during influenza season. But Jackson et al., (2006) found greater effect
before the influenza season, suggesting that the observed effect is due to
unmeasured confounding.

Jackson et al., (2006) seems the most convincing one since the influenza-
related outcomes before and during the influenza season should have similar
confounding patterns. Cornfield et al., (1959)’s additional evidence seems
weaker since car accident and lung cancer have very different causal
mechanisms with respect to cigarette smoking. In fact, Fisher, (1957)’s
critique was that the relationship between cigarette smoking on lung cancer
may be due to an unobserved genetic factor. Such a genetic factor might affect
cigarette smoking and lung cancer simultaneously, but it seems unlikely that
it also affects car accident.

Lipsitch et al., (2010) is a recent article on negative outcomes. Rosenbaum,
(1989) discussed the role of known effects in causal inference.

#####  81.2 Using negative exposures

Negative exposures are duals of negative outcomes. Assume $Z^{\textup{n}}$ is
a treatment variable similar to $Z$ and shares the same confounding structure
as $Z$. If we believe $Z\begin{picture}(9.0,8.0)\put(0.0,0.0){\line(1,0){9.0}}
\put(3.0,0.0){\line(0,1){8.0}} \put(6.0,0.0){\line(0,1){8.0}}
\end{picture}Y(z)\mid X$, then we tend to believe
$Z^{\textup{n}}\begin{picture}(9.0,8.0)\put(0.0,0.0){\line(1,0){9.0}}
\put(3.0,0.0){\line(0,1){8.0}} \put(6.0,0.0){\line(0,1){8.0}}
\end{picture}Y(z)\mid X.$ Moreover, we know, a priori, the effect of
$Z^{\textup{n}}$ on $Y$

$\tau(Z^{\textup{n}}\rightarrow
Y)=E\\{Y(1^{\textup{n}})-Y(0^{\textup{n}})\\}.$

An important example is that $\tau(Z^{\textup{n}}\rightarrow Y)=0.$ A causal
diagram satisfying these requirements is below:

$\textstyle{Z^{\textup{n}}}$$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{Z\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{Y}$

###### Example 16.4

Sanderson et al., (2017) give many examples of negative exposures in
determining the effect of intrauterine exposure on later outcomes by comparing
the association of a maternal exposure during pregnancy with the outcome of
interest, with the association of the paternal exposure with the same outcome.
They review studies on the effect of maternal and paternal smoking on
offspring outcomes, the effect of maternal and paternal BMI on later offspring
BMI and autism spectrum disorder. In these examples, we expect the the
association of the maternal exposure with the outcome is larger than that of
the paternal exposure with the outcome.

#####  81.3 Summary

The unconfoundedness assumption is fundamentally untestable without additional
assumptions. Although negative outcomes and negative controls in observational
studies cannot prove or disprove unconfoundedness, using them in supplementary
analyses can strengthen the evidence for causation. However, it is often non-
trivial to conduct this type of supplementary analyses because it involves
more data and more importantly, deeper understanding of the causal problems in
order to find convincing negative outcomes and negative controls.

####  82 Problems of over adjustment

We have discussed many methods for estimating causal effects under
ignorability:

$Z\begin{picture}(9.0,8.0)\put(0.0,0.0){\line(1,0){9.0}}
\put(3.0,0.0){\line(0,1){8.0}} \put(6.0,0.0){\line(0,1){8.0}}
\end{picture}\\{Y(1),Y(0)\\}\mid X.$

This is an assumption conditioning on $X$. It is crucial to select the right
set of $X$ that ensure the conditional independence. Rosenbaum, 2002b  wrote
that“there is no reason to avoid adjustment for a variable describing subjects
before treatment.” Similarly, Rubin, (2007) wrote that “typically, the more
conditional an assumption, the more acceptable it is.” Both argued that we
should control for all observed pretreatment covariate. VanderWeele and
Shpitser, (2011) called it the pretreatment criterion. Pearl disagreed with
this recommendation and gave two counterexamples below.

#####  82.1 M-bias

M-bias appears in the following causal diagram with an M-structure:

$\textstyle{U_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{U_{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X}$$\textstyle{Z}$$\textstyle{Y}$

We can read from the diagram the data generating process:

$\left\\{\begin{aligned}
&U_{1}\begin{picture}(9.0,8.0)\put(0.0,0.0){\line(1,0){9.0}}
\put(3.0,0.0){\line(0,1){8.0}} \put(6.0,0.0){\line(0,1){8.0}}
\end{picture}U_{2},\\\ &X=f_{X}(U_{1},U_{2},\varepsilon_{X}),\\\
&Z=f_{Z}(U_{1},\varepsilon_{Z}),\\\
&Y=Y(z)=f_{Y}(U_{2},\varepsilon_{Y}),\end{aligned}\right.$

where $(\varepsilon_{X},\varepsilon_{Z},\varepsilon_{Y})$ are independent
random error terms. In the above causal diagram, $X$ is observed, but $U_{1}$
and $U_{2}$ are unobserved. If we change the value of $Z$, the value of $Y$
will not change at all. So the true causal effect of $Z$ on $Y$ must be $0$.
From the data-generating equations, we can easy read that
$Z\begin{picture}(9.0,8.0)\put(0.0,0.0){\line(1,0){9.0}}
\put(3.0,0.0){\line(0,1){8.0}} \put(6.0,0.0){\line(0,1){8.0}} \end{picture}Y$,
so the association between $Z$ and $Y$ is $0$, and, in particular,

$\tau_{\textsc{PF}}=E(Y\mid Z=1)-E(Y\mid Z=0)=0.$

This means that without adjusting for the covariate $X$, the simple estimator
is unbiased for the true parameter.

However, if we condition on $X$, then
$U_{1}\begin{picture}(9.0,8.0)\put(0.0,0.0){\line(1,0){9.0}}
\put(3.0,0.0){\line(0,1){8.0}} \put(6.0,0.0){\line(0,1){8.0}}
\put(1.0,0.0){{\it/}} \end{picture}U_{2}\mid X$, and consequently,
$Z\begin{picture}(9.0,8.0)\put(0.0,0.0){\line(1,0){9.0}}
\put(3.0,0.0){\line(0,1){8.0}} \put(6.0,0.0){\line(0,1){8.0}}
\put(1.0,0.0){{\it/}} \end{picture}Y\mid X$ and

$\int\\{E(Y\mid Z=1,X=x)-E(Y\mid Z=0,X=x)\\}F(\textnormal{d}x)\neq 0$

in general. To gain intuition, we consider the case with Gaussian linear
models202020It is not ideal for our discussion of binary $Z$, but it
simplifies the derivations. Ding and Miratrix, (2015) gave detailed discussion
with more natural models for binary $Z$. :

$\left\\{\begin{aligned} &X=aU_{1}+bU_{2}+\varepsilon_{X},\\\
&Z=cU_{1}+\varepsilon_{Z},\\\
&Y=Y(z)=dU_{2}+\varepsilon_{Y},\end{aligned}\right.$

where
$(U_{1},U_{2},\varepsilon_{X},\varepsilon_{Z},\varepsilon_{Y})\stackrel{{\scriptstyle\text{IID}}}{{\sim}}\textsc{N}(0,1)$.
We have

$\textup{cov}(Z,Y)=\textup{cov}(cU_{1}+\varepsilon_{Z},dU_{2}+\varepsilon_{Y})=0,$

but by the result in Problem 1.2, the partial correlation coefficient between
$Z$ and $Y$ given $X$ is

$\displaystyle\rho_{ZY\mid
X}=\frac{\rho_{ZY}-\rho_{ZX}\rho_{YX}}{\sqrt{1-\rho_{ZX}^{2}}\sqrt{1-\rho_{YX}^{2}}}\propto-\rho_{ZX}\rho_{YX}\propto-\textup{cov}(Z,X)\textup{cov}(Y,X)=-abcd,$

the product of the coefficients on the path from $Z$ to $Y$. So the unadjusted
estimator is unbiased but the adjusted estimator has bias proportional to
$abcd$.

The following simple example illustrates M-bias.

⬇

> n = 10^6

>

> ## M bias

> U1 = rnorm(n)

> U2 = rnorm(n)

> X = U1 + U2 + rnorm(n)

> Z = U1 + rnorm(n)

> Y = U2 + rnorm(n)

>

> round(summary(lm(Y ~ Z))$coef[2, 1], 3)

[1] -0.001

> round(summary(lm(Y ~ Z + X))$coef[2, 1], 3)

[1] -0.201

>

> Z = (Z >= 0)

> round(summary(lm(Y ~ Z))$coef[2, 1], 3)

[1] -0.002

> round(summary(lm(Y ~ Z + X))$coef[2, 1], 3)

[1] -0.421

#####  82.2 Z-bias

Consider the following causal diagram:

$\textstyle{U\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{b}$$\scriptstyle{c}$$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{a}$$\textstyle{Z\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\tau}$$\textstyle{Y}$

with the data generating process212121Again, we generate continuous $Z$ from a
linear model to simplify the derivations. Ding et al., 2017b  extended the
theory to more general causal models, especially for binary $Z$.

