16. Instrumental Variables with Heterogeneous Effects

The basic idea is that there is a causal chain between \(Z\), \(D\) and \(Y\). We can argue that \(Z\) affects \(D\) which in turn affects \(Y\), but there is no effect of \(Z\) directly on \(Y\), which makes \(Z\) a good instrument variable (IV) to measure \(D\)'s effect on \(Y\).

For example, in the binary treatment and single binary instrument case, we write the problem in potential outcome notation as $$Y_i=\underbrace{\mathbb E[Y_{0,i}]}_{\alpha}+\underbrace{(Y_{1,i}-Y_{0,i})}_{\beta_i}D_i+\underbrace{Y_{0,i}-\mathbb E[Y_{0,i}]}_{U_i}\Rightarrow Y_i=\alpha+\beta D_i+U_i$$ $$D_i=D_{1,i}Z_i+D_{0,i}(1-Z_i)$$ where \(\beta\) is the average of the heterogeneous effect \(\beta_i\), \(Y_{0,i}\) and \(Y_{1,i}\) are the potential outcomes of treated and untreated individual \(i\), and finally \(D_{1,i}\) and \(D_{0,i}\) are the two potential treatment decisions of the person with and without the instrument.

The fundamental question is: what does linear IV identify when treatment effects are heterogeneous? A typical approach is to find an instrumental variable first, and then reverse engineer what it identifies.

• \(D_z\) is the treatment status of individual \(i\) at instrument value \(Z=z\in\{0,1\}\) (we drop the \(i\) subscript, and this applies to all the variables below).
• \(Y_{d,z}\) is the outcome of individual \(i\) if he receives treatment \(D=d\) and instrument value is \(Z=z\).

By the structure of this problem, we can now define various causal effects to be identified, for example: - \(Y_{1,z}-Y_{0,z}\): the effect of treatment on the potential outcome for each instrument value. - \(Y_{D,1}-Y_{D,0}\): the effect of instrument on the potential outcome for each treatment decision. - \(D_1-D_0\): the effect of instrument on the treatment decision.

Notice that for these potential outcomes with IV we need to jointly specify the value of both the instrument and the treatment — this stands in contrast to our baseline model of potential outcomes without IV where we only distinguish between the outcomes under treatment and without treatment.

The variable \(Z\) is an instrumental variable for the causal effect of \(D\) on \(Y\) if the following four assumptions hold: 1. Random assignment: \(Y_{d,z},D_z\perp Z\) for all \(d,z\). In words, we need that \(D\) is as good as randomly assigned as in an experimental or quasi-experimental setting. This is sufficient to identify the average causal effect of \(Z\) on \(Y\). To see this, notice that $$\mathbb E[Y\mid Z=1]-\mathbb E[Y\mid Z=0]=\mathbb E[Y_{D_1,1}\mid Z=1]-\mathbb E[Y_{D_0,0}\mid Z=0]\overset{Y_{d,z},D_z\perp Z}{=}\mathbb E[Y_{D_1,1}-Y_{D_0,0}]\tag{16.1}$$ 2. Exclusion restriction: \(Y_{d,1}=Y_{d,0}\) for all \(d\), which states that any effect of \(Z\) on \(Y\) must be via an effect of \(Z\) on \(D\). If you fix \(D\), there's no more effect of the instrument on outcome. - One example of where this might be violated is a tuition subsidy (instrument): a tuition subsidy will make college (treatment) cheaper, but might also allow people to buy a tutor, spend a shorter time working for a job, etc. It's typically harder to argue for exclusion for random assignment. - The exclusion restriction is often expressed by omitting \(Z\) in the equation of interest, i.e. \(Y=\alpha+\beta D+U\), which means outcomes don't depend on \(Z\) when we control for treatment \(D\). - Exclusion restriction + random assignment = instrument exogeneity. It is good to break down instrument exogeneity into exclusion restriction + random assignment because they are conceptually different objects, so it's better to argue separately. 3. Instrument relevance: \(\mathbb E[D_1-D_0]\ne0\). We need that \(Z\) has some effect on the average probability of treatment. In the population this means there is at least one guy whose decision changes based on the value of the instrument. In practice, you probably need more than a single person for estimation. 4. Monotonicity: \(D_1\ge D_0\) for all \(i\), or \(D_1\le D_0\) for all \(i\). In words, the instrument should have weakly same direction influence on all agent's choice of treatment. - Monotonicity is in some sense a bad word for this assumption: we really want to capture uniformity of response to the instrument — everyone is affected the same way. - Some intuition: only seeing flows in the data, i.e. \(D=\alpha_0+\alpha_1Z+\epsilon\), \(\alpha_1=\mathbb E[D\mid Z=1]-\mathbb E[D\mid Z=0]=\mathbb P(D=1\mid Z=1)-\mathbb P(D=1\mid Z=0)\) measures the net change in the fraction of treatment due to the instrument. Under monotonicity, \(\alpha_1\) measures (not net any more) the fraction of people moved towards treatment by instrument, i.e. the fraction of compliers to the instrument \(Z\). If monotonicity is violated, \(\alpha_1\) becomes the net effect, and it's not clear how to interpret \(\alpha_1\) in a meaningful way.

