Lecture notes. Written from the material of Berkeley Stat 156, licensed CC BY-NC 4.0. These are notes, not a transcript: the material has been reorganised and rewritten. This adaptation carries the same licence, and the original is linked above.

5. Instrumental Variables and Noncompliance¶

What this covers¶

This chapter works through the identification arguments behind Homework 5 of Stat 156: recovering a causal effect of treatment from data on an instrument $Z$, actual treatment taken $D$, and an outcome $Y$, when not everyone assigned to treatment actually takes it up. It assumes the potential-outcomes framework and the vocabulary of Relevance, Exclusion Restriction and Exchangeability from earlier in the course — those are used here but not re-derived, since the lecture that introduced them is not part of the material supplied for this chapter. What follows is the setup the homework's ten problems ask you to work with; the problems themselves, not their solutions, are the point.

The noncompliance problem¶

Write $Z \in \{0, 1\}$ for the instrument — an assignment, encouragement, or eligibility signal — and $D \in \{0, 1\}$ for whether treatment is actually taken. $D$ need not equal $Z$: someone assigned $Z = 1$ can decline treatment. One-sided noncompliance is the restriction that deviation only runs one way: $Z = 0$ implies $D = 0$. Nobody gets treated without being assigned to be; the only slack is that some of the assigned do not take it up.

Potential outcomes carry both arguments before anything is observed: $D(z)$ is the treatment a unit would take under assignment $z$, and $Y(z, d)$ is the outcome under assignment $z$ and treatment $d$. The three assumptions from the course, restated as they appear in the homework:

Relevance. $Z$ is associated with $D$ — in the classical linear case, $\operatorname{cov}(D, Z) \neq 0$.
Exclusion restriction. $Y(Z = z, D = d) = Y(D = d)$: the instrument affects the outcome only through the treatment it induces, never directly.
Exchangeable IV. $Z \perp\!\!\!\perp \big(D(Z = z), Y(Z = z)\big)$: the instrument is independent of the potential values of treatment and outcome, exactly as if it had been assigned at random, whether or not it actually was.

The instrument moves the outcome only by moving the treatment (no arrow straight from Z to Y), and nothing feeds into Z itself (no arrow into Z from the unobserved U driving both D and Y) — that second absence is what exchangeability buys.

Two identification targets under one-sided noncompliance¶

Problem 1 asks for two causal contrasts, each written purely in terms of quantities you could estimate from data — no potential outcomes on the right-hand side. The effect of removing treatment on the whole population: $$E[Y - Y(D = 0)] = E[Y] - E[Y \mid Z = 0]$$ and the effect of removing treatment on the treated: $$E[Y - Y(D = 0) \mid D = 1] = \frac{E[Y] - E[Y \mid Z = 0]}{\mathbb{P}(D = 1)}.$$ Both hold under Relevance, Exclusion Restriction, Exchangeability, and one-sided noncompliance. The lever in both proofs is the same: one-sided noncompliance forces everyone observed at $Z = 0$ to also have $D = 0$, so their observed outcome already is $Y(D=0)$; exchangeability then lets you drop the conditioning on $Z = 0$ to get an unconditional statement about $Y(D=0)$. Working out exactly how that step goes, and how the second identity follows from the first by conditioning on $D = 1$, is Exercise 1 below.

The structural-equations picture and the two-stage model¶

Problem 2 gives the same setup a linear, structural-equation form. At the level of potential outcomes: $$D(Z = z) = \alpha_0 + \alpha_1 z + \nu, \qquad Y(Z = z, D = d) = \beta_0 + \beta_1 d + \epsilon.$$ These are statements about what would happen under a hypothetical assignment $z$. What is actually fit to data is the corresponding equation in the observed $Z$ and $D$: $$D = \alpha_0 + \alpha_1 Z + \nu, \qquad Y = \beta_0 + \beta_1 D + \epsilon.$$ The first equation is the first stage — regress the treatment taken on the instrument. The second is the second stage — regress the outcome on the treatment. Moving from the potential-outcome model to this observed-data model, under Relevance, Exclusion Restriction and Exchangeable IV, is supposed to also hand you three more familiar-looking conditions on the observed-data equations:

(a) Relevance: $\operatorname{cov}(D, Z) \neq 0$ (b) Exclusion restriction: $Z$ does not appear in the second-stage equation for $Y$ (c) Exchangeable IV: $\operatorname{cov}(Z, \nu) = \operatorname{cov}(Z, \epsilon) = 0$

Showing that the potential-outcome assumptions really do collapse to exactly these three, once $z$ is set to the realized $Z$, is Exercise 2.