$\left\\{\begin{aligned} &Z=aX+bU+\varepsilon_{Z},\\\ &Y(z)=\tau
z+cU+\varepsilon_{Y},\end{aligned}\right.$

where
$(U,X,\varepsilon_{Z},\varepsilon_{Y})\stackrel{{\scriptstyle\text{IID}}}{{\sim}}\textsc{N}(0,1)$.
In this data generating process, we have
$X\begin{picture}(9.0,8.0)\put(0.0,0.0){\line(1,0){9.0}}
\put(3.0,0.0){\line(0,1){8.0}} \put(6.0,0.0){\line(0,1){8.0}} \end{picture}U,$
$X\begin{picture}(9.0,8.0)\put(0.0,0.0){\line(1,0){9.0}}
\put(3.0,0.0){\line(0,1){8.0}} \put(6.0,0.0){\line(0,1){8.0}}
\put(1.0,0.0){{\it/}} \end{picture}Z$, and $X$ affects $Y$ only through $Z$.

The unadjusted estimator is

$\displaystyle\tau_{\textup{unadj}}=\frac{\textup{cov}(Z,Y)}{\textup{var}(Z)}=\frac{\textup{cov}(Z,\tau
Z+cU)}{\textup{var}(Z)}=\tau+\frac{c\textup{cov}(aX+bU,U)}{\textup{var}(Z)}=\tau+\frac{cb}{a^{2}+b^{2}+1},$

which has bias $bc/(a^{2}+b^{2}+1)$. The adjusted estimator from the OLS of
$Y$ on $(Z,X)$ satisfies

$\left\\{\begin{aligned} E\\{Z(Y-\tau_{\textup{adj}}Z-\alpha X)\\}&=0,\\\
E\\{X(Y-\tau_{\textup{adj}}Z-\alpha X)\\}&=0,\end{aligned}\right.$

which is equivalent to

$\left\\{\begin{aligned} E(ZY)&=\tau_{\textup{adj}}\textup{var}(Z)+\alpha
E(XZ),\\\
E(XY)&=\tau_{\textup{adj}}E(XZ)+\alpha\textup{var}(X).\end{aligned}\right.$

We need to solve for $(\tau_{\textup{adj}},\alpha)$ from the above two linear
equations:

$\displaystyle\tau_{\textup{adj}}$ $\displaystyle=$
$\displaystyle\frac{\begin{vmatrix}E(ZY)&E(XZ)\\\
E(XY)&\textup{var}(X)\end{vmatrix}}{\begin{vmatrix}\textup{var}(Z)&E(XZ)\\\
E(XZ)&\textup{var}(X)\end{vmatrix}}=\frac{E(ZY)\textup{var}(X)-E(XZ)E(XY)}{\textup{var}(Z)\textup{var}(X)-E(XZ)^{2}}$
$\displaystyle=$ $\displaystyle\frac{\tau(a^{2}+b^{2}+1)+bc-a\tau
a}{(a^{2}+b^{2}+1)-a^{2}}=\frac{\tau(b^{2}+1)+bc}{b^{2}+1}=\tau+\frac{bc}{b^{2}+1},$

which has bias $bc/(b^{2}+1)$.

So the unadjusted estimator has smaller bias than the adjusted estimator. More
interestingly, the stronger the association between $X$ and $Z$ is (measured
by $a$), the larger the bias of the adjusted estimator is.

The mathematical derivation is not extremely hard. But this type of bias seems
rather mysterious. Here is the intuition. The treatment is a function of $X$,
$U$, and other random errors. If we condition on $X$, it is merely a function
of $U$ and other random errors. Therefore, conditioning makes $Z$ less random,
and more critically, makes the unmeasured confounder $U$ play a more important
role in $Z$. Consequently, the confounding bias due to $U$ is amplified by
conditioning on $X$. This idealized example illustrates the danger of over
adjusting for some covariates.

Heckman and Navarro-Lozano, (2004) observed the phenomenon in simulation
studies, and Wooldridge, (2016, technical report in 2006) verified it in
linear models. Pearl, (2010, 2011) explained it using causal diagrams. This
type of bias is called Z-bias because in Pearl’s original papers, he used the
symbol $Z$ for our variable $X$. Throughout the book, however, $Z$ is used for
the treatment variable. In Part V of this book, we will call $Z$ an
instrumental variable if it satisfies the causal diagram presented in this
subsection. This justifies the instrumental variable bias as another name of
this type of bias.

The following simple example illustrates Z-bias.

⬇

> X = rnorm(n)

> U = rnorm(n)

> Z = X + U + rnorm(n)

> Y = U + rnorm(n)

>

> round(summary(lm(Y ~ Z))$coef[2, 1], 3)

[1] 0.334

> round(summary(lm(Y ~ Z + X))$coef[2, 1], 3)

[1] 0.501

>

> Z = 2*X + U + rnorm(n)

> round(summary(lm(Y ~ Z))$coef[2, 1], 3)

[1] 0.167

> round(summary(lm(Y ~ Z + X))$coef[2, 1], 3)

[1] 0.5

>

> Z = 10*X + U + rnorm(n)

> round(summary(lm(Y ~ Z))$coef[2, 1], 3)

[1] 0.01

> round(summary(lm(Y ~ Z + X))$coef[2, 1], 3)

[1] 0.5

#####  82.3 What covariates should we adjust for in observational studies?

We never know the true underlying data generating process which can be quite
complicated. However, the following causal diagram helps to clarify many
ideas. It already rules out the possibility of $M$-bias discussed in Section
82.1.

|  |
---|---|---
|  |
$\textstyle{X_{R}}$$\textstyle{X_{Z}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{Y}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{Z\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{Y\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{X_{I}}$

The covariates above have different features:

  1. 1.

$X$ affects both the treatment and the outcome. Conditioning on $X$ ensures
ignorability, so we should control for $X$.

  2. 2.

$X_{R}$ is pure random noise not affecting either the treatment or the
outcome. Including it in analysis does not bias the estimate but it introduces
unnecessary variability in finite sample.

  3. 3.

$X_{Z}$ is an instrumental variable that affects the outcome only through the
treatment. In the diagram above, including it in analysis does not bias the
estimate although it increases variability. However, with unmeasured
confounding, including it in analysis amplifies the bias as shown in Section
82.1.

  4. 4.

$X_{Y}$ affects the outcome only but not the treatment. Without conditioning
on it, the ignorability still holds. Since they are predictive to the outcome,
including them in analysis often improves precision.

  5. 5.

$X_{I}$ is affected by the treatment and outcome. It is a post-treatment
variable, not a pretreatment covariate. We should not include it if the goal
is to infer the effect of the treatment on the outcome. We will discuss issues
with post-treatment variables in causal inference in Part VI of this book.

If we believe the above causal diagram, we should adjust for at least $X$ to
remove bias and more ideally, further adjust for $X_{Y}$ to reduce variance.

####  83 Homework Problems

######  16.1 Cochran’s formula or the omitted variable bias formula

Sir David Cox calls the following result Cochran’s formula (Cochran,, 1938;
Cox,, 2007) and econometricians call it the omitted variable bias formula
(Angrist and Pischke,, 2008). A special case appeared in Fisher, (1925). It is
also a sister of the Frisch–Waugh–Lovell Theorem in Chapter 160.

The formula has two versions. All vectors below are column vectors.

  1. [(1)]

  2. 1.

(Population version) Assume $(y_{i},x_{1i},x_{2i})_{i=1}^{n}$ are iid, where
$y_{i}$ is a scalar, $x_{i1}$ has dimension $K$, and $x_{i2}$ has dimension
$L$.

We have the following OLS decompositions of random variables

$\displaystyle y_{i}$ $\displaystyle=$
$\displaystyle\beta_{1}^{\mkern-1.5mu\mathsf{T}}x_{i1}+\beta_{2}^{\mkern-1.5mu\mathsf{T}}x_{2i}+\varepsilon_{i},$
(74) $\displaystyle y_{i}$ $\displaystyle=$
$\displaystyle\gamma^{\mkern-1.5mu\mathsf{T}}x_{i1}+e_{i},$ (75)
$\displaystyle x_{i2}$ $\displaystyle=$
$\displaystyle\delta^{\mkern-1.5mu\mathsf{T}}x_{i1}+v_{i}.$ (76)

Equation (74) is called the long regression, and Equation (75) is called the
short regression. In Equation (76), $\delta$ is a matrix because it is a
regression of a vector on a vector. You can view (76) as regression of each
component of $x_{i2}$ on $x_{i1}$.

Show that $\gamma=\beta_{1}+\delta\beta_{2}.$

  3. 2.

(Sample version) We have an $n\times 1$ vector $Y$, an $n\times K$ matrix
$X_{1}$, and an $n\times L$ matrix $X_{2}$. We do not assume any randomness.
All results below are purely linear algebra.

We can obtain the following OLS fits:

$\displaystyle Y$ $\displaystyle=$ $\displaystyle
X_{1}\hat{\beta}_{1}+X_{2}\hat{\beta}_{2}+\hat{\varepsilon},$ $\displaystyle
Y$ $\displaystyle=$ $\displaystyle X_{1}\hat{\gamma}+\hat{e},$ $\displaystyle
X_{2}$ $\displaystyle=$ $\displaystyle X_{1}\hat{\delta}+\hat{v},$

where $\hat{\varepsilon},\hat{e},\hat{v}$ are the residuals. Again, the last
OLS fit means the OLS fit of each column of $X_{2}$ on $X_{1}$, and therefore
the residual $\hat{v}$ is an $n\times L$ matrix.

Show that $\hat{\gamma}=\hat{\beta}_{1}+\hat{\delta}\hat{\beta}_{2}$.