Wald Estimand. Consider the case where we use a single binary $(0,1)$ instrument \(Z_i\) to estimate a model for outcome \(Y_i\) with one endogenous treatment regressor \(D_i\) and no covariates. The causal regression model is \(Y_i=\alpha+\beta D_i+\eta_i\), and the instrument (first stage) regression is \(D_i=\psi+\gamma Z_i+v_i\). From (5.14) in §5.2.5, $$\beta^{\text{IV}}=\frac{\operatorname{Cov}(Z_i,Y_i)}{\operatorname{Cov}(Z_i,D_i)}$$ Computing \(\operatorname{Cov}(Z_i,Y_i)=(\mathbb E[Y_i\mid Z_i=1]-\mathbb E[Y_i\mid Z_i=0])\mathbb E[Z_i](1-\mathbb E[Z_i])\), and similarly \(\operatorname{Cov}(Z_i,D_i)=(\mathbb E[D_i\mid Z_i=1]-\mathbb E[D_i\mid Z_i=0])\mathbb E[Z_i](1-\mathbb E[Z_i])\), so $$\beta^{\text{IV}}=\frac{\mathbb E[Y_i\mid Z_i=1]-\mathbb E[Y_i\mid Z_i=0]}{\mathbb E[D_i\mid Z_i=1]-\mathbb E[D_i\mid Z_i=0]}\tag{16.2}$$ The last line is the Wald Estimand, and its sample analog is the Wald Estimator.

The essential idea of an over-identifying test is that in a constant effect (\(\beta_i=\beta\) for all \(i\)) model, you should get the same estimate (subject to some sampling error) from different instruments. For example, suppose you have two instruments \(Z_1\) and \(Z_2\) and the corresponding Wald Estimands are \(\beta_1^{\text{IV}}\) and \(\beta_2^{\text{IV}}\). Then, under the constant effect assumption, ATE among any group of compliers are the ATE among the whole population, so we should have \(\beta_1^{\text{IV}}=\beta_2^{\text{IV}}\). Suppose we have confidence that \(Z_1\) is a valid instrument, then we can do the following test $$H_0:\ Z_2\text{ is valid},\qquad H_1:\ Z_2\text{ is not valid}$$ by testing the distance between \(\beta_1^{\text{IV}}\) and \(\beta_2^{\text{IV}}\).

However, under the heterogeneous effect assumption, different instruments define different groups of compliers, so \(\beta_1^{\text{IV}}\) and \(\beta_2^{\text{IV}}\) can be different when both instruments are valid. So, the instrument rejected by such an over-identifying test might not be invalid.

So far we have considered an IV model with heterogeneous treatment effect in the simple case of: (1) average effects (for compliers): LATE; (2) binary treatment, and single binary instrument; (3) no covariates. Now we relax these assumptions.

