 2. Drop $P(x \mid z)$ and fix $X = x$ :

P(z, m, y \mid \operatorname{do}(x)) = P(z)\,P(m \mid x)\,P(y \mid z, m).

Exercise 3 · Linear seeing versus doing

$Z \sim N(0, 1)$ , $X := Z + U_X$ with $\operatorname{Var}(U_X) = 1$ , $Y := 2X + 3Z + U_Y$ . Find $E[Y \mid \operatorname{do}(x)]$ and $E[Y \mid X = x]$ .

Show solution

Step 1. Doing: $Z$ 's equation is untouched, so $E[Y \mid \operatorname{do}(x)] = 2x + 3E[Z] = 2x$ .

Step 2. Seeing: $E[Z \mid X = x] = \dfrac{\operatorname{Cov}(Z, X)}{\operatorname{Var}(X)}x = \dfrac{1}{2}x$ .

Step 3. $E[Y \mid X = x] = 2x + 3 \cdot \tfrac12 x = 3.5x$ .

Answer: slope 2 for doing, 3.5 for seeing.

Exercise 4 · No parents, no difference

If $X$ has no parents (a coin flip), show $P(y \mid \operatorname{do}(x)) = P(y \mid x)$ using the adjustment formula.

Show solution

Step 1. With $\operatorname{pa}(X) = \varnothing$ the sum has one term: $P(y \mid \operatorname{do}(x)) = P(y \mid x)$ .

Answer: randomisation makes $X$ parentless; that is why experiments identify causal effects.

Exercise 5 · Upstream variables do not move

In the graph of Exercise 1, is $P(z \mid \operatorname{do}(x)) = P(z)$ ? Is $P(z \mid x) = P(z)$ ?

Show solution

Step 1. Doing: $Z$ is not a descendant of $X$ , and the truncated factorisation keeps $P(z)$ . Summing out $m, y$ gives $P(z \mid \operatorname{do}(x)) = P(z)$ .

Step 2. Seeing: $Z \to X$ , so observing $X$ is informative about $Z$ : in general $P(z \mid x) \ne P(z)$ .

Part 2 · Proofs

Exercise 6 · Truncated factorisation from the SCM

Prove the truncated factorisation for a Markovian SCM.

Show solution

Step 1. $\mathcal M_x$ is again a Markovian SCM: same equations and disturbances for all $V_i \ne X$ , and $X := x$ (a constant with no parents).

Step 2. By Day 50 (Exercise 6), its distribution factorises over its graph $G_{\overline X}$ : $\prod_{V_i \ne X} P_{\mathcal M_x}(v_i \mid \operatorname{pa}(v_i)) \times \mathbf 1[X = x]$ .

Step 3. For $V_i \ne X$ , the conditional distribution of $V_i$ given its parents comes from the same $f_i$ and $U_i$ as in $\mathcal M$ , so $P_{\mathcal M_x}(v_i \mid \operatorname{pa}) = P(v_i \mid \operatorname{pa})$ . $\blacksquare$

Exercise 7 · The parent adjustment formula

Derive $P(y \mid \operatorname{do}(x)) = \sum_{\mathbf{pa}} P(y \mid x, \mathbf{pa})P(\mathbf{pa})$ from the truncated factorisation.

Show solution

Step 1. Multiply and divide the truncated product by $P(x \mid \mathbf{pa})$ : $P(\mathbf v \mid \operatorname{do}(x)) = \dfrac{P(\mathbf v)}{P(x \mid \mathbf{pa})}$ for $\mathbf v$ consistent with $x$ .

Step 2. Write $P(\mathbf v) = P(\mathbf{pa})P(x \mid \mathbf{pa})P(\mathbf r \mid x, \mathbf{pa})$ , where $\mathbf R$ is all remaining variables, including $Y$ .

Step 3. So $P(\mathbf v \mid \operatorname{do}(x)) = P(\mathbf{pa})P(\mathbf r \mid x, \mathbf{pa})$ .