Remark: The product terms $\delta\beta_{2}$ and $\hat{\delta}\hat{\beta}_{2}$
are often referred to as the omitted-variable bias at the population level and
sample level, respectively.

######  16.2 Recommended reading

Imbens, (2020) reviews and compares the roles of potential outcomes and causal
diagrams for causal inference.

###  Chapter 17 E-Value: Evidence for Causation in Observational Studies with
Unmeasured Confounding

E-Value

All the methods discussed in Part III rely crucially on the ignorability
assumption. They require controlling for all confounding between the treatment
and outcome. However, we cannot use the data to validate the ignorability
assumption. Observational studies are often criticized due to the possibility
of unmeasured confounding. The famous Yule–Simpson Paradox demonstrates that
an unmeasured binary confounder can completely overturn an observed
association between the treatment and outcome. However, to overturn a larger
observed association, this unmeasured confounder must have stronger
association with the treatment and the outcome. In other words, not all
observational studies are created equal. Some provide stronger evidence for
causation than others.

The following three chapters will discuss various sensitivity analysis
techniques that can quantify the evidence of causation based on observational
studies in the presence of unmeasured confounding. This chapter starts with
the E-value, introduced by VanderWeele and Ding, (2017) based on the theory in
Ding and VanderWeele, (2016). It is more useful for observational studies
using logistic regressions. Chapter 18 discusses sensitivity analysis for the
average causal effect based on inverse probability weighting, outcome
regression, and doubly robust estimation. Chapter 19 discusses Rosenbaum’s
framework for sensitivity analysis for matched observational studies.

####  84 Cornfield-type sensitivity analysis

Although we do not assume ignorability given $X$:

$Z\begin{picture}(9.0,8.0)\put(0.0,0.0){\line(1,0){9.0}}
\put(3.0,0.0){\line(0,1){8.0}} \put(6.0,0.0){\line(0,1){8.0}}
\put(1.0,0.0){{\it/}} \end{picture}\\{Y(1),Y(0)\\}\mid X,$

we still assume latent ignorability given $X$ and another unmeasured
confounder $U$:

$Z\begin{picture}(9.0,8.0)\put(0.0,0.0){\line(1,0){9.0}}
\put(3.0,0.0){\line(0,1){8.0}} \put(6.0,0.0){\line(0,1){8.0}}
\end{picture}\\{Y(1),Y(0)\\}\mid(X,U).$

The technique in this chapter works the best for a binary outcome $Y$ although
it can be extended to other non-negative outcomes (Ding and VanderWeele,,
2016). Focus on binary $Y$ now. The true conditional causal effect on the risk
ratio scale is defined as

$\textsc{rr}_{ZY\mid x}^{\textup{true}}=\frac{\textup{pr}\\{Y(1)=1\mid
X=x\\}}{\textup{pr}\\{Y(0)=1\mid X=x\\}},$

and the observed conditional risk ratio equals

$\textsc{rr}_{ZY\mid x}^{\textup{obs}}=\frac{\textup{pr}(Y=1\mid
Z=1,X=x)}{\textup{pr}(Y=1\mid Z=0,X=x)}.$

In general, with an unmeasured confounder $U$,

$\textsc{rr}_{ZY\mid x}^{\textup{true}}\neq\textsc{rr}_{ZY\mid
x}^{\textup{obs}}$

because

$\textsc{rr}_{ZY\mid x}^{\textup{true}}=\frac{\int\textup{pr}(Y=1\mid
Z=1,X=x,U=u)F(\textnormal{d}u\mid X=x)}{\int\textup{pr}(Y=1\mid
Z=0,X=x,U=u)F(\textnormal{d}u\mid X=x)}$

and

$\textsc{rr}_{ZY\mid x}^{\textup{obs}}=\frac{\int\textup{pr}(Y=1\mid
Z=1,X=x,U=u)F(\textnormal{d}u\mid Z=1,X=x)}{\int\textup{pr}(Y=1\mid
Z=0,X=x,U=u)F(\textnormal{d}u\mid Z=0,X=x)}$

are averaged over different distributions of $U$.

Doll and Hill, (1950) found that the risk ratio of cigarette smoking on lung
cancer was $9$ even after adjusting for many observed covariates
$X$222222Their original analysis was based on a case-control study and
estimated the odds ratio of cigarette smoking on lung cancer. But the risk
ratio is close to the odds ratio since lung cancer is a rare outcome.. Fisher,
(1957) criticized their result to be noncausal because it is possible that a
hidden gene simultaneously causes cigarette smoking and lung cancer although
the true causal effect of cigarette smoking on lung cancer is absent. This is
the common cause hypothesis, also discussed by Reichenbach, (1957). Cornfield
et al., (1959) took a more constructive perspective and asked: how strong this
unmeasured confounder must be in order to explain away the observed
association between cigarette smoking and lung cancer? Below we will use Ding
and VanderWeele, (2016)’s general formulation of the problem.

Consider the following causal diagram:

$\textstyle{U\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{Z}$$\textstyle{Y}$

which conditions on $X$. So
$Z\begin{picture}(9.0,8.0)\put(0.0,0.0){\line(1,0){9.0}}
\put(3.0,0.0){\line(0,1){8.0}} \put(6.0,0.0){\line(0,1){8.0}}
\end{picture}Y\mid(X,U)$. Conditioning on $X$ and $U$, we observe no
association between $Z$ and $Y$; but conditioning on only $X$, we observe
association between $Z$ and $Y$. Although we can allow $U$ to be general as
Ding and VanderWeele, (2016), we assume that $U$ is binary to simplify the
presentation.

Define two sensitivity parameters:

$\textsc{rr}_{ZU\mid x}=\frac{\textup{pr}(U=1\mid
Z=1,X=x)}{\textup{pr}(U=1\mid Z=0,X=x)}\equiv\frac{f_{1,x}}{f_{0,x}}$

measures the treatment-confounder association, and

$\textsc{rr}_{UY\mid x}=\frac{\textup{pr}(Y=1\mid
U=1,X=x)}{\textup{pr}(Y=1\mid U=0,X=x)},$

measures the confounder-outcome association, conditional on covariates $X=x$.
Without loss of generality, we assume that
$\textsc{rr}_{x}^{\textup{obs}}>1,\textsc{rr}_{ZU\mid x}>1$, and
$\textsc{rr}_{UY\mid x}>1$. We can show the main result below.

###### Theorem 17.1

Under $Z\begin{picture}(9.0,8.0)\put(0.0,0.0){\line(1,0){9.0}}
\put(3.0,0.0){\line(0,1){8.0}} \put(6.0,0.0){\line(0,1){8.0}}
\end{picture}Y\mid(X,U)$, we have

$\textsc{rr}_{ZY\mid x}^{\textup{obs}}\leq\frac{\textsc{rr}_{ZU\mid
x}\textsc{rr}_{UY\mid x}}{\textsc{rr}_{ZU\mid x}+\textsc{rr}_{UY\mid x}-1}.$

Theorem 17.1 shows the upper bound of the observed risk ratio of the treatment
on the outcome if the conditional independence
$Z\begin{picture}(9.0,8.0)\put(0.0,0.0){\line(1,0){9.0}}
\put(3.0,0.0){\line(0,1){8.0}} \put(6.0,0.0){\line(0,1){8.0}}
\end{picture}Y\mid(X,U)$ holds. Under this conditional independence
assumption, the association between the treatment and the outcome is purely
due to the association between the treatment, $\textsc{rr}_{ZU\mid x}$, and
the confounder and the association between the confounder and the outcome,
$\textsc{rr}_{UY\mid x}$. The upper bound equals $\textsc{rr}_{ZU\mid
x}\textsc{rr}_{UY\mid x}/(\textsc{rr}_{ZU\mid x}+\textsc{rr}_{UY\mid x}-1)$. A
similar inequality appeared in Lee, (2011). It is also related to Cochran’s
formula or the omitted-variable bias formula for linear models, which was
reviewed in Problem 16.1.

Reversely, to generate a certain value of the observed risk ratio
$\textsc{rr}_{x}^{\textup{obs}}$, the two confounding measures
$\textsc{rr}_{ZU\mid x}$ and $\textsc{rr}_{UY\mid x}$ cannot be arbitrary.
Their function $\textsc{rr}_{ZU\mid x}\textsc{rr}_{UY\mid
x}/(\textsc{rr}_{ZU\mid x}+\textsc{rr}_{UY\mid x}-1)$ must be at least at
large as $\textsc{rr}_{x}^{\textup{obs}}$.

I will give the proof of Theorem 17.1 below.

