Instrumental variable estimation of weighted local average treatment effects

Abstract

Instrumental variable (IV) analysis addresses bias owing to unmeasured confounding when comparing two nonrandomized treatment groups. To date, studies in the statistical and biomedical literature have focused on the local average treatment effect (LATE), the average treatment effect for compliers. In this article, we study the weighted local average treatment effect (WLATE), which represents the weighted average treatment effect for compliers. In the WLATE, the population of interest is determined by either the instrumental propensity score or compliance score, or both. The LATE is a special case of the proposed WLATE, where the target population is the entire population of compliers. Here, we discuss the interpretation of a few special cases of the WLATE, identification results, inference methods, and optimal weights. We demonstrate the proposed methods with two published examples in which considerations of local causal estimands that deviate from the LATE are beneficial.

Appendices

Appendix A: Proof for Eq. (11)

Let $f_z(x)=\text{ pr }(X=x\mid Z=z)$, which is equal to $f(x)\text{ pr }(Z=z\mid X=x)/\text{pr }(Z=z)$. Based on Bayes’ theorem, for $z=\{0,1\}$, $f_z^c(x) = \text{ pr }(X=x\mid Z=z, U=c)$ can be written as

$$\begin{aligned} f_z^c(x)= & {} \frac{\text{ pr }(U=c\mid X=x,Z=z)}{\text{ pr }(U=c\mid Z=z)}f_z(x), \nonumber \\= & {} \frac{\text{ pr }(U=c\mid X=x)}{\text{ pr }(U=c\mid Z=z)}\frac{f(x)\text{ pr }(Z=z\mid X=x)}{\text{ pr }(Z=z)}, \end{aligned}$$

(A1)

$$\begin{aligned}\propto & {} \text{ pr }(U=c\mid X=x)f(x)\text{ pr }(Z=z\mid X=x). \end{aligned}$$

(A2)

Equation (A1) holds because U is independent of Z conditional on X by Assumption 1. Using Bayes’ theorem, $f^c(x) = \text{ pr }(U=c\mid X=x)f(x)/\text{pr }(U=c)$. Therefore, Eq. (A2) becomes

$$\begin{aligned} f_z^c(x) \propto f^c(x)\text{ pr }(Z=z\mid X=x), \end{aligned}$$

which implies Eq. (11).

Appendix B: Proof for Theorem 2

We seek to determine h(x) that minimizes

$$\begin{aligned} \frac{1}{E\{\delta (X)h(X)\}^2}E\left[ h(X)^2\left\{ \frac{\sigma ^{2}_{h,1}(X)}{e(X)} + \frac{\sigma ^{2}_{h,0}(X)}{1-e(X)}\right\} \right] . \end{aligned}$$

Let $k(X) = \sigma ^{2}_{h,1}(X)/e(X) + \sigma ^{2}_{h,0}(X)/\{1-e(X)\}$ and $g(X) = \delta (X)h(X)$. Then we attempt to find g(x) that minimizes

$$\begin{aligned} \frac{1}{(\int g(x)f(x)dx)^2} \int g(x)^2 \{k(x)/\delta (x)^2\} f(x)dx. \end{aligned}$$

We can normalize g(x) to satisfiy $\int g(x)f(x)dx=1$. Then, our problem becomes the minimization of

$$\begin{aligned} \int g(x)^2 \{k(x)/\delta (x)^2\} f(x)dx\ \text{ such } \text{ that }\ \int g(x)f(x)dx=1. \end{aligned}$$

The solution should satisfy $0 = 2\,g(x) \{k(x)/\delta (x)^2\} f(x) - \lambda f(x)$; thus, the solution g(x) is proportional to $\delta (x)^2/k(x)$. Therefore, the solution of h(x) is $\delta (x)/k(x)$.

Appendix C: Proof for Theorem 3

Under the conditions of Theorems 1 and 3, based on the results of Section 4.2. of Hirano et al. (2003), we have $\sqrt{n}({\hat{\tau _h}} - \tau _h) = (1/\sqrt{n})\sum _{i=1}^n t(Y_i,D_i,Z_i,X_i)+o_p(1)$ and $\sqrt{n}({\hat{\delta _h}} - \delta _h) = (1/\sqrt{n})\sum _{i=1}^n \pi (Y_i,D_i,Z_i,X_i)+o_p(1)$, where

$$\begin{aligned} t(Y_i,D_i,Z_i,X_i)= & {} \frac{h(X_i)}{E\{h(X_i)\}}\left\{ \frac{Z_iY_i}{e(X_i)} - \frac{(1-Z_i)Y_i}{1-e(X_i)} - \tau _h \right\} \nonumber \\- & {} \frac{h(X_i)}{E\{h(X_i)\}}\left\{ \frac{m_1(X_i)}{e(X_i)} + \frac{m_0(X_i)}{1-e(X_i)} \right\} (Z_i - e(X_i)), \end{aligned}$$

(C1)

$$\begin{aligned} \pi (Y_i,D_i,Z_i,X_i)= & {} \frac{h(X_i)}{E\{h(X_i)\}}\left\{ \frac{Z_iD_i}{e(X_i)} - \frac{(1-Z_i)D_i}{1-e(X_i)} - \delta _h \right\} \nonumber \\- & {} \frac{h(X_i)}{E\{h(X_i)\}}\left\{ \frac{q_1(X_i)}{e(X_i)} + \frac{q_0(X_i)}{1-e(X_i)} \right\} (Z_i - e(X_i)). \end{aligned}$$

(C2)