Imbens & Rubin (1997) show that we can use IV to estimate more than the average causal effect for compliers (LATE). In particular, we can recover the complete marginal distributions of the outcome among the four groups of agents. First, introduce shorthand notation: \(n=\) Never Taker, \(a=\) Always Taker, \(c=\) Complier, \(d=\) Defier.

For the different combinations of \(Z\) and \(D\), we know the following (defiers ruled out by monotonicity):

Since \(Z\) is random, the distribution of types \(a,n,c\) is the same for each value of \(Z\). Denote \(p_s\) (\(s\in\{a,n,c\}\)) as the fraction of type \(s\) agent in the population; then the table implies $$p_a=\mathbb P(D=1\mid Z=0),\qquad p_n=\mathbb P(D=0\mid Z=1),\qquad p_c=1-p_a-p_n$$ i.e. the share of always takers = the share of treated among those who get \(Z=0\); the share of never takers = the share of untreated among those who get \(Z=1\); the share of compliers by subtracting the sum of these two shares from 1.

We can exploit this to calculate the marginal distributions for each of these groups, in particular $$f_{10}(y)=g_n(y)\tag{16.6}$$ $$f_{01}(y)=g_a(y)\tag{16.7}$$ $$f_{00}(y)=g_{a0}(y)\cdot\frac{p_c}{p_c+p_n}+g_n(y)\cdot\frac{p_n}{p_c+p_n}\tag{16.8}$$ $$f_{11}(y)=g_{a1}(y)\cdot\frac{p_c}{p_c+p_a}+g_a(y)\cdot\frac{p_a}{p_c+p_a}\tag{16.9}$$ where: \(f_{zd}(y)\) is the p.d.f. of outcome \(y\) among those \(Z=z,D=d\); \(g_n(y)\) is the p.d.f. of outcome \(y\) among never takers; \(g_a(y)\) is the p.d.f. of outcome \(y\) among always takers; \(g_{a0}(y)\) is the p.d.f. of outcome \(y\) among untreated compliers; \(g_{a1}(y)\) is the p.d.f. of outcome \(y\) among treated compliers. \(f_{00},f_{01},f_{10},f_{11},p_a,p_n,p_c\) can all be directly estimated using non-parametric methods from data, so we can obtain \(g_n(y),g_a(y),g_{a0}(y),g_{a1}(y)\), which are the outcome distributions of never takers, always takers, untreated compliers, and treated compliers respectively.

Suppose we have more than one binary instrument. For example, with two binary instruments, based on the result in §16.1.3, we can estimate two different LATEs: for \(j=1,2\), $$\beta_j=\frac{\operatorname{Cov}(Y,Z_j)}{\operatorname{Cov}(D,Z_j)}=\mathbb E[Y_1-Y_0\mid\underbrace{D_{Z_j=1}-D_{Z_j=0}=1}_{\text{Complier to instrument }j}]$$ In practice, researchers often use the TSLS method, and do the first-stage regression $$D=\pi_1Z_1+\pi_2Z_2+\varepsilon$$ (equivalently \(D=Z_1(1-Z_2)D_{1,0}+Z_2(1-Z_1)D_{0,1}+Z_1Z_2D_{1,1}+(1-Z_1)(1-Z_2)D_{0,0}\).) Then use \(\hat D=\pi_1Z_1+\pi_2Z_2\) to obtain the Wald estimand $$\beta_{\text{TSLS}}=\frac{\operatorname{Cov}(Y,\hat D)}{\operatorname{Cov}(D,\hat D)}$$ Notice that this gives us $$\begin{aligned}\beta_{\text{TSLS}}&=\pi_1\frac{\operatorname{Cov}(Y,Z_1)}{\operatorname{Cov}(D,\hat D)}+\pi_2\frac{\operatorname{Cov}(Y,Z_2)}{\operatorname{Cov}(D,\hat D)}=\underbrace{\pi_1\frac{\operatorname{Cov}(D,Z_1)}{\operatorname{Cov}(D,\hat D)}}_{\equiv\psi}\frac{\operatorname{Cov}(Y,Z_1)}{\operatorname{Cov}(D,Z_1)}+\underbrace{\pi_2\frac{\operatorname{Cov}(D,Z_2)}{\operatorname{Cov}(D,\hat D)}}_{=1-\psi}\frac{\operatorname{Cov}(Y,Z_2)}{\operatorname{Cov}(D,Z_2)}\\&=\psi\beta_1+(1-\psi)\beta_2\end{aligned}$$ where $$\psi\equiv\pi_1\frac{\operatorname{Cov}(D,Z_1)}{\operatorname{Cov}(D,\hat D)}=\frac{\pi_1\operatorname{Cov}(D,Z_1)}{\pi_1\operatorname{Cov}(D,Z_1)+\pi_2\operatorname{Cov}(D,Z_2)}$$ is the relative strength of \(Z_1\) (i.e. the fraction of population moved towards treatment by \(Z_1\) in the first stage). So \(\beta_{\text{TSLS}}\) is the weighted average of the instrument-specific LATEs. Notice that the term \(\operatorname{Cov}(D,Z_i)\) means that higher weight is assigned to the instrument that has a high covariance with \(D\), i.e. we give a higher weight to the instrument that shifts more people around.

