Instrumental Variables

Let $A$ be a treatment, $Y$ be an outcome with potential outcomes $Y(a)$ and $U$ be an unobserved confounder of $A$ and $Y.$ When $U$ is unobserved the treatment is not ignorable given the observed covariates, and the adjustment formula does not identify the causal effect.

Let $A$ be a treatment and $Y$ be an outcome. An instrument for the effect of $A$ on $Y$ is a random variable $Z$ that satisfies the following assumptions:

• relevance: $Z$ is associated with $A,$ so that $\mathrm{cov}\left[Z,A\right]\neq 0;$
• exclusion restriction: $Z$ affects $Y$ only through $A,$ so that the potential outcome $Y(a,z)$ under treatment $a$ and instrument $z$ does not depend on $z$ and is written $Y(a);$
• independence or instrument exogeneity: $Z$ is independent of the potential outcomes and potential treatments, $Z\perp\!\!\!\perp\left(Y(a),A(z)\right)$ for all $a,z,$ possibly given covariates.

In the causal graph an instrument is a vertex with an edge into $A,$ no other directed path to $Y$ and no open backdoor path to $Y.$ Relevance is testable from the data, whereas the exclusion restriction and independence are not.

Let $Z$ be an instrument for the effect of $A$ on $Y.$ The linear structural model is the structural equation for $Y$ with a constant treatment effect and an error term correlated with $A$ but uncorrelated with $Z.$

$$Y=\alpha+\beta A+U,\qquad\mathrm{cov}\left[Z,U\right]=0,\qquad\mathrm{cov}\left[A,U\right]\neq 0$$

Under the linear structural model the treatment effect is identified by the ratio of covariances, since $\mathrm{cov}\left[Z,Y\right]=\beta\,\mathrm{cov}\left[Z,A\right].$

$$\beta=\frac{\mathrm{cov}\left[Z,Y\right]}{\mathrm{cov}\left[Z,A\right]}$$

Least squares regression of $Y$ on $A$ estimates $\beta+\frac{\mathrm{cov}\left[A,U\right]}{\mathrm{var}\left[A\right]}$ rather than $\beta,$ and the difference is the endogeneity bias.

Let $Z$ be a binary instrument for the effect of $A$ on $Y.$ The Wald estimator of the treatment effect is the ratio of the difference in mean outcomes to the difference in mean treatments between the two instrument values.

$$\hat{\beta}_{\mathrm{W}}:=\frac{\overline{Y}_{Z=1}-\overline{Y}_{Z=0}}{\overline{A}_{Z=1}-\overline{A}_{Z=0}}$$

The Wald estimator is the sample analogue of the ratio of covariances for a binary instrument.

Let $\underline{a}\in\mathbb{R}^n$ be the vector of treatments, $\underline{y}\in\mathbb{R}^n$ be the vector of outcomes, $Z\in\mathbb{R}^{\,n\times q}$ be the matrix of $q\geq 1$ instruments and exogenous covariates, $X\in\mathbb{R}^{\,n\times p}$ be the matrix of the treatment and exogenous covariates with $p\leq q,$ and $P_Z:=Z\left(Z^\intercal Z\right)^{-1}Z^\intercal$ be the projection onto the column space of $Z.$ Two-stage least squares (2SLS) is the method that regresses $X$ on $Z$ in a first stage, then regresses $\underline{y}$ on the fitted values $\hat{X}=P_ZX$ in a second stage.

$$\hat{\underline{\beta}}_{\mathrm{2SLS}}:=\left(X^\intercal P_ZX\right)^{-1}X^\intercal P_Z\underline{y}$$

Two-stage least squares has the following properties:

• with one instrument and one treatment it equals the ratio of covariances, and with a binary instrument it equals the Wald estimator;
• it is consistent and asymptotically normal under relevance and exogeneity, but biased in finite samples towards the least squares estimator;
• the standard errors must be computed from the second-stage residuals $\underline{y}-X\hat{\underline{\beta}},$ not from $\underline{y}-\hat{X}\hat{\underline{\beta}};$
• the model is just identified if $p=q$ and overidentified if $p<q.$

Let $Z$ be a binary instrument and $A$ be a binary treatment with potential treatments $A(1),A(0)$ under the two instrument values. The compliance type of a unit is the pair $\left(A(0),A(1)\right),$ which takes one of four values:

• always-takers have $A(0)=A(1)=1;$
• never-takers have $A(0)=A(1)=0;$
• compliers have $A(0)=0$ and $A(1)=1;$
• defiers have $A(0)=1$ and $A(1)=0.$

The compliance type is not observed, since only one potential treatment is observed for each unit.

