Disentangle Mixture Distributions and Instrumental Variable Inequalities

$$ \begin{aligned} \ \end{aligned} $$

Introduction

• The IV model in last chapter imposes three assumptions

  • (randomization) $Z \perp\!\!\!\perp\{D(1), D(0), Y(1), Y(0)\}$
  • (monotonicity) $\operatorname{pr}(U=\mathrm{d})=0$ or $D_i(1)\geq D_i(0)$
  • (exclusionrestriction) $Y(1)=Y(0) \text { for } U=\mathrm{a} \text { or } \mathrm{n}$

$$ {\tiny U_i= \begin{cases} a, & \text { if } D_i(1)=1 \text { and } D_i(0)=1 \\ c, & \text { if } D_i(1)=1 \text { and } D_i(0)=0 \\ d, & \text { if } D_i(1)=0 \text { and } D_i(0)=1 \\ n, & \text { if } D_i(1)=0 \text { and } D_i(0)=0 \end{cases} } $$

• Observed groups and latent groups under the assumptions

$$ {\small \begin{array}{cccl} Z=1 & D=1 & D(1)=1 & U=\mathrm{c} \text { or a } \\ Z=1 & D=0 & D(1)=0 & U=\mathrm{n} \\ Z=0 & D=1 & D(0)=1 & U=\mathrm{a} \\ Z=0 & D=0 & D(0)=0 & U=\mathrm{c} \text { or } \mathrm{n} \end{array} } $$

• Interestingly, the assumptions have some testable implications. Balke and Pearl (1997) called them the instrumental variable inequalities.

Outline

• Disentangle Mixture Distributions and Instrumental Variable Inequalities

• Testable implications

• Examples

Disentangle Mixture Distributions and IV Inequalities

• Recall $\pi_u$ as the proportion of type $U=u$, and define

$$ \mu_{z u}=E\{Y(z) \mid U=u\}, \quad(z=0,1 ; u=\mathrm{a}, \mathrm{n}, \mathrm{c}) . $$

Theorem 22.1 Under the three assumptions, we can identify the proportions of the latent types by $$ \begin{aligned} & \pi_{\mathrm{n}}=\operatorname{pr}(D=0 \mid Z=1), \\ & \pi_{\mathrm{a}}=\operatorname{pr}(D=1 \mid Z=0), \\ & \pi_{\mathrm{c}}=E(D \mid Z=1)-E(D \mid Z=0), \end{aligned} $$ and the type-specific means of the potential outcomes by $$ \begin{aligned} \mu_{1 \mathrm{n}}=\mu_{0 \mathrm{n}} \equiv \mu_{\mathrm{n}} & =E(Y \mid Z=1, D=0) \\ \mu_{1 \mathrm{a}}=\mu_{0 \mathrm{a}} \equiv \mu_{\mathrm{a}} & =E(Y \mid Z=0, D=1) \\ \mu_{1 \mathrm{c}} & =\pi_{\mathrm{c}}^{-1}\{E(D Y \mid Z=1)-E(D Y \mid Z=0)\} \\ \mu_{0 \mathrm{c}} & =\pi_{\mathrm{c}}^{-1}[E\{(1-D) Y \mid Z=0\}-E\{(1-D) Y \mid Z=1\}] \end{aligned} $$

Proof of Theorem 22.1 (Part I)

• We can identify the proportion of the never takers by

$$ \begin{aligned} \operatorname{pr}(D=0 \mid Z=1) & =\operatorname{pr}(U=\mathrm{n} \mid Z=1)=\operatorname{pr}(U=\mathrm{n})=\pi_{\mathrm{n}}, \end{aligned} $$

• the proportion of the always takers by

$$ \begin{aligned} \operatorname{pr}(D=1 \mid Z=0) & =\operatorname{pr}(U=\mathrm{a} \mid Z=0)=\operatorname{pr}(U=\mathrm{a})=\pi_{\mathrm{a}} . \end{aligned} $$

• the proportion of compliers is

$$ \begin{aligned} \pi_{\mathrm{c}} & =\operatorname{pr}(U=\mathrm{c})=1-\pi_{\mathrm{n}}-\pi_{\mathrm{a}} \\ & =1-\operatorname{pr}(D=0 \mid Z=1)-\operatorname{pr}(D=1 \mid Z=0) \\ & =E(D \mid Z=1)-E(D \mid Z=0)=\tau_D, \end{aligned} $$

Remark: Although we do not know individual latent compliance types for all units, we can identify the proportions of never takers, always takers, and compliers.

Proof of Theorem 22.1 (Part II)

Under the three assumptions, we have

$$ \mu_{1 \mathrm{a}}=\mu_{0 \mathrm{a}} \equiv \mu_{\mathrm{a}}, \quad \mu_{1 \mathrm{n}}=\mu_{0 \mathrm{n}} \equiv \mu_{\mathrm{n}} . $$

• The observed group $(Z=1, D=0)$ only has never takers, so

$$ E(Y \mid Z=1, D=0)=E\{Y(1) \mid Z=1, U=\mathrm{n}\}=E\{Y(1) \mid U=\mathrm{n}\}=\mu_{\mathrm{n}} . $$

• The observed group $(Z=0, D=1)$ only has always takers, so

$$ E(Y \mid Z=0, D=1)=E\{Y(0) \mid Z=0, U=\mathrm{a}\}=E\{Y(0) \mid U=\mathrm{a}\}=\mu_{\mathrm{a}} $$

$$~~$$

How about the observed groups $(Z=1, D=1)$ and $(Z=0, D=0)$ ?