Monotonicity becomes more difficult to impose and more restrictive with multiple binary instruments. What does monotonicity even mean with multiple instruments? Intuitively, monotonicity in the multiple binary instrument case means that moving from any combination of instrument values to another combination of instrument values, agents should respond (in terms of taking treatment or not) in a weakly same direction. Precisely, for the vector of instruments \(\mathbf Z=(Z_1,\dots,Z_m)'\), for any pair of values of \(\mathbf Z\), i.e. \(\mathbf z,\mathbf z'\), either \(D_\mathbf z\ge D_{\mathbf z'}\) for all \(i\) or \(D_\mathbf z\le D_{\mathbf z'}\) for all \(i\). Equivalently, denote the set of instrument values that makes individual \(i\) take treatment by \(\mathbb Z_i=\{\mathbf z\in\mathbf Z:D_\mathbf z=1\}\); then for any pair of individuals \(i\) and \(j\), either \(\mathbb Z_i\subseteq\mathbb Z_j\) or \(\mathbb Z_j\subseteq\mathbb Z_i\).

Suppose in the world where treatment varies in intensity. Assume treatment \(S\) is no longer binary but varies in its level, i.e. $$S=\{0,1,2,\dots,J\}$$ For example, the treatment \(S\) now can be the years of schooling. We can then define potential outcomes indexed by the level of treatment, i.e. \(Y_s\) for \(s\in S\), and potential treatments are as before indexed by the value of the instrument, i.e. \(S_Z\). In the case of binary instruments (conditional on \(X=x\)), $$\beta(x)=\frac{\mathbb E[Y\mid Z=1,X=x]-\mathbb E[Y\mid Z=0,X=x]}{\mathbb E[D\mid Z=1,X=x]-\mathbb E[D\mid Z=0,X=x]}=\mathbb E[Y_1-Y_0\mid D_1>D_0,X=x]$$ $$S=S_1Z+S_0(1-Z)$$ The observed outcome is $$Y=\sum_{s=0}^J Y_s\cdot\mathbf 1\{S=s\}=Y_0+\sum_{s=1}^J(Y_s-Y_{s-1})\cdot\mathbf 1\{S\ge s\}$$ and the average effect of the \(s\)-th year of schooling is \(\mathbb E[Y_s-Y_{s-1}]\), with \(J\) different treatment effects. Assumptions: 1. Random assignment: \(Y_{s,z},S_z\perp Z\) for all \(s,z\). 2. Exclusion restriction: \(Y_{s,z}=Y_s\). 3. Monotonicity: \(S_1\ge S_0\) for all \(i\) or \(S_1\le S_0\) for all \(i\). 4. Instrument relevance: \(\mathbb E[S_1-S_0]\ne0\).