Proof of Theorem 17.1: We can decompose $\textsc{rr}_{ZY\mid
x}^{\textup{obs}}$ as

$\displaystyle\textsc{rr}_{ZY\mid x}^{\textup{obs}}$ $\displaystyle=$
$\displaystyle\frac{\textup{pr}(Y=1\mid Z=1,X=x)}{\textup{pr}(Y=1\mid
Z=0,X=x)}$ $\displaystyle=$
$\displaystyle\frac{\left[\begin{multlined}\textup{pr}(U=1\mid
Z=1,X=x)\textup{pr}(Y=1\mid Z=1,U=1,X=x)\\\ +\textup{pr}(U=0\mid
Z=1,X=x)\textup{pr}(Y=1\mid Z=1,U=0,X=x)\end{multlined}\textup{pr}(U=1\mid
Z=1,X=x)\textup{pr}(Y=1\mid Z=1,U=1,X=x)\\\ +\textup{pr}(U=0\mid
Z=1,X=x)\textup{pr}(Y=1\mid
Z=1,U=0,X=x)\right]}{\left[\begin{multlined}\textup{pr}(U=1\mid
Z=0,X=x)\textup{pr}(Y=1\mid Z=0,U=1,X=x)\\\ +\textup{pr}(U=0\mid
Z=0,X=x)\textup{pr}(Y=1\mid Z=0,U=0,X=x)\end{multlined}\textup{pr}(U=1\mid
Z=0,X=x)\textup{pr}(Y=1\mid Z=0,U=1,X=x)\\\ +\textup{pr}(U=0\mid
Z=0,X=x)\textup{pr}(Y=1\mid Z=0,U=0,X=x)\right]}$ $\displaystyle=$
$\displaystyle\frac{\left[\begin{multlined}\textup{pr}(U=1\mid
Z=1,X=x)\textup{pr}(Y=1\mid U=1,X=x)\\\ +\textup{pr}(U=0\mid
Z=1,X=x)\textup{pr}(Y=1\mid U=0,X=x)\end{multlined}\textup{pr}(U=1\mid
Z=1,X=x)\textup{pr}(Y=1\mid U=1,X=x)\\\ +\textup{pr}(U=0\mid
Z=1,X=x)\textup{pr}(Y=1\mid
U=0,X=x)\right]}{\left[\begin{multlined}\textup{pr}(U=1\mid
Z=0,X=x)\textup{pr}(Y=1\mid U=1,X=x)\\\ +\textup{pr}(U=0\mid
Z=0,X=x)\textup{pr}(Y=1\mid U=0,X=x)\end{multlined}\textup{pr}(U=1\mid
Z=0,X=x)\textup{pr}(Y=1\mid U=1,X=x)\\\ +\textup{pr}(U=0\mid
Z=0,X=x)\textup{pr}(Y=1\mid U=0,X=x)\right]}$ $\displaystyle=$
$\displaystyle\frac{f_{1,x}\textsc{rr}_{UY\mid
x}+1-f_{1,x}}{f_{0,x}\textsc{rr}_{UY\mid x}+1-f_{0,x}}$ $\displaystyle=$
$\displaystyle\frac{(\textsc{rr}_{UY\mid
x}-1)f_{1,x}+1}{\frac{\textsc{rr}_{UY\mid x}-1}{\textsc{rr}_{ZU\mid
x}}f_{1,x}+1}.$

We can verify that $\textsc{rr}_{ZY\mid x}^{\textup{obs}}$ is increasing in
$f_{1,x}$. So letting $f_{1,x}=1$, we have

$\textsc{rr}_{ZY\mid x}^{\textup{obs}}\leq\frac{(\textsc{rr}_{UY\mid
x}-1)+1}{\frac{\textsc{rr}_{UY\mid x}-1}{\textsc{rr}_{ZU\mid
x}}+1}=\frac{\textsc{rr}_{ZU\mid x}\textsc{rr}_{UY\mid x}}{\textsc{rr}_{ZU\mid
x}+\textsc{rr}_{UY\mid x}-1}.$

$\square$

In the proof of Theorem 17.1, we have obtain an identity

$\textsc{rr}_{ZY\mid x}^{\textup{obs}}=\frac{(\textsc{rr}_{UY\mid
x}-1)f_{1,x}+1}{\frac{\textsc{rr}_{UY\mid x}-1}{\textsc{rr}_{ZU\mid
x}}f_{1,x}+1}.$

But this identity involves three parameters

$\\{f_{1,x},\textsc{rr}_{ZU\mid x},\textsc{rr}_{UY\mid x}\\};$

see Problem 17.2 for a related formula. In contrast, the upper bound in
Theorem 17.1 involves only two parameters

$\\{\textsc{rr}_{ZU\mid x},\textsc{rr}_{UY\mid x}\\}$

which measure the strength of the confounder.

####  85 E-value

Lemma 17.1 below is useful for deriving interesting corollaries of Theorem
17.1.

###### Lemma 17.1

Define $\beta(w_{1},w_{2})=w_{1}w_{2}/(w_{1}+w_{2}-1)$ for $w_{1}>1$ and
$w_{2}>1$.

  1. [(1)]

  2. 1.

$\beta(w_{1},w_{2})$ is symmetric in $w_{1}$ and $w_{2}$;

  3. 2.

$\beta(w_{1},w_{2})$ increasing in both $w_{1}$ and $w_{2}$;

  4. 3.

$\beta(w_{1},w_{2})\leq w_{1}$ and $\beta(w_{1},w_{2})\leq w_{2}$;

  5. 4.

$\beta(w_{1},w_{2})\leq w^{2}/(2w-1)$, where $w=\max(w_{1},w_{2})$.

Using Theorem 17.1 and Lemma 17.1(3), we have

$\textsc{rr}_{ZU\mid x}\geq\textsc{rr}_{ZY\mid
x}^{\textup{obs}},\quad\textsc{rr}_{UY\mid x}\geq\textsc{rr}_{ZY\mid
x}^{\textup{obs}},$

or, equivalently,

$\min(\textsc{rr}_{ZU\mid x},\textsc{rr}_{UY\mid x})\geq\textsc{rr}_{ZY\mid
x}^{\textup{obs}}.$

Therefore, to explain away the observed relative risk, both confounding
measures $\textsc{rr}_{ZU\mid x}$ and $\textsc{rr}_{UY\mid x}$ must be at
least as large as $\textsc{rr}_{ZY\mid x}^{\textup{obs}}$ Cornfield et al.,
(1959) first derived the inequality $\textsc{rr}_{ZU\mid
x}\geq\textsc{rr}_{ZY\mid x}^{\textup{obs}}$, also called the Cornfield
inequality (Gastwirth et al.,, 1998). Schlesselman, (1978) derived the
inequality $\textsc{rr}_{UY\mid x}\geq\textsc{rr}_{ZY\mid x}^{\textup{obs}}$.
These are related to to the data processing inequality in information
theory232323In information theory, the mutual information $I(A,B)=\iint
p(a,b)\log_{2}\frac{p(a,b)}{p(a)p(b)}\text{d}a\text{d}b$ measures the
dependence between two random variables $A$ and $B$, where $p(\cdot)$ denotes
the joint or marginal density of $(A,B)$. The data processing inequality is a
famous result: if $Z\begin{picture}(9.0,8.0)\put(0.0,0.0){\line(1,0){9.0}}
\put(3.0,0.0){\line(0,1){8.0}} \put(6.0,0.0){\line(0,1){8.0}}
\end{picture}Y\mid U$, then $I(Z,Y)\geq I(Z,U)$ and $I(Z,Y)\geq I(U,Y)$. Lihua
Lei and Bin Yu pointed out to me the connection between Cornfield’s inequality
and the data processing inequality. .

If we define $w=\max(\textsc{rr}_{ZU\mid x},\textsc{rr}_{UY\mid x})$, then we
can use Theorem 17.1 and Lemma 17.1(4) to obtain

$\displaystyle w^{2}/(2w-1)\geq\beta(\textsc{rr}_{ZU\mid
x},\textsc{rr}_{UY\mid x})\geq\textsc{rr}_{x}^{\textup{obs}}$
$\displaystyle\Longrightarrow$ $\displaystyle
w^{2}-2\textsc{rr}_{x}^{\textup{obs}}w+\textsc{rr}_{x}^{\textup{obs}}\geq 0,$

which is a quadratic inequality. One root $\textsc{rr}_{ZY\mid
x}^{\textup{obs}}-\sqrt{\textsc{rr}_{ZY\mid
x}^{\textup{obs}}(\textsc{rr}_{ZY\mid x}^{\textup{obs}}-1)}$ is always smaller
than or equal to $1$, so we have

$w=\max(\textsc{rr}_{ZU\mid x},\textsc{rr}_{UY\mid x})\geq\textsc{rr}_{ZY\mid
x}^{\textup{obs}}+\sqrt{\textsc{rr}_{ZY\mid
x}^{\textup{obs}}(\textsc{rr}_{ZY\mid x}^{\textup{obs}}-1)}.$

Therefore, to explain away the observed relative risk, the maximum of the
confounding measures $\textsc{rr}_{ZU\mid x}$ and $\textsc{rr}_{UY\mid x}$
must be at least as large as $\textsc{rr}_{ZY\mid
x}^{\textup{obs}}+\sqrt{\textsc{rr}_{ZY\mid
x}^{\textup{obs}}(\textsc{rr}_{ZY\mid x}^{\textup{obs}}-1)}$. Based on this
result, VanderWeele and Ding, (2017) introduced the following notion of
E-value for measuring the evidence of causation with observational studies.

######  Definition 17.1 (E-Value)

With the observed conditional risk ratio $\textsc{rr}_{ZY\mid
x}^{\textup{obs}}$, define the E-Value as

$\textsc{rr}_{ZY\mid x}^{\textup{obs}}+\sqrt{\textsc{rr}_{ZY\mid
x}^{\textup{obs}}(\textsc{rr}_{ZY\mid x}^{\textup{obs}}-1)}$

The E-value is defined for the parameter $\textsc{rr}_{ZY\mid
x}^{\textup{obs}}$. In practice, $\textsc{rr}_{ZY\mid x}^{\textup{obs}}$ is
estimated with sampling error. We can calculate the E-value based on the
estimated $\textsc{rr}_{ZY\mid x}^{\textup{obs}}$, as well as the
corresponding E-values for the lower and upper confidence limits of
$\textsc{rr}_{ZY\mid x}^{\textup{obs}}$.