The Observed Group $(Z=1, D=1)$

The observed group $(Z=1, D=1)$ has both compliers and always takers, so $$ \small \begin{aligned} E(Y \mid Z=1, D=1)= & E\{Y(1) \mid Z=1, D(1)=1\} \\ = & E\{Y(1) \mid D(1)=1\} \\ = & \operatorname{pr}\{D(0)=1 \mid D(1)=1\} E\{Y(1) \mid D(1)=1, D(0)=1\} \\ & +\operatorname{pr}\{D(0)=0 \mid D(1)=1\} E\{Y(1) \mid D(1)=1, D(0)=0\} \\ = & \frac{\pi_{\mathrm{c}}}{\pi_{\mathrm{c}}+\pi_{\mathrm{a}}} \mu_{1 \mathrm{c}}+\frac{\pi_{\mathrm{a}}}{\pi_{\mathrm{c}}+\pi_{\mathrm{a}}} \mu_{\mathrm{a}} . \end{aligned} $$

Solve the linear equation above to obtain $$ \small \begin{aligned} \mu_{1 \mathrm{c}}= & \pi_{\mathrm{c}}^{-1}\left\{\left(\pi_{\mathrm{c}}+\pi_{\mathrm{a}}\right) E(Y \mid Z=1, D=1)-\pi_{\mathrm{a}} E(Y \mid Z=0, D=1)\right\} \\ = & \pi_{\mathrm{c}}^{-1}\{\operatorname{pr}(D=1 \mid Z=1) E(Y \mid Z=1, D=1) \\ & \quad-\operatorname{pr}(D=1 \mid Z=0) E(Y \mid Z=0, D=1)\} \\ = & \pi_{\mathrm{c}}^{-1}\{E(D Y \mid Z=1)-E(D Y \mid Z=0)\} . \end{aligned} $$

Based on the formulas of $\mu_{1 \mathrm{c}}$ and $\mu_{0 \mathrm{c}}$ in Theorem 22.1, we have

$$ \tau_{\mathrm{c}}=\mu_{1 \mathrm{c}}-\mu_{0 \mathrm{c}}=\{E(Y \mid Z=1)-E(Y \mid Z=0)\} / \pi_{\mathrm{c}}. $$

Testable implications

• Is there any additional value of the this detour for deriving the formula of $\tau_c$ ?
• For binary outcome, the following inequalities must be true:

$$ 0 \leq \mu_{1 \mathrm{c}} \leq 1, \quad 0 \leq \mu_{0 \mathrm{c}} \leq 1, $$

• It implies four inequalities

$$ \small \begin{aligned} E(D Y \mid Z=1)-E(D Y \mid Z=0) & \geq 0, \\ E(D Y \mid Z=1)-E(D Y \mid Z=0) & \leq E(D \mid Z=1)-E(D \mid Z=0), \\ E\{(1-D) Y \mid Z=0\}-E\{(1-D) Y \mid Z=1\} & \geq 0, \\ E\{(1-D) Y \mid Z=0\}-E\{(1-D) Y \mid Z=1\} & \leq E(D \mid Z=1)-E(D \mid Z=0) . \end{aligned} $$

Theorem 22.2 (Instrumental Variable Inequalities) With a binary outcome $Y$, the three assumptions imply

$$ E(Q \mid Z=1)-E(Q \mid Z=0) \geq 0, $$

where $Q=D Y, D(1-Y),(D-1) Y$ and $D+Y-D Y$.

Remarks

• Under the three assumptions, the difference in means for $Q=$ $D Y, D(1-Y),(D-1) Y$ and $D+Y-D Y$ must all be non-negative.

• Importantly, these implications only involve the distribution of the observed variables.

• Rejection of the IV inequalities leads to rejection of the IV assumptions.

• Balke and Pearl (1997) derived more general IV inequalities without assuming monotonicity.

Eample 1.

• Investigators et al. (2014) assess the effectiveness of the emergency endovascular versus the open surgical repair strategies for patients with a clinical diagnosis of ruptured aortic aneurism.
• Patients are randomized to either the emergency endovascular or the open repair strategy.
• The primary outcome is the survival status after 30 days
• Let $Z$ be the treatment assigned, with $Z=1$ for the endovascular strategy and $Z=0$ for the open repair.
• Let $D$ be the treatment received.
• Let $Y$ be the survival status, with $Y=1$ for dead, and $Y=0$ for alive.
• The estimate of $\tau_{\mathrm{c}}$ is 0.131 with $95 \%$ confidence interval $(-0.036,0.298)$ including 0 .
• Using the function “IVbinary” above, we can obtain $\hat\mu_{1c}=0.708$ and $\hat\mu_{0c}=0.629$
• There is no evidence of violating the IV assumptions.

Example 2.

• In Hirano et al. (2000), physicians are randomly selected to receive a letter encouraging them to inoculate patients at risk for flu.
• The treatment is the actual flu shot, and the outcome is an indicator for flu-related hospital visits.
• However, some patients do not comply with their assignments. Let $Z_i$ be the indicator of encouragement to receive the flu shot, with $Z=1$ if the physician receives the encouragement letter, and $Z=0$ otherwise.
• Let $D$ be the treatment received.
• Let $Y$ be the outcome, with $Y=0$ if for a flu-related hospitalization during the winter, and $Y=1$ otherwise.
• The estimate of $\tau_{\mathrm{c}}$ is 0.116 with $95 %$ confidence interval $(-0.061,0.293)$ including 0 .
• Using the function above, we can obtain $\hat\mu_{1c}=-0.0045$ and $\hat\mu_{0c}=0.1200$
• Since $\hat\mu_{1c}<0$, there is evidence of violating the IV assumptions.