Suppose \(S=\{0,1,2\}\) and \(S_1\ge S_0\) for all \(i\). Monotonicity \(S_1\ge S_0\) implies that for all \(s\in S\), \(\mathbf 1\{S_1\ge s\}-\mathbf 1\{S_0\ge s\}\in\{0,1\}\), so that $$\mathbb P(\mathbf 1\{S_1\ge s\}>\mathbf 1\{S_0\ge s\})=\mathbb P(S_1\ge s>S_0)$$ i.e. if \(\mathbb P(S_1\ge s>S_0)\) is greater than zero, then the instrument \(Z\) affects the incidence of treatment level \(s\). In particular $$\mathbb E[S\mid Z=1]-\mathbb E[S\mid Z=0]=\mathbb E[S_1-S_0]=\mathbb P(S_1\ge1>S_0)\times1+\mathbb P(S_1\ge2>S_0)\times1=\mathbb P(S_1\ge1>S_0)+\mathbb P(S_1\ge2>S_0)$$ Instrument relevance requires \(\mathbb E[S\mid Z=1]-\mathbb E[S\mid Z=0]>0\). The outcome \(Y=Y_0+(Y_1-Y_0)\mathbf 1\{S\ge1\}+(Y_2-Y_1)\mathbf 1\{S\ge2\}\). Expanding the reduced form (regressing outcome \(Y\) directly on instrument \(Z\)) gives $$\mathbb E[Y\mid Z=1]-\mathbb E[Y\mid Z=0]=\mathbb E[Y_1-Y_0\mid S_1\ge1>S_0]\mathbb P(S_1\ge1>S_0)+\mathbb E[Y_2-Y_1\mid S_1\ge2>S_0]\mathbb P(S_1\ge2>S_0)$$ which gives us the Wald estimand $$\beta^{\text{IV}}=\sum_{s=1}^2\omega_s\mathbb E[Y_s-Y_{s-1}\mid S_1\ge s>S_0]\tag{16.10}$$ $$\omega_s=\frac{\mathbb P(S_1\ge s>S_0)}{\sum_{j=1}^2\mathbb P(S_1\ge j>S_0)}$$

By exactly the same argument as for (16.10), the Wald estimand in the general case where \(S=\{0,1,2,\dots,J\}\) is $$\beta^{\text{IV}}=\frac{\mathbb E[Y\mid Z=1]-\mathbb E[Y\mid Z=0]}{\mathbb E[S\mid Z=1]-\mathbb E[S\mid Z=0]}=\sum_{s=1}^J\omega_s\mathbb E[Y_s-Y_{s-1}\mid S_1\ge s>S_0]$$ $$\omega_s=\frac{\mathbb P(S_1\ge s>S_0)}{\sum_{j=1}^J\mathbb P(S_1\ge j>S_0)}$$ which we call the average causal response (ACR). We can estimate \(\omega_s\) since $$\begin{aligned}\mathbb P(S_1\ge s>S_0)&=\mathbb P(S_1\ge s,S_0

In this section we focus on binary treatment and single binary instrument. Under most non-experimental set-ups, we cannot justify unconditional random assignment. Instead, we would argue for independence conditional on covariates \(X\) (as a vector in general), i.e. \((Y_1,Y_0,D_1,D_0)\perp Z\mid X\).

16.5.1 Adjusted Assumptions

For the model with covariates, we assume: - Conditional random assignment (or exogeneity): \((Y_1,Y_0,D_1,D_0)\perp Z\mid X\). This means in each cell \(X=x\), instrument \(Z\) is as good as randomly assigned. - Relevance: \(\mathbb P(D=1\mid Z=1,X)\ne\mathbb P(D=1\mid Z=0,X)\). This means in any cell \(X=x\), instrument \(Z\) has non-zero effect on people's probability of taking treatment. - Monotonicity: \(\mathbb P(D_1\ge D_0\mid X)=1\). This means in any cell \(X=x\), instrument affects people's probability of taking treatment in weakly same direction. - Overlap: \(\mathbb P(Z=1\mid X)\in(0,1)\). This means in any cell \(X=x\), we have people with both \(Z=1\) and \(Z=0\), so that the LATE\((x)\) within each cell is well-defined.