Fisher’s $p$-value measures the evidence for causal effects in randomized
experiment. We have discussed the $p$-value based on the FRT in Part II of
this book. However, in observational studies with large sample sizes,
$p$-values can be a poor measure of evidence for causal effects. Even if the
true causal effects are $0$, a tiny amount of unmeasured confounding can bias
the estimate, which can result in extremely small $p$-values given the small
sampling uncertainty. The sampling uncertainty is usually secondary in
observational studies with large sample sizes, but the uncertainty due to
unmeasured confounding is often the first order problem that does not diminish
with increased sample sizes. VanderWeele and Ding, (2017) argued that the
E-value is a better measure of the evidence for causal effects in
observational studies.

####  86 A classic example

I revisit a classic example below.

###### Example 17.1

Hammond and Horn, (1958) used the U.S. population to study the cigarette
smoking and lung cancer relationship. Ignoring covariates, their data can be
represented by a $2\times 2$ table:

| Lung cancer | No lung cancer
---|---|---
Smoker | 397 | 78557
Non-smoker | 51 | 108778

Based on the data, they obtained an estimate of the risk ratio $10.73$ with a
95% confidence interval $[8.02,14.36]$. To explain away the point estimate,
the E-value is

$10.73+\sqrt{10.73\times(10.73-1)}=20.95;$

to explain away the lower confidence limit, the E-value is

$8.02+\sqrt{8.02\times(8.02-1)}=15.52.$

Figure 19 shows the joint values of the two confounding measures to explain
away the point estimate and lower confidence limit of the risk ratio. In
particular, to explain way the point estimate, they must lie in the area above
the solid curve; to explain away the lower confidence limit, they must lie in
the area above the dashed curve.

Figure 19: Magnitude of confounding to explain away the observed risk ratio in
Hammond and Horn, (1958)

####  87 Extensions

#####  87.1 E-value and Bradford Hill’s criteria for causation

The E-value provides evidence for causation. But evidence is not a proof. With
a larger E-value, we need a stronger unmeasured confounder to explain away the
observed risk ratio; the evidence for causation is stronger. With a smaller
E-value, we need a weaker unmeasured confounder to explain away the observed
risk ratio; the evidence for causation is weaker. Coupled with the discussion
in Section 88.1, a larger observed risk ratio have stronger evidence for
causation. This is closely related to Sir Bradford Hill’s first criterion for
causation: strength of the association (Bradford Hill,, 1965). Theorem 17.1
provides a mathematical quantification of his heuristic argument.

In a famous paper, Bradford Hill, (1965) proposed a set of nine criteria to
provide evidence for causation between a presumed cause and outcome. His
criteria are

  1. 1.

strength;

  2. 2.

consistency;

  3. 3.

specificity;

  4. 4.

temporality;

  5. 5.

biological gradient;

  6. 6.

plausibility;

  7. 7.

coherence;

  8. 8.

experiment;

  9. 9.

analogy.

The E-value is a way to justify his first criterion. That is, stronger
association often provides stronger evidence for causation because to explain
way stronger association, we need stronger confounding measures. We have
discussed randomized experiments in Part II, which corroborates his eighth
criterion. Due to the space limit, I omit the detailed discussion of his other
criteria and encourage the readers to read (Bradford Hill,, 1965). Recently,
this paper is re-printed as Bradford Hill, (2020) with insightful comments
from many leading researchers in causal inference.

#####  87.2 E-value after logistic regression

With a binary outcome, it is common for epidemiologists to use a logistic
regression of the outcome $Y_{i}$ on the treatment indicator $Z_{i}$ and
covariates $X_{i}$:

$\textup{pr}(Y_{i}=1\mid
Z_{i},X_{i})=\frac{e^{\beta_{0}+\beta_{1}Z_{i}+\beta_{2}^{\mkern-1.5mu\mathsf{T}}X_{i}}}{1+e^{\beta_{0}+\beta_{1}Z_{i}+\beta_{2}^{\mkern-1.5mu\mathsf{T}}X_{i}}}.$

In the logistic model above, the coefficient of $Z_{i}$ is the log of the
conditional odds ratio between the treatment and the outcome given the
covariates:

$\beta_{1}=\log\frac{\textup{pr}(Y_{i}=1\mid
Z_{i}=1,X_{i}=x)/\textup{pr}(Y_{i}=0\mid
Z_{i}=1,X_{i}=x)}{\textup{pr}(Y_{i}=1\mid
Z_{i}=0,X_{i}=x)/\textup{pr}(Y_{i}=0\mid Z_{i}=0,X_{i}=x)}.$

Importantly, the logistic model assumes a common odds ratio across all values
of the covariates. Moreover, when the outcome is rare in that
$\textup{pr}(Y_{i}=1\mid Z_{i}=1,X_{i}=x)$ and $\textup{pr}(Y_{i}=1\mid
Z_{i}=0,X_{i}=x)$ are close to $0$, the conditional odds ratio approximates
the conditional risk ratio (see Proposition 1.1(3)):

$\beta_{1}\approx\log\frac{\textup{pr}(Y_{i}=1\mid
Z_{i}=1,X_{i}=x)}{\textup{pr}(Y_{i}=1\mid
Z_{i}=0,X_{i}=x)}=\log\textsc{rr}_{ZY\mid x}^{\textup{obs}}.$

Therefore, based on the estimated logistic regression coefficient and the
corresponding confidence limits, we can calculated the E-value immediately.
This is the leading application of the E-value.

###### Example 17.2

The NCHS2003.txt contains the National Center for Health Statistics birth
certificate data, with the following binary indicator variables useful for us:

PTbirth | pre-term birth
---|---
preeclampsia |  pre-eclampsia242424a multisystem hypertensive disorder of pregnancy
ageabove35 |  an older mother with age $\geq 35$ (the treatment)
somecollege | college education
mar | marital status
smoking | smoking status
drinking | drinking status
hispanic | mother’s ethnicity
black | mother’s ethnicity
nativeamerican | mother’s ethnicity
asian | mother’s ethnicity

This version of the data is from Valeri and Vanderweele, (2014). This example
focuses on the outcome PTbirth and Problem 17.3. The following R code computes
the E-values after fitting a logistic regression. Based on the E-values, we
conclude that to explain away the point estimate, the maximum confounding
measure must be larger than 1.94, and to explain away the lower confidence
limit, the maximum confounding measure must be larger than 1.91. Although
these confounding measures are not as strong as those in Section 86, they
appear to be fairly large in epidemiologic studies.

⬇

> evalue = function(rr)

+ {

+ rr + sqrt(rr*(rr - 1))

+ }

>

> NCHS2003 = read.table("NCHS2003.txt", header = TRUE, sep = "\t")

>

> ## outcome: PTbirth

> y_logit = glm(PTbirth ~ ageabove35 +

+ mar + smoking + drinking + somecollege +

+ hispanic + black + nativeamerican + asian,

+ data = NCHS2003,

+ family = binomial)

> log_or = summary(y_logit)$coef[2, 1:2]

> est = exp(log_or[1])

> lower.ci = exp(log_or[1] - 1.96*log_or[2])

> est

Estimate

1.305982

> evalue(est)

Estimate

1.938127

>

> lower.ci

Estimate

1.294619

> evalue(lower.ci)

Estimate

1.912211

#####  87.3 Non-zero true causal effect

Theorem 17.1 assumes no true causal effect of the treatment on the outcome.
Ding and VanderWeele, (2016) proved a general theorem allowing for non-zero
true causal effect.

###### Theorem 17.2

Modify the definition of $\textsc{rr}_{UY\mid x}$ as

$\textsc{rr}_{UY\mid x}=\max_{z=0,1}\frac{\textup{pr}(Y=1\mid
Z=z,U=1,X=x)}{\textup{pr}(Y=1\mid Z=z,U=0,X=x)}.$

We have

$\textsc{rr}_{ZY\mid x}^{\textup{true}}\geq\textsc{rr}_{ZY\mid
x}^{\textup{obs}}\Big{/}\frac{\textsc{rr}_{ZU\mid x}\textsc{rr}_{UY\mid
x}}{\textsc{rr}_{ZU\mid x}+\textsc{rr}_{UY\mid x}-1}.$

Theorem 17.1 is a special case of Theorem 17.2 with $\textsc{rr}_{ZY\mid
x}^{\textup{true}}=1.$ See the original paper of Ding and VanderWeele, (2016)
for the proof of Theorem 17.2. Without assuming any additional assumptions,
Theorem 17.2 states a lower bound of the true risk ratio $\textsc{rr}_{ZY\mid
x}^{\textup{true}}$ given the observed risk ratio $\textsc{rr}_{ZY\mid
x}^{\textup{obs}}$ and the two sensitivity parameters $\textsc{rr}_{ZU\mid x}$
and $\textsc{rr}_{UY\mid x}$.

When the treatment is apparently preventive to the outcome, the observed risk
ratio is smaller than 1. In this case, Theorems 17.1 and 17.2 are not directly
useful, and we must re-label the treatment levels and calculate the E-value
based on $1/\textsc{rr}_{ZY\mid x}^{\textup{obs}}$.