Step 4. Sum over $\mathbf{pa}$ and over all components of $\mathbf r$ other than $y$ : $P(y \mid \operatorname{do}(x)) = \sum_{\mathbf{pa}} P(\mathbf{pa})P(y \mid x, \mathbf{pa})$ . $\blacksquare$

Exercise 8 · Non-descendants are unaffected

Prove that for any set $\mathbf N$ of non-descendants of $X$ , $P(\mathbf n \mid \operatorname{do}(x)) = P(\mathbf n)$ .

Show solution

Step 1. Order the variables topologically with all non-descendants of $X$ first. This is possible: no non-descendant has a descendant of $X$ as an ancestor.

Step 2. The equations of non-descendants involve only other non-descendants and their own disturbances.

Step 3. The intervention changes only $X$ 's equation, and $X$ is not an input to any non-descendant's equation (directly or indirectly).

Step 4. So the non-descendants are the same functions of the same disturbances in $\mathcal M$ and $\mathcal M_x$ : their joint distribution is unchanged. $\blacksquare$

Exercise 9 · Conditional ignorability from the SCM

In a Markovian SCM, prove that the potential outcome $Y(x)$ is independent of $X$ given $\operatorname{pa}(X)$ .

Show solution

Step 1. In $\mathcal M_x$ , $X$ 's equation is gone, so $Y(x)$ is a function of the disturbances other than $U_X$ : $Y(x) = g(\mathbf U_{-X})$ .

Step 2. $\operatorname{pa}(X)$ is a function of the disturbances of its ancestors, also in $\mathbf U_{-X}$ .

Step 3. $X = f_X(\operatorname{pa}(X), U_X)$ . Given $\operatorname{pa}(X) = \mathbf p$ , $X = f_X(\mathbf p, U_X)$ depends only on $U_X$ .

Step 4. $U_X$ is independent of $\mathbf U_{-X}$ , also after conditioning on an event defined by $\mathbf U_{-X}$ (the event $\operatorname{pa}(X) = \mathbf p$ ).

Step 5. So given $\operatorname{pa}(X)$ , $X$ is independent of $Y(x)$ . $\blacksquare$ This is Phase 3's unconfoundedness, derived from a graph.

Exercise 10 · Conditioning is a weighted adjustment

Show that $P(y \mid x) - P(y \mid \operatorname{do}(x)) = \sum_{\mathbf{pa}} P(y \mid x, \mathbf{pa})\big[P(\mathbf{pa} \mid x) - P(\mathbf{pa})\big]$ , and give two conditions under which it is zero.

Show solution

Step 1. By the law of total probability, $P(y \mid x) = \sum_{\mathbf{pa}} P(y \mid x, \mathbf{pa})P(\mathbf{pa} \mid x)$ .

Step 2. Subtract the adjustment formula term by term.

Step 3. Zero if $P(\mathbf{pa} \mid x) = P(\mathbf{pa})$ for all values ( $X$ independent of its parents, as under randomisation).

Step 4. Zero if $P(y \mid x, \mathbf{pa})$ does not depend on $\mathbf{pa}$ (the parents do not affect $Y$ except through $X$ ). $\blacksquare$

Part 3 · Applied

Exercise 11 · Vaccination: seeing versus doing

Use Day 47's flu model: 30% are older; $P(V = 1 \mid \text{old}) = 0.8$ , $P(V = 1 \mid \text{young}) = 0.4$ ; infection risks young unvaccinated 0.10, young vaccinated 0.04, old unvaccinated 0.30, old vaccinated 0.12. Compute $P(I = 1 \mid \operatorname{do}(V = 1))$ and $P(I = 1 \mid \operatorname{do}(V = 0))$ and compare with the observed 0.077 and 0.125.

Show solution

Step 1. Age is the only parent of $V$ . Adjust: $P(I = 1 \mid \operatorname{do}(V = 1)) = 0.7(0.04) + 0.3(0.12) = 0.028 + 0.036 = 0.064$ .

Step 2. $P(I = 1 \mid \operatorname{do}(V = 0)) = 0.7(0.10) + 0.3(0.30) = 0.07 + 0.09 = 0.16$ .

Step 3. Causal risk ratio $0.064/0.16 = 0.40$ ; risk difference $-0.096$ .