16.5.2 Stratified LATEs

The stratified LATE \(\beta(x)\) is defined by $$\text{LATE}(x)\equiv\beta(x)=\frac{\mathbb E[Y\mid Z=1,X=x]-\mathbb E[Y\mid Z=0,X=x]}{\mathbb E[D\mid Z=1,X=x]-\mathbb E[D\mid Z=0,X=x]}=\mathbb E[Y_1-Y_0\mid D_1>D_0,X=x]\tag{16.11}$$ which is proved by re-doing the proof of Proposition 16.1 in §16.1.3 within the cell \(X=x\).

So we can identify LATE\((x)=\beta(x)\) within each cell \(X=x\), and then estimate \(\beta(x)\) by non-parametric methods.

In the discrete \(X\) case, \(X\) can only take \(K\) finite number of discrete values, so we can break it down to \(K\) binary dummy variables. Do the following TSLS regression $$Y=\beta D+\sum_x\alpha_x\cdot\mathbf 1\{X=x\}+e\tag{16.15}$$ $$D=\pi_x Z+\gamma_x+u$$ which assumes separability in the effect on outcome \(Y\) of covariates \(X\) and treatment \(D\). Angrist and Imbens (1995) show that $$\beta^{\text{TSLS}}=\mathbb E[\beta(x)\omega(x)]$$ where $$\omega(x)=\frac{\sigma_D^2(x)}{\mathbb E[\sigma_D^2(x)]}=\frac{\pi_x^2\sigma_Z^2(x)}{\mathbb E[\pi_x^2\sigma_Z^2(x)]}$$

Abadie's (2003) \(\kappa\) is a more elegant approach to the conditional independence case. The idea is to run regressions only on the compliers. Although compliers cannot be directly observed, we can still use the correct weights of compliers. Abadie showed that, in the case with binary \(D\) and \(Z\), for any function \(g(Y,X,D)\), $$\mathbb E[g(Y,X,D)\mid\text{Complier}]=\frac{\mathbb E[\kappa\times g(Y,X,D)]}{\mathbb P(\text{Complier})}\tag{16.18}$$ where $$\kappa=1-\frac{D(1-Z)}{\mathbb P(Z=0\mid X)}-\frac{(1-D)Z}{\mathbb P(Z=1\mid X)}$$

Use Abadie's \(\kappa\). For example, for the OLS regression among compliers \(Y=\alpha D+X'\beta+\varepsilon\), we want to solve $$(\hat\alpha,\hat\beta)=\arg\max_{\alpha,\beta}\mathbb E\big[\underbrace{(Y-\alpha D-X'\beta)^2}_{\equiv g(Y,X,D)}\mid\text{Complier}\big]$$ (16.18) tells us that we can estimate \((\hat\alpha,\hat\beta)\) directly among compliers even though we don't really know who the compliers are: $$\mathbb E[(Y-\alpha D-X'\beta)^2\mid\text{Complier}]=\frac{\mathbb E[\kappa\times(Y-\alpha D-X'\beta)^2]}{\mathbb P(\text{Complier})}\Rightarrow(\hat\alpha,\hat\beta)=\arg\max_{\alpha,\beta}\mathbb E[\kappa\times(Y-\alpha D-X'\beta)^2]$$

In some cases, individuals may choose from a treatment with multiple values but in an unordered manner, which is different from ordered multiple treatments discussed in §16.4. For example, multiple treatment could be the different occupations, education types and locations etc., which are hard to be ranked in terms of effect strength on outcome.