####  88 Critiques and responses

Since the original paper was published, E-value has been a standard number
reported by many epidemiologic studies. Nevertheless, it also attracted
critiques (Ioannidis et al.,, 2019). I will review some limitations of
E-values below.

#####  88.1 E-value is just a monotone transformation of the risk ratio

Figure 20: E-value as a monotone transformation of the risk ratio: three
figures have different ranges of the risk ratio.

From Figure 20, we can see that if the risk ratio is large, then the E-value
$\textsc{rr}_{ZY\mid x}^{\textup{obs}}+\sqrt{\textsc{rr}_{ZY\mid
x}^{\textup{obs}}(\textsc{rr}_{ZY\mid x}^{\textup{obs}}-1)}$ is nearly
$2\textsc{rr}_{ZY\mid x}^{\textup{obs}}$ which is linear in the risk ratio.
For small risk ratio, the E-value is more nonlinear. Critics often say that
the E-value is merely a monotone transformation of the point estimator or the
confidence limits of the risk ratio. So it does not provide any additional
information.

This is partially true. Indeed, the E-value is entirely based on the point
estimator or the confidence limits of the risk ratio. It has a meaningful
interpretation based on Theorem 17.1: to explain away the observed risk ratio,
the maximum of the confounding measures must be at least as large as the
E-value.

#####  88.2 Calibration of the E-value

The E-value equals the maximum value of the association between the confounder
and the treatment and that between the confounder and the outcome to
completely explain aways an observed association. An obvious problem is that
this confounder is fundamentally latent. So it is not trivial to decide
whether a certain E-value is large or small. Another related problem is that
the E-value depends on how many observed covariates $X$ we have controlled for
since it quantifies the strength of the residual confounding given $X$.
Therefore, E-values across studies are not directly comparable. The E-value
provides evidence for causation but this evidence should be assessed carefully
based on background knowledge of the problem of interest.

The following leave-one-covariate-out approach is an intuitive approach to
calibrate the E-value. With $X=(X_{1},\ldots,X_{p})$, we can pretend that the
component $X_{j}$ were not observed and compute the $Z$-$X_{j}$ and
$X_{j}$-$Y$ risk ratios given other observed covariates $(j=1,\ldots,p)$.
These risk ratios provide the range for the confounding measures due to $U$ if
we believe that the unmeasured $U$ is not as strong as all of the observed
covariates. However, I am not aware of any formal justification of this
approach.

#####  88.3 It works the best for a binary outcome and the risk ratio

Theorem 17.1 works well for a binary outcome and the risk ratio. Ding and
VanderWeele, (2016) also proposed sensitivity analysis methods for other
causal parameters, but they are not as elegant as the E-value for binary
outcome based on the risk ratio. The next chapter will propose a simple
sensitivity analysis method for the average causal effect that include several
methods in Part III as special cases.

####  89 Homework Problems

######  17.1 Lemma 17.1

Prove Lemma 17.1.

######  17.2 Schlesselman, (1978)’s formula

For simplicity, we condition on $X$ implicitly in the following discussion.
With binary treatment $Z$, outcome $Y$, and unmeasured confounder $U$, show
that

$\frac{\textsc{rr}_{ZY}^{\textup{obs}}}{\textsc{rr}_{ZY}^{\textup{true}}}=\frac{1+(\gamma-1)\textup{pr}(U=1\mid
Z=1)}{1+(\gamma-1)\textup{pr}(U=1\mid Z=0)}$

assuming a common risk ratio of the treatment on the outcome within both $U=0$
and $U=1$:

$\textsc{rr}_{ZY|U=0}=\textsc{rr}_{ZY|U=1},$

and also a common risk ratio of the confounder on the outcome within both
$Z=0$ and $Z=1$:

$\textsc{rr}_{UY|Z=0}=\textsc{rr}_{UY|Z=1},\text{ denoted by }\gamma.$

Hint: First verify that if $\textsc{rr}_{ZY|U=0}=\textsc{rr}_{ZY|U=1}$ then

$\textsc{rr}_{ZY}^{\textup{true}}=\textsc{rr}_{ZY|U=0}=\textsc{rr}_{ZY|U=1}.$

This identity shows the collapsibility of the risk ratio. In epidemiology, the
risk ratio is a collapsible measure of association.

Remark: Schlesselman, (1978)’s formula does not assume conditional
independence $Z\begin{picture}(9.0,8.0)\put(0.0,0.0){\line(1,0){9.0}}
\put(3.0,0.0){\line(0,1){8.0}} \put(6.0,0.0){\line(0,1){8.0}}
\end{picture}Y\mid U$, but assumes homogeneity of the $Z$-$Y$ and $U$-$Y$ risk
ratios. It is a classic formula for sensitivity analysis. It is an identity
that is simple to implement with pre-specified

$\\{\gamma,\textup{pr}(U=1\mid Z=1),\textup{pr}(U=1\mid Z=0)\\}.$

However, it involves more sensitivity parameters than Theorem 17.1. Even
though Theorem 17.1 only gives an inequality, it is not a loose inequality
compared to Schlesselman, (1978)’s formula under stronger assumptions. With
Theorem 17.1, Schlesselman, (1978)’s formula is only of historical interest.

######  17.3 E-value after logistic regression: data analysis

This problem uses the same dataset as Example 17.2.

Report the E-value for the outcome preeclampsia.

######  17.4 Cornfield-type inequalities for the risk difference

Consider binary $Z,Y,U$, and condition on $X$ implicitly. Assume latent
ignorability given $U$. Show that under
$Z\begin{picture}(9.0,8.0)\put(0.0,0.0){\line(1,0){9.0}}
\put(3.0,0.0){\line(0,1){8.0}} \put(6.0,0.0){\line(0,1){8.0}}
\end{picture}Y\mid U$, we have

$\displaystyle\textsc{rd}_{ZY}^{\textup{obs}}=\textsc{rd}_{ZU}\times\textsc{rd}_{UY}$
(79)

where $\textsc{rd}_{ZY}^{\textup{obs}}$ is the observed risk difference of $Z$
on $Y$, and $\textsc{rd}_{ZU}$ and $\textsc{rd}_{UY}$ are the treatment-
confounder and confounder-outcome risk differences, respectively (recall the
definition of the risk difference in Chapter 2.2).

Remark: Without loss of generality, assume that
$\textsc{rd}_{ZY}^{\textup{obs}},\textsc{rd}_{ZU},\textsc{rd}_{UY}$ are all
positive. Then (79) implies that

$\min(\textsc{rd}_{ZU},\textsc{rd}_{UY})\geq\textsc{rd}_{ZY}^{\textup{obs}}$

and

$\max(\textsc{rd}_{ZU},\textsc{rd}_{UY})\geq\sqrt{\textsc{rd}_{ZY}^{\textup{obs}}}.$

These are the Cornfield inequalities for the risk difference with a binary
confounder. They show that for an unmeasured confounder to explain away an
observed risk difference $\textsc{rd}_{ZY}^{\textup{obs}}$, the treatment-
confounder and confounder-outcome risk differences must both be larger than
$\textsc{rd}_{ZY}^{\textup{obs}}$, and the maximum of them must be larger than
the square root of $\textsc{rd}_{ZY}^{\textup{obs}}$.

Cornfield et al., (1959) obtained, but did not appreciate the significance of
(79). Gastwirth et al., (1998) and Poole, (2010) discussed the first Cornfield
condition for the risk difference, and Ding and VanderWeele, (2014) discussed
the second.

Ding and VanderWeele, (2014) also derived more general results without
assuming a binary $U$. Unfortunately, the results for a general $U$ are weaker
than those above for a binary $U$, that is, the inequalities become looser
with more levels of $U$. This motivated Ding and VanderWeele, (2016) to focus
on the Cornfield inequalities for the risk ratio, which do not deteriorate
with more levels of $U$.

######  17.5 Recommended reading

Ding and VanderWeele, (2016) extended and unified the Cornfield-type
sensitivity analysis, which is the theoretical basis for the notion of
E-value.

###  Chapter 18 Sensitivity Analysis for the Average Causal Effect with
Unmeasured Confounding

Sensitivity Analysis for Average Causal Effect

Cornfield-type sensitivity analysis works the best for binary outcomes on the
risk ratio scale, conditioning on the observed covariates. Although Ding and
VanderWeele, (2016) also proposed Cornfield-type sensitivity analysis methods
for the average causal effect, they are not general enough and are not
convenient to apply. Below I give a more direct approach to sensitivity
analysis based on the conditional expectations of the potential outcomes. The
idea appeared in early work of Robins, (1999) and Scharfstein et al., (1999).
This chapter is based on Lu and Ding, (2023)’s recent formulation.

The approach is closely related to the idea of deriving worse-case bounds on
the average potential outcomes. I will first review the simpler idea of
bounds, and then discuss the approach to sensitivity analysis.