Let $Z$ be a binary instrument and $A$ be a binary treatment. Monotonicity is the assumption that the instrument does not decrease the treatment for any unit, so that there are no defiers.

$$A(1)\geq A(0)$$

Let $Z$ be a binary instrument and $A$ be a binary treatment. The intention-to-treat effect is the average effect of the instrument on the outcome, $\mathbb{E}\left[Y\mid Z=1\right]-\mathbb{E}\left[Y\mid Z=0\right],$ and the first-stage effect is the average effect of the instrument on the treatment, $\mathbb{E}\left[A\mid Z=1\right]-\mathbb{E}\left[A\mid Z=0\right].$ Under independence the first-stage effect is the proportion of compliers when monotonicity holds.

Let $Z$ be a binary instrument and $A$ be a binary treatment. The local average treatment effect (LATE) or complier average causal effect is the average treatment effect among the compliers.

$$\tau_{\mathrm{LATE}}:=\mathbb{E}\left[Y(1)-Y(0)\mid A(1)=1,A(0)=0\right]$$

Let $Z$ be a binary instrument satisfying relevance, the exclusion restriction, independence and monotonicity. The LATE theorem states that the local average treatment effect is identified by the ratio of the intention-to-treat effect to the first-stage effect.

$$\tau_{\mathrm{LATE}}=\frac{\mathbb{E}\left[Y\mid Z=1\right]-\mathbb{E}\left[Y\mid Z=0\right]}{\mathbb{E}\left[A\mid Z=1\right]-\mathbb{E}\left[A\mid Z=0\right]}$$

The LATE theorem has the following properties:

• the Wald estimator is a consistent estimator of the local average treatment effect without assuming a constant treatment effect;
• the always-takers and never-takers contribute nothing to the numerator, since the instrument does not change their treatment, and monotonicity removes the defiers, whose contribution would enter with the opposite sign;
• the local average treatment effect is specific to the instrument, since different instruments have different compliers, and it equals the average treatment effect only under additional assumptions such as a constant treatment effect;
• the distributions of $Y(1)$ and $Y(0)$ among the compliers are also identified.

Let $Z$ be an instrument for the effect of $A$ on $Y.$ The instrument is weak if and only if its association with the treatment is small relative to the sampling variability of the first stage. The first-stage $F$-statistic is the $F$-statistic for the hypothesis that the instruments have zero coefficients in the first-stage regression. Weak instruments have the following properties:

• the two-stage least squares estimator is biased towards the least squares estimator, with bias approximately $\frac{1}{F}$ of the endogeneity bias;
• the normal approximation to its sampling distribution fails and confidence intervals undercover;
• a first-stage $F$-statistic below about $10$ is the conventional indication of a weak instrument.

Let $Z$ be an instrument for the effect of $A$ on $Y$ under the linear structural model and $\beta_0$ be a hypothesised value. The Anderson–Rubin test of $H_0:\beta=\beta_0$ is the test of whether $Z$ has zero coefficient in the regression of $Y-\beta_0A$ on $Z.$ The Anderson–Rubin test is valid whatever the strength of the instrument, and inverting it gives a confidence set for $\beta$ that may be unbounded when the instrument is weak.

Let the linear structural model be overidentified with $q>p$ and $\hat{\underline{u}}$ be the two-stage least squares residuals. The Sargan–Hansen test or $J$-test of overidentifying restrictions is the test of the hypothesis that all instruments are exogenous, with statistic $J=n\,R^2$ from the regression of $\hat{\underline{u}}$ on $Z,$ which has an approximate $\chi^2_{q-p}$ distribution under the null hypothesis. The $J$-test can detect that some instruments are invalid but cannot identify which, and it has no power when all instruments are invalid in the same way.