Example with 3 treatment choices. Suppose \(D\in\{0,1,2\}\). Break unordered \(D\) into two separate binary treatment variables \(D_1,D_2\) where \(D_1=\mathbf 1\{D=1\}\) and \(D_2=\mathbf 1\{D=2\}\). The regression $$Y=\beta_0+\beta_1D_1+\beta_2D_2+\varepsilon$$ or equivalently, define \(\Delta^1\equiv Y^1-Y^0\) and \(\Delta^2\equiv Y^2-Y^0\), $$Y=Y^0+D_1(Y^1-Y^0)+D_2(Y^2-Y^0)=\underbrace{\mathbb E[Y^0]}_{=\beta_0}+\beta_1D_1+\beta_2D_2+\underbrace{Y_0-\mathbb E[Y^0]+(\Delta^1-\beta_1)D_1+(\Delta^2-\beta_2)D_2}_{=\varepsilon}$$ The instrument \(Z\in\{0,1,2\}\) is replaced by \(Z_1=\mathbf 1\{Z=1\}\) and \(Z_2=\mathbf 1\{Z=2\}\). Assumptions: 1. Exclusion: \(Y^{d,z}=Y^d\) for all \(d,z\). 2. Independence: \(Y^0,Y^1,Y^2,D^0,D^1,D^2\perp Z\). 3. Rank condition: \(\operatorname{rank}(\mathbb E[ZD'])=3\). 4. Monotonicity: \(D_1^1\ge D_1^0\) and \(D_2^2\ge D_2^0\).

Exclusion and independence together imply exogeneity, i.e. \(\mathbb E[\varepsilon Z_1]=\mathbb E[\varepsilon Z_2]=\mathbb E[\varepsilon]=0\). The first-stage regressions are \(D_1=\alpha_1+\gamma_1Z_1+\eta_1Z_2+v_1\), \(D_2=\alpha_2+\gamma_2Z_1+\eta_2Z_2+v_2\). From \(\mathbb E[\varepsilon Z_1]=0\) and \(\mathbb E[\varepsilon Z_2]=0\) respectively we get $$\mathbb E[D_1^1(\Delta^1-\beta_1)+D_2^1(\Delta^2-\beta_2)]=0\tag{16.19}$$ $$\mathbb E[D_1^2(\Delta^1-\beta_1)+D_2^2(\Delta^2-\beta_2)]=0\tag{16.20}$$ Under the constant effect assumption \(\Delta^1,\Delta^2\) are constants, so (16.19) and (16.20) give \(\beta_1=\Delta^1\), \(\beta_2=\Delta^2\). Under heterogeneous effects, further assume \(D_1^2=D_2^1=0\) (e.g. students apply to a baseline field 0 but can always apply to baseline field 0, which corresponds to \(D_1^1=0\) and \(D_2^2=0\)); then $$\beta_1=\mathbb E[\Delta^1\mid D_1^1=1],\qquad\beta_2=\mathbb E[\Delta^2\mid D_2^2=1]$$ i.e. the LATE for some subgroup of people.

§16.7.1 Weak instruments. An instrumental variable is weak if its correlation with the included endogenous regressor is small. In the potential outcome model we are talking about, the instrument \(Z\) is weak if \(\gamma\approx0\) in the first-stage regression \(D=\alpha+\gamma Z+v\). A weak instrument is a problem because the denominator of the Wald estimand \(\beta^{\text{IV}}=\frac{\operatorname{Cov}(Z,Y)}{\operatorname{Cov}(Z,D)}\) would be approximately zero, which makes \(\beta^{\text{IV}}\) very volatile. - To identify a weak instrument: check the \(F\) statistic for the instrument in the first-stage regression. - If the \(F\) statistic exceeds 10 (or some other arbitrary number), proceed with the standard IV or TSLS methods. - If the \(F\) statistic is below 10: think hard to explain why the instrument is weak; if it's still a theoretically good instrument but just weak, then just report the weak-instrument-robust confidence sets — they give the correct coverage even when the instrument is weak, do not require screening on the first stage so avoid pretesting bias, avoid throwing away good instruments just because they are weak, and can still be very informative.