####  90 Introduction

Recall the canonical setup of an observational study with
$\\{Z_{i},X_{i},Y_{i}(1),Y_{i}(0)\\}_{i=1}^{n}\stackrel{{\scriptstyle\text{IID}}}{{\sim}}\\{Z,X,Y(1),Y(0)\\}$
and focus on the average causal effect

$\tau=E\\{Y(1)-Y(0)\\}.$

It decomposes to

$\displaystyle\tau$ $\displaystyle=$ $\displaystyle\left[E(Y\mid
Z=1)\textup{pr}(Z=1)+E\\{Y(1)\mid Z=0\\}\textup{pr}(Z=0)\right]$
$\displaystyle-\left[E\\{Y(0)\mid Z=1\\}\textup{pr}(Z=1)+E(Y\mid
Z=0)\textup{pr}(Z=0)\right].$

So the fundamental difficulty is to estimate the counterfactual means

$E\\{Y(1)\mid Z=0\\},\quad\quad E\\{Y(0)\mid Z=1\\}.$

There are in general two extreme strategies to estimate them.

We have discussed the first strategy in Part III, which relies on
ignorability. Assuming

$\displaystyle E\\{Y(1)\mid Z=1,X\\}$ $\displaystyle=$ $\displaystyle
E\\{Y(1)\mid Z=0,X\\},$ $\displaystyle E\\{Y(0)\mid Z=1,X\\}$ $\displaystyle=$
$\displaystyle E\\{Y(0)\mid Z=0,X\\},$

we can identify the counterfactual means by the observables:

$E\\{Y(1)\mid Z=0\\}=E\left\\{E(Y\mid Z=1,X)\mid Z=0\right\\}$

and, similarly,

$E\\{Y(0)\mid Z=1\\}=E\left\\{E(Y\mid Z=0,X)\mid Z=1\right\\}.$

The second strategy in the next section assumes nothing except that the
outcomes are bounded between $\ell$ and $u$. This is natural for binary
outcomes with $\ell=0$ and $u=1$. With this assumption, the two counterfactual
means are also bounded between $\ell$ and $u$, which implies the worse-case
bounds on $\tau$. I will review this strategy below.

####  91 Manski-type worse-case bounds on the average causal effect without
assumptions

Table 6: Science Table with bounded outcome $[\ell,u]$, where $\ell$ and $u$ are two constants $Z$ | $Y(1)$ | $Y(0)$ | Lower $Y(1)$ | Upper $Y(1)$ | Lower $Y(0)$ | Upper $Y(0)$
---|---|---|---|---|---|---
$1$ | $Y_{1}(1)$ | ? | $Y_{1}(1)$ | $Y_{1}(1)$ | $\ell$ | $u$
$\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$
$1$ | $Y_{n_{1}}(1)$ | ? | $Y_{n_{1}}(1)$ | $Y_{n_{1}}(1)$ | $\ell$ | $u$
$0$ | ? | $Y_{n_{1}+1}(0)$ | $\ell$ | $u$ | $Y_{n_{1}+1}(0)$ | $Y_{n_{1}+1}(0)$
$\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$ | $\vdots$
$0$ | ? | $Y_{n}(0)$ | $\ell$ | $u$ | $Y_{n}(0)$ | $Y_{n}(0)$

Assume that the outcome is bounded between $\ell$ and $u$. From the
decomposition

$E\\{Y(1)\\}=E\\{Y(1)\mid Z=1\\}\textup{pr}(Z=1)+E\\{Y(1)\mid
Z=0\\}\textup{pr}(Z=0),$

we can derive that $E\\{Y(1)\\}$ has lower bound

$E\\{Y\mid Z=1\\}\textup{pr}(Z=1)+\ell\textup{pr}(Z=0)$

and upper bound

$E\\{Y\mid Z=1\\}\textup{pr}(Z=1)+u\textup{pr}(Z=0).$

Similarly, from the decomposition

$E\\{Y(0)\\}=E\\{Y(0)\mid Z=1\\}\textup{pr}(Z=1)+E\\{Y(0)\mid
Z=0\\}\textup{pr}(Z=0),$

we can derive that $E\\{Y(0)\\}$ has lower bound

$\ell\textup{pr}(Z=1)+E\\{Y\mid Z=0\\}\textup{pr}(Z=0)$

and upper bound

$u\textup{pr}(Z=1)+E\\{Y\mid Z=0\\}\textup{pr}(Z=0).$

Combining these bounds, we can derive that the average causal effect
$\tau=E\\{Y(1)\\}-E\\{Y(0)\\}$ has lower bound

$E\\{Y\mid
Z=1\\}\textup{pr}(Z=1)+\ell\textup{pr}(Z=0)-u\textup{pr}(Z=1)-E\\{Y\mid
Z=0\\}\textup{pr}(Z=0)$

and upper bound

$E\\{Y\mid
Z=1\\}\textup{pr}(Z=1)+u\textup{pr}(Z=0)-\ell\textup{pr}(Z=1)-E\\{Y\mid
Z=0\\}\textup{pr}(Z=0).$

The length of the bounds is $u-\ell$, which is not informative but is better
than the a priori bounds $[\ell-u,u-\ell]$ with length $2(u-\ell)$. Without
further assumptions, the observed data distribution does not uniquely
determine $\tau$. In this case, we say that $\tau$ is partially identified,
with the formal definition below.

######  Definition 18.1 (partial identification)

A parameter $\theta$ is partially identified if the observed data distribuion
is compatible with multiple values of $\theta$.

Compare Definitions 10.1 and 18.1. If the parameter $\theta$ is uniquely
determined by the observed data distribution, then it is identifiable;
otherwise, it is partially identifiable. Therefore, $\tau$ is identifiable
with the ignorability assumption, but only partially identifiable without the
ignorability assumption.

Cochran, (1953) used the idea of worse-case bounds in surveys with missing
data, but abandoned the idea because it often gives very conservative results.
Similarly, the worst-case bounds above are often uninteresting from a
practical perspective because they often cover $0$. Moreover, this strategy is
not applicable to the settings with unbounded outcomes.

Manski applied the idea to causal inference (Manski,, 1990) and many other
econometric models (Manski,, 2003). This idea of bounding causal parameters
with minimal assumptions is powerful when coupled with other qualitative
assumptions. Manski, (2003) surveyed many strategies. For instance, we may
believe that the treatment does not harm any units, so the monotonicity
assumption holds: $Y(1)\geq Y(0).$ Then the lower bound on $\tau$ is zero but
the upper bound is unchanged. Another type of assumption is $Z=I\\{Y(1)\geq
Y(0)\\}$, that is, the treatment selection is based on the difference between
the latent potential outcomes. This assumption can also improve the bounds on
$\tau$.

####  92 Sensitivity analysis for the average causal effect

The first strategy is optimistic which assumes that the potential outcomes do
not differ across treatment and control groups, conditioning on the observed
covariates. The second strategy is pessimistic which does not infer the
counterfactual means based on the observed data at all. The following strategy
is in-between.

#####  92.1 Identification formulas

Define

$\displaystyle\frac{E\\{Y(1)\mid Z=1,X\\}}{E\\{Y(1)\mid
Z=0,X\\}}=\varepsilon_{1}(X),$ $\displaystyle\frac{E\\{Y(0)\mid
Z=1,X\\}}{E\\{Y(0)\mid Z=0,X\\}}=\varepsilon_{0}(X),$

which are the sensitivity parameters. For simplicity, we can further assume
that they are constant independent of $X$. In practice, we need to fix them or
vary them in a pre-specified range. Recall that $\mu_{1}(X)=E(Y\mid Z=1,X)$
and $\mu_{0}(X)=E(Y\mid Z=0,X)$. We can identify the two counterfactual means
and the average causal effect as follows.

###### Theorem 18.1

With known $\varepsilon_{1}(X)$ and $\varepsilon_{0}(X)$, we have

$\displaystyle E\\{Y(1)\mid Z=0\\}$ $\displaystyle=$ $\displaystyle
E\left\\{\mu_{1}(X)/\varepsilon_{1}(X)\mid Z=0\right\\},$ $\displaystyle
E\\{Y(0)\mid Z=1\\}$ $\displaystyle=$ $\displaystyle
E\left\\{\mu_{0}(X)\varepsilon_{0}(X)\mid Z=1\right\\}$

and therefore

$\displaystyle\tau$ $\displaystyle=$ $\displaystyle
E\\{ZY+(1-Z)\mu_{1}(X)/\varepsilon_{1}(X)\\}$ (81)
$\displaystyle-E\\{Z\mu_{0}(X)\varepsilon_{0}(X)+(1-Z)Y\\}$ $\displaystyle=$
$\displaystyle E\\{Z\mu_{1}(X)+(1-Z)\mu_{1}(X)/\varepsilon_{1}(X)\\}$
$\displaystyle-E\\{Z\mu_{0}(X)\varepsilon_{0}(X)+(1-Z)\mu_{0}(X)\\}.$

I leave the proof of Theorem 18.1 to Problem 18.1. With the fitted outcome
model, (81) and (81) motivate the following predictive and projective
estimators for $\tau$:

$\displaystyle\hat{\tau}^{\textup{pred}}$ $\displaystyle=$
$\displaystyle\left\\{n^{-1}\sum_{i=1}^{n}Z_{i}Y_{i}+n^{-1}\sum_{i=1}^{n}(1-Z_{i})\hat{\mu}_{1}(X_{i})/\varepsilon_{1}(X_{i})\right\\}$
$\displaystyle-\left\\{n^{-1}\sum_{i=1}^{n}Z_{i}\hat{\mu}_{0}(X_{i})\varepsilon_{0}(X_{i})+n^{-1}\sum_{i=1}^{n}(1-Z_{i})Y_{i}\right\\},$

and

$\displaystyle\hat{\tau}^{\textup{proj}}$ $\displaystyle=$
$\displaystyle\left\\{n^{-1}\sum_{i=1}^{n}Z_{i}\hat{\mu}_{1}(X_{i})+n^{-1}\sum_{i=1}^{n}(1-Z_{i})\hat{\mu}_{1}(X_{i})/\varepsilon_{1}(X_{i})\right\\}$
$\displaystyle-\left\\{n^{-1}\sum_{i=1}^{n}Z_{i}\hat{\mu}_{0}(X_{i})\varepsilon_{0}(X_{i})+n^{-1}\sum_{i=1}^{n}(1-Z_{i})\hat{\mu}_{0}(X_{i})\right\\}.$

The terminologies “predictive” and “projective” are from the survey sampling
literature (Firth and Bennett,, 1998; Ding and Li,, 2018). The estimators
$\hat{\tau}^{\textup{pred}}$ and $\hat{\tau}^{\textup{proj}}$ differ slightly:
the former uses the observed outcomes when available; in contrast, the latter
replaces the observed outcomes with the fitted values.