The Wald estimator and two-stage least squares¶

Problems 3 and 4 are about two other ways of writing the same coefficient $\beta_1$. The first is a ratio of observable differences — the classical Wald estimator for a binary instrument — set equal to a ratio of covariances: $$\frac{E[Y \mid Z = 1] - E[Y \mid Z = 0]}{E[D \mid Z = 1] - E[D \mid Z = 0]} = \frac{\operatorname{cov}(Y, Z)}{\operatorname{cov}(D, Z)}.$$ The left side is what you would compute directly by comparing the two instrument arms; the right side is the same quantity in covariance form, which is the form that generalizes once $Z$ is not just binary. Their equality is Exercise 3.

The second rewrites $\beta_1$ yet again: it is also the coefficient from a population regression of $Y$ on the fitted values from a population regression of $D$ on $Z$. That two-step recipe — regress $D$ on $Z$, keep only the predicted $D$, then regress $Y$ on the prediction — is exactly two-stage least squares (2SLS). The predicted $D$ keeps only the part of the treatment variation driven by the instrument and discards the part correlated with the second-stage error $\epsilon$, which is what makes the second regression identify $\beta_1$ rather than something confounded. Proving that this 2SLS coefficient equals $\beta_1$ under $\operatorname{cov}(D, Z) \neq 0$ and $\operatorname{cov}(Z, \nu) = \operatorname{cov}(Z, \epsilon) = 0$ is Exercise 4.

Exercises¶

Due 2024-11-12, 11:59pm PT. (Submit code for any coding parts, and assign Gradescope pages to the matching problem — the assignment notes that points are deducted otherwise.)

Under Relevance, Exclusion Restriction, Exchangeability, and one-sided noncompliance ($Z = 0$ implies $D = 0$), prove $$E[Y - Y(D = 0)] = E[Y] - E[Y \mid Z = 0]$$ and $$E[Y - Y(D = 0) \mid D = 1] = \frac{E[Y] - E[Y \mid Z = 0]}{\mathbb{P}(D = 1)}.$$
Under the standard IV assumptions — relevance ($Z$ is associated with $D$), exclusion restriction ($Y(Z = z, D = d) = Y(D = d)$), and exchangeable IV ($Z \perp\!\!\!\perp (D(Z = z), Y(Z = z))$) — show that the structural equation model $$D(Z = z) = \alpha_0 + \alpha_1 z + \nu, \qquad Y(Z = z, D = d) = \beta_0 + \beta_1 d + \epsilon$$ implies the observed two-stage model $$D = \alpha_0 + \alpha_1 Z + \nu, \qquad Y = \beta_0 + \beta_1 D + \epsilon$$ together with the three classical IV assumptions: (a) $\operatorname{cov}(D, Z) \neq 0$; (b) $Z$ is not included in the second-stage $Y$ equation; (c) $\operatorname{cov}(Z, \nu) = \operatorname{cov}(Z, \epsilon) = 0$.
Prove $$\frac{E[Y \mid Z = 1] - E[Y \mid Z = 0]}{E[D \mid Z = 1] - E[D \mid Z = 0]} = \frac{\operatorname{cov}(Y, Z)}{\operatorname{cov}(D, Z)}.$$
Assume a classical two-stage IV model with $\operatorname{cov}(D, Z) \neq 0$ and $\operatorname{cov}(Z, \nu) = \operatorname{cov}(Z, \epsilon) = 0$. Show that $\beta_1$ equals the coefficient from a population regression of $Y$ on the predicted value from a population regression of $D$ on $Z$.
Problem 20.2 from A First Course in Causal Inference.
Problem 21.7 from A First Course in Causal Inference.
Problem 17.4 from A First Course in Causal Inference.
Problem 17.6 from A First Course in Causal Inference.
Problem 18.1 from A First Course in Causal Inference.
Problem 18.2 from A First Course in Causal Inference.

Sources¶

Everything above is drawn from a single source: HW5.md (Stat 156, Berkeley, Fall 2024), converted from HW5.pdf. That conversion is itself a model's reconstruction of a PDF with no extractable text layer, so its equations are marked unverified at the source and are reproduced here as given, not independently re-checked. No slide deck or lecture transcript was supplied for this chapter — the homework's own phrasing ("recall from lecture," "in lecture we presented") points to a lecture on instrumental variables and noncompliance that this chapter does not have access to, so the exposition above states only the definitions and claims the homework itself states, and stops short of the proofs, which are exactly what the ten problems ask for. Problems 5–10 point to A First Course in Causal Inference, a textbook not included in the supplied material; their statements are not reproduced here for that reason.

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