A first-order Taylor expansion of ${\hat{\tau _h}}/{\hat{\delta _h}}$ around the point $(\tau _h, \delta _h)$ yields

$$\begin{aligned} \sqrt{n}({\hat{\theta _h^c}} - \theta _h^c) = \sqrt{n}\left( \frac{{\hat{\tau _h}}}{{\hat{\delta _h}}}-\frac{\tau _h}{\delta _h} \right) =\frac{1}{\delta _h}\left\{ \sqrt{n}({\hat{\tau _h}} - \tau _h) - \theta _h^c \sqrt{n}({\hat{\delta _h}} - \delta _h) \right\} + o_p(1).\nonumber \\ \end{aligned}$$

(C3)

Applying Eqs. (C1) and (C2) to Eq. (C3) gives

$$\begin{aligned} \sqrt{n}({\hat{\theta _h^c}} - \theta _h^c) = \frac{1}{\sqrt{n}} \sum _{i=1}^n \psi (Y_i,D_i,Z_i,X_i) + o_p(1). \end{aligned}$$

(C4)

Applying the Lindeberg–Levy central limit theorem to Eq. (C4) gives the asymptotic normal result in Theorem 3.

Appendix D: Proof for Theorem 5

It suffices to show that ${\hat{\tau _h}}(1)$ and ${\hat{\tau _h}}(0)$ are consistent estimators for $\tau _h(1)$ and $\tau _h(0)$. The following equalities hold because $Y = DY(1)+(1-D)Y(0)$, $D=ZD(1)+(1-Z)D(0)$, and Z is independent of all potential outcome and treatment variables conditional on X.

$$\begin{aligned} E(YD\mid X=x,Z=1)= & {} E\{Y(1)D(1)\mid X=x\},\\ E(YD\mid X=x,Z=0)= & {} E\{Y(1)D(0)\mid X=x\},\\ E\{Y(1-D) \mid X=x,Z=0\}= & {} E\{Y(0)(1-D(0))\mid X=x\},\\ E\{Y(1-D)\mid X=x,Z=1\}= & {} E\{Y(0)(1-D(1))\mid X=x\}. \end{aligned}$$

Hence, we can write $\tau _h(1)$ as

$$\begin{aligned}{} & {} \frac{\int E\left\{ Y(1)D(1)Z\frac{h(x)}{e(x)} \mid X=x\right\} dF(x)}{\int E\left\{ Z\frac{h(x)}{e(x)}\mid X=x\right\} dF(x)} \\{} & {} \quad - \frac{\int E\left\{ Y(1)D(0)(1-Z)\frac{h(x)}{1-e(x)}\mid X=x\right\} dF(x)}{\int E\left\{ (1-Z)\frac{h(x)}{1-e(x)}\mid X=x\right\} dF(x)}. \end{aligned}$$

Therefore, we obtain the following consistent estimator for $\tau _h(1)$:

$$\begin{aligned} {\hat{\tau _h}}(1) = \frac{\sum _i W_iZ_iY_i(1)D_i(1)}{\sum _i W_iZ_i} - \frac{\sum _i W_i(1-Z_i)Y_i(1)D_i(0)}{\sum _i W_i(1-Z_i)}. \end{aligned}$$

We can show that $Z_iD_i(1)Y_i(1) = Z_iD_iY_i$ and $(1-Z_i)D_i(0)Y_i(1) = (1-Z_i)D_iY_i$. This gives the estimator ${\hat{\tau _h}}(1)$ in Eq. (18).

We can write $\tau _h(0)$ as

$$\begin{aligned}{} & {} \frac{\int E\left\{ \frac{Y(0)(1-D(0))(1-Z)h(x)}{1-e(x)} \mid X=x\right\} dF(x)}{\int E\left\{ \frac{(1-Z)h(x)}{1-e(x)}\mid X=x\right\} dF(x)} \\{} & {} \quad - \frac{\int E\left\{ \frac{Y(0)(1-D(1))Zh(x)}{e(x)}\mid X=x\right\} dF(x)}{\int E\left\{ \frac{Zh(x)}{e(x)}\mid X=x\right\} dF(x)}. \end{aligned}$$

Therefore, we obtain the following consistent estimator for $\tau _h(0)$:

$$\begin{aligned} {\hat{\tau _h}}(0) = \frac{\sum _i W_i(1-Z_i)Y_i(0)(1-D_i(0))}{\sum _i W_i(1-Z_i)} - \frac{\sum _i W_iZ_iY_i(0)(1-D_i(1))}{\sum _i W_iZ_i}. \end{aligned}$$

It can be shown that $(1-Z_i)(1-D_i(0))Y_i(0) = (1-Z_i)(1-D_i)Y_i$ and $Z_i(1-D_i(1))Y_i(0) = Z_i(1-D_i)Y_i$. This gives the estimator ${\hat{\tau _h}}(0)$ in Eq. (19).

Appendix E: Figures

See Figures 6, 7, and 8.

Fig. 6

Percentage bias of various local average treatment effect estimators when there is a good overlap in IPS distributions $(\gamma =1)$. The columns indicate the simulations when $P(U=c)$ is 0.1, 0.3, and 0.5. The rows indicate the simulations when $\eta$ is 1, 2, and 3. The red vertical line indicates the value of 0

Fig. 7

Percentage bias of various local average treatment effect estimators when there is a limited overlap in IPS distributions $(\gamma =2)$. The rows indicate the simulations when $\eta$ is 1, 2, and 3. The red vertical line indicates the value of 0

Fig. 8

Percentage bias of various local average treatment effect estimators when there is a severely limited overlap in IPS distributions $(\gamma =3)$. The rows indicate the simulations when $\eta$ is 1, 2, and 3. The red vertical line indicates the value of 0