§16.7.2 Many instruments. When people are faced with the weak instrument problem, they may typically add in more instruments to solve the "weak", but it induces another problem: many instruments (or over-identifying). Over-identifying is indeed a problem because when we are including many instruments (or too powerful an instrument), the instruments, combined together, look almost like the endogenous regressor, so the exogeneity assumption is broken. We may suspect this problem when \(\beta^{\text{IV}}\approx\beta^{\text{OLS}}\). - Example: \(Y=\beta_0+\beta_1D+\varepsilon\), \(D=\alpha+\gamma Z+v\). The instrument \(Z\) is using the exogenous part of \(D\), i.e. \(\hat D=\alpha+\gamma Z\), to unbiasedly estimate \(\beta_1\). The instrument can neither be too strong, nor too weak: if too weak, then the estimator \(\beta^{\text{IV}}\) or \(\beta^{\text{TSLS}}\) will have too large variance; if too strong, the \(\hat D\) will be almost the same as \(D\), which makes \(\hat D\) also contain some endogenous parts.

• Single binary \(D\) and single binary \(Z\) case: the IV estimand \(\beta^{\text{IV}}\) is the LATE, i.e. the ATE among compliers to instrument \(Z\). This LATE could be relevant for a policy intervention that affects these compliers.
• Multiple binary \(Z\)'s for a single binary \(D\) case: \(\mathbf Z=(Z_1,\dots,Z_m)'\) is a vector. IV won't work because \(\mathbb E[\mathbf Z D]\) is not invertible. We instead use TSLS to estimate \(D=\alpha+\gamma'\mathbf Z+v\) and obtain \(\hat D=\alpha+\gamma'\mathbf Z\), and then regress \(Y\) on \(\hat D\) to obtain an unbiased estimate of \(\beta_1\) in \(Y=\beta_0+\beta_1D+\varepsilon\). It's hard to intuitively interpret the \(\beta^{\text{TSLS}}\) here. But in fact, it is a weighted average of LATEs of compliers group to \(Z_1,\dots,Z_m\).
• Ordered multiple treatment \(S=\{0,1,\dots,J\}\): the Wald estimand is the average causal response (ACR), \(\beta^{\text{IV}}=\sum_s\omega_s\mathbb E[Y_s-Y_{s-1}\mid S_1\ge s>S_0]\). Notice that although \(\omega_s\) is positive and sums up to 1, the Wald estimand is not a weighted average of effects of mutually exclusive subgroups of people (overlapping group, counted multiple times).
• Unordered multiple treatment: break down the different values of treatments into several binary treatments, and use at least one binary instrument for each of these binary treatments. Under constant treatment effect assumption, OLS gives us the unbiased estimates; under heterogeneous effect assumption, IV gives us the LATE in some subgroup of people.
• Conditional random assignment case: we can first obtain the stratified LATEs \(\beta(x)\) by non-parametric method; then we can either further use TSLS to obtain a weighted average of stratified LATEs \(\beta(x)\), where the weights are related to the local variance of treatment in the cell; or, we can use Abadie's (2003) \(\kappa\) to run OLS among the compliers group.

Some general advice on using IV: - Reverse engineering could be useful: start with an estimator with a particular IV, and then think about who the compliers are for that IV, and think about policy relevance of that estimation. - Motivate instruments: motivate exclusion and independence for the IV, which requires good theoretical or logical justification. - exclusion: why does \(Z\) only affects \(Y\) through \(D\)? Is there any possibility that \(Z\) has other channels to affect \(Y\) (i.e. violates exclusion)? - independence: how is \(Z\) generated? Is \(Z\) as good as randomly assigned to agents? If not, what could be the covariates to control for conditional independence and what the weighted average stratified LATEs across \(X=x\) cells mean? - interpret who the compliers are and argue for policy relevance. - Check the instruments: - Always report the first stage result (discuss whether or not the magnitude and signs are as expected; report \(F\) statistic on instruments and check whether the instruments are weak, and if so, report weak instrument robust confidence intervals). - Inspect the reduced form regression (directly regress outcome \(Y\) on instruments \(Z\) without endogenous regressor \(D\)): check the magnitude and sign (the coefficient on instrument in reduced form regression is proportional to the causal effect of \(D\) on \(Y\) by the coefficient in the first stage, so the magnitude and sign put together should be consistent with expectation).