Answer: universal vaccination would cut the infection rate from 16% to 6.4%. The observed comparison (7.7% vs 12.5%, ratio 0.62) understates the benefit, because the vaccinated are older.

Exercise 12 · A vaccination subsidy

A subsidy raises vaccination among the young from 40% to 70%, leaving the old at 80%. Using Exercise 11's model, what infection rate does the policy produce, compared with 10% now?

Show solution

Step 1. Stochastic intervention: replace $P(V = 1 \mid \text{young}) = 0.4$ by $q = 0.7$ .

Step 2. Young: $0.7 \times [0.7(0.04) + 0.3(0.10)] = 0.7 \times 0.058 = 0.0406$ .

Step 3. Old (unchanged): $0.3 \times [0.8(0.12) + 0.2(0.30)] = 0.3 \times 0.156 = 0.0468$ .

Step 4. Total: $0.0406 + 0.0468 = 0.0874$ .

Answer: about 8.7% instead of 10%, a 1.3-point reduction. The policy effect depends on whom the subsidy reaches, which a single ATE does not tell you.

Exercise 13 · Fertiliser and yields

On farms, rainfall $R \sim N(0, 1)$ affects both fertiliser use $F := 0.5R + U_F$ ( $\operatorname{Var}(U_F) = 0.75$ ) and yields $Y := 2F + 3R + U_Y$ . A regression of yields on fertiliser alone finds what slope? What is the effect of a fertiliser subsidy that raises $F$ by one unit on every farm?

Show solution

Step 1. $\operatorname{Var}(F) = 0.25 + 0.75 = 1$ and $\operatorname{Cov}(F, R) = 0.5$ .

Step 2. Seeing: $E[Y \mid F = f] = 2f + 3E[R \mid F = f] = 2f + 3(0.5f) = 3.5f$ . Slope 3.5.

Step 3. Doing: $E[Y \mid \operatorname{do}(F = f + 1)] - E[Y \mid \operatorname{do}(F = f)] = 2$ .

Answer: the regression overstates the yield response by 75%, because farmers use more fertiliser in wet years, which are also good years.

Exercise 14 · Should hospitals close intensive care units?

20% of patients are severe. $P(\text{ICU} \mid \text{severe}) = 0.9$ and $P(\text{ICU} \mid \text{mild}) = 0.1$ . Death risks: mild without ICU 0.02, mild in ICU 0.01, severe without ICU 0.40, severe in ICU 0.25. Compute observed mortality with and without ICU care, and the mortality under do(ICU) and do(no ICU).

Show solution

Step 1. $P(\text{ICU}) = 0.8(0.1) + 0.2(0.9) = 0.26$ .

Step 2. Seeing, ICU: $\dfrac{0.8(0.1)(0.01) + 0.2(0.9)(0.25)}{0.26} = \dfrac{0.0008 + 0.045}{0.26} \approx 0.176$ .

Step 3. Seeing, no ICU: $\dfrac{0.8(0.9)(0.02) + 0.2(0.1)(0.40)}{0.74} = \dfrac{0.0144 + 0.008}{0.74} \approx 0.030$ .

Step 4. Doing: $P(D \mid \operatorname{do}(\text{ICU})) = 0.8(0.01) + 0.2(0.25) = 0.058$ ; $P(D \mid \operatorname{do}(\text{no ICU})) = 0.8(0.02) + 0.2(0.40) = 0.096$ .

Answer: ICU patients die six times as often (17.6% vs 3.0%), yet giving everyone ICU care would cut mortality from 9.6% to 5.8%. Severity is a parent of both admission and death.

Exercise 15 · Seeing or doing?

Classify each question as a conditional (seeing) or interventional (doing) query: (a) "What share of microfinance borrowers start a business?"; (b) "How much would business creation rise if every household got a loan?"; (c) "Do workers who join a union earn more?"; (d) "What would happen to wages if union membership became mandatory?"; (e) "Which patients with a positive test actually have the disease?"

Show solution

Step 1. (a) Seeing: $P(B \mid L = 1)$ .

Step 2. (b) Doing: $P(B \mid \operatorname{do}(L = 1)) - P(B)$ .

Step 3. (c) Seeing: $E[W \mid U = 1] - E[W \mid U = 0]$ .

Step 4. (d) Doing: $E[W \mid \operatorname{do}(U = 1)]$ .

Step 5. (e) Seeing: $P(D \mid T = +)$ , a diagnostic question answered by Bayes' rule (Day 2); no intervention is involved.

Verify in Python

import numpy as np

# Exercises 11-12: vaccination
pOld = .3; pV = {"young": .4, "old": .8}
risk = {("young", 0): .10, ("young", 1): .04, ("old", 0): .30, ("old", 1): .12}
w = {"young": 1 - pOld, "old": pOld}
do = {v: sum(w[a] * risk[a, v] for a in w) for v in (0, 1)}
see = {v: sum(w[a] * (pV[a] if v else 1 - pV[a]) * risk[a, v] for a in w) / sum(w[a] * (pV[a] if v else 1 - pV[a]) for a in w) for v in (0, 1)}
print("do:", {k: round(x, 4) for k, x in do.items()}, " see:", {k: round(x, 4) for k, x in see.items()})
q = {"young": .7, "old": .8}
print("infection with subsidy:", round(sum(w[a] * (q[a] * risk[a, 1] + (1 - q[a]) * risk[a, 0]) for a in w), 4))

# Exercise 13: fertiliser, by simulation of the SCM and of the intervened SCM
rng = np.random.default_rng(51); n = 1_000_000
R = rng.normal(size=n); F = .5 * R + np.sqrt(.75) * rng.normal(size=n); Y = 2 * F + 3 * R + rng.normal(size=n)
print("observational slope:", round(np.polyfit(F, Y, 1)[0], 3))
Y1 = 2 * (F + 1) + 3 * R + rng.normal(size=n)       # do(F := F + 1) for every farm
print("effect of +1 unit:", round((Y1 - Y).mean(), 3))

# Exercise 14: ICU
pS = .2; pICU = {0: .1, 1: .9}; d = {(0, 0): .02, (0, 1): .01, (1, 0): .40, (1, 1): .25}   # key (severe, icu)
ws = {0: 1 - pS, 1: pS}
for icu in (0, 1):
    num = sum(ws[s] * (pICU[s] if icu else 1 - pICU[s]) * d[s, icu] for s in (0, 1))
    den = sum(ws[s] * (pICU[s] if icu else 1 - pICU[s]) for s in (0, 1))
    print(f"ICU={icu}: seeing {num / den:.4f}  doing {sum(ws[s] * d[s, icu] for s in (0, 1)):.4f}")

Key takeaways

$\operatorname{do}(X = x)$ deletes $X$ 's equation and the arrows into $X$ ; everything else stays the same.
With observed parents, $P(y \mid \operatorname{do}(x)) = \sum_{\mathbf{pa}} P(y \mid x, \mathbf{pa})P(\mathbf{pa})$ ; conditioning instead weights by $P(\mathbf{pa} \mid x)$ .
Policies are often stochastic interventions: they shift $P(x \mid \operatorname{pa})$ rather than fix $x$ .

Day 51. Interventions & the do-Operator

Theory

Intervening on a model

Seeing versus doing

The truncated factorisation

Adjusting for the parents

Causal effects

Stochastic (policy) interventions

Part 1 · Warm-up

Exercise 1 · Mutilated graphs

Exercise 2 · Truncated factorisation

Exercise 3 · Linear seeing versus doing

Exercise 4 · No parents, no difference

Exercise 5 · Upstream variables do not move

Part 2 · Proofs

Exercise 6 · Truncated factorisation from the SCM

Exercise 7 · The parent adjustment formula

Exercise 8 · Non-descendants are unaffected

Exercise 9 · Conditional ignorability from the SCM

Exercise 10 · Conditioning is a weighted adjustment

Part 3 · Applied

Exercise 11 · Vaccination: seeing versus doing

Exercise 12 · A vaccination subsidy

Exercise 13 · Fertiliser and yields

Exercise 14 · Should hospitals close intensive care units?

Exercise 15 · Seeing or doing?

Verify in Python

Key takeaways

Simulation Lab: Seeing vs Doing: the ICU