More interesting, we can also identify $\tau$ by an inverse probability
weighting formula.

###### Theorem 18.2

With known $\varepsilon_{1}(X)$ and $\varepsilon_{0}(X)$, we have

$\displaystyle E\\{Y(1)\\}=E\left\\{w_{1}(X)\frac{Z}{e(X)}Y\right\\},\quad
E\\{Y(0)\\}=E\left\\{w_{0}(X)\frac{1-Z}{1-e(X)}Y\right\\},$

where

$w_{1}(X)=e(X)+\\{1-e(X)\\}/\varepsilon_{1}(X),\quad
w_{0}(X)=e(X)\varepsilon_{0}(X)+1-e(X).$

I leave the proof of Theorem 18.2 to Problem 18.2. Theorem 18.2 modifies the
classic inverse probability weighting formulas with two extra factors
$w_{1}(X)$ and $w_{0}(X)$ depending on both the propensity score and the
sensitivity parameters. With the fitted propensity score model, Theorem 18.2
motivates the following estimators for $\tau$:

$\displaystyle\hat{\tau}^{\textup{ht}}$ $\displaystyle=$ $\displaystyle
n^{-1}\sum_{i=1}^{n}\frac{\\{\hat{e}(X_{i})\varepsilon_{1}(X_{i})+1-\hat{e}(X_{i})\\}Z_{i}Y_{i}}{\varepsilon_{1}(X_{i})\hat{e}(X_{i})}$
$\displaystyle-n^{-1}\sum_{i=1}^{n}\frac{\\{\hat{e}(X_{i})\varepsilon_{0}(X_{i})+1-\hat{e}(X_{i})\\}(1-Z_{i})Y_{i}}{1-\hat{e}(X_{i})}$

and

$\displaystyle\hat{\tau}^{\textup{haj}}$ $\displaystyle=$
$\displaystyle\sum_{i=1}^{n}\frac{\\{\hat{e}(X_{i})\varepsilon_{1}(X_{i})+1-\hat{e}(X_{i})\\}Z_{i}Y_{i}}{\varepsilon_{1}(X_{i})\hat{e}(X_{i})}\Big{/}\sum_{i=1}^{n}\frac{Z_{i}}{\hat{e}(X_{i})}$
$\displaystyle-n^{-1}\sum_{i=1}^{n}\frac{\\{\hat{e}(X_{i})\varepsilon_{0}(X_{i})+1-\hat{e}(X_{i})\\}(1-Z_{i})Y_{i}}{1-\hat{e}(X_{i})}\Big{/}\sum_{i=1}^{n}\frac{1-Z_{i}}{1-\hat{e}(X_{i})}.$

More interestingly, with fitted propensity score and outcome models, the
following estimator for $\tau$ is doubly robust:

$\displaystyle\hat{\tau}^{\textup{ht}}=\hat{\tau}^{\textup{ipw}}-n^{-1}\sum_{i=1}^{n}\\{Z_{i}-\hat{e}(X_{i})\\}\left\\{\frac{\hat{\mu}_{1}(X_{i})}{\hat{e}(X_{i})\varepsilon_{1}(X_{i})}+\frac{\hat{\mu}_{0}(X_{i})\varepsilon_{0}(X_{i})}{1-\hat{e}(X_{i})}\right\\}.$

That is, with known $\varepsilon_{1}(X_{i})$ and $\varepsilon_{0}(X_{i})$, the
estimator $\hat{\tau}^{\textup{dr}}$ is consistent for $\tau$ if either the
propensity score model or the outcome model is correctly specified. We can use
the bootstrap to approximate the variance of the above estimators. See Lu and
Ding, (2023) for technical details.

When $\varepsilon_{1}(X_{i})=\varepsilon_{0}(X_{i})=1$, the above estimators
reduce to the predictive estimator, inverse probability weighting estimator,
and the doubly robust estimators introduced in Part III.

####  93 Example

We revisit Example 10.3. With

$\varepsilon_{1}(X)=\varepsilon_{0}(X)\in\\{1/2,1/1.7,1/1.5,1/1.3,1,1.3,1.5,1.7,2\\},$

we obtain an array of doubly robust estimates of $\tau$.

⬇

1/2 1/1.7 1/1.5 1/1.3 1 1.3 1.5 1.7 2

1/2 11.62 10.44 9.40 8.03 4.96 0.97 -1.69 -4.35 -8.34

1/1.7 9.22 8.05 7.00 5.64 2.57 -1.42 -4.08 -6.75 -10.74

1/1.5 7.63 6.45 5.41 4.05 0.97 -3.02 -5.68 -8.34 -12.33

1/1.3 6.03 4.86 3.81 2.45 -0.62 -4.61 -7.27 -9.94 -13.93

1 3.64 2.47 1.42 0.06 -3.01 -7.01 -9.67 -12.33 -16.32

1.3 1.80 0.63 -0.42 -1.78 -4.85 -8.85 -11.51 -14.17 -18.16

1.5 0.98 -0.19 -1.24 -2.60 -5.67 -9.66 -12.33 -14.99 -18.98

1.7 0.36 -0.82 -1.86 -3.23 -6.30 -10.29 -12.95 -15.61 -19.60

2 -0.35 -1.52 -2.57 -3.93 -7.00 -10.99 -13.65 -16.32 -20.31

The signs of the estimates are not sensitive to sensitivity parameters larger
than 1, but they are quite sensitivity to sensitivity parameters smaller than
1. When the participants of the meal plan tend to have higher BMI, the average
causal effect of the meal plan on BMI is negative. However, this conclusion
can be quite sensitive if the participants of the meal plan tend to have lower
BMI.

####  94 Homework Problems

######  18.1 Proof of Theorem 18.1

Prove Theorem 18.1.

######  18.2 Proof of Theorem 18.2

Prove Theorem 18.2.

######  18.3 Sensitivity analysis for the average causal effect on the treated
units $\tau_{\textsc{T}}$

This problem extends Chapter 13 to allow for unmeasured confounding for
estimating

$\tau_{\textsc{T}}=E\\{Y(1)-Y(0)\mid Z=1\\}=E(Y\mid Z=1)-E\\{Y(0)\mid Z=1\\}.$

We can easily estimate $E(Y\mid Z=1)$ by the sample moment. The only
counterfactual term is $E\\{Y(0)\mid Z=1\\}$. Therefore, we only need the
sensitivity parameter $\varepsilon_{0}(X)$. We have the following two
identification formulas with a known $\varepsilon_{0}(X).$

###### Theorem 18.3

With known $\varepsilon_{0}(X)$, we have

$\displaystyle E\\{Y(0)\mid Z=1\\}$ $\displaystyle=$ $\displaystyle
E\left\\{Z\mu_{0}(X)\varepsilon_{0}(X)\right\\}/e$ $\displaystyle=$
$\displaystyle E\left\\{e(X)\varepsilon_{0}(X)\frac{1-Z}{1-e(X)}Y\right\\}/e,$

where $e=\textup{pr}(Z=1)$

Prove Theorem 18.3.

Remark: Theorem 18.3 motivates using
$\hat{\tau}_{\textsc{t}}^{*}=\hat{\mu}_{\textsc{t}1}-\hat{\mu}_{\textsc{t}0}^{*}$
to estimate $\tau_{\textsc{t}}$, where
$\hat{\mu}_{\textsc{t}1}=\sum_{i=1}^{n}Z_{i}Y_{i}/\sum_{i=1}^{n}Z_{i}$ and

$\displaystyle\hat{\mu}_{\textsc{t}0}^{\text{reg}}$ $\displaystyle=$
$\displaystyle
n_{1}^{-1}\sum_{i=1}^{n}Z_{i}\varepsilon_{0}(X_{i})\hat{\mu}_{0}(X_{i}),$
$\displaystyle\hat{\mu}_{\textsc{t}0}^{\text{ht}}$ $\displaystyle=$
$\displaystyle
n_{1}^{-1}\sum_{i=1}^{n}\varepsilon_{0}(X_{i})\hat{o}(X_{i})(1-Z_{i})Y_{i},$
$\displaystyle\hat{\mu}_{\textsc{t}0}^{\text{haj}}$ $\displaystyle=$
$\displaystyle\sum_{i=1}^{n}\varepsilon_{0}(X_{i})\hat{o}(X_{i})(1-Z_{i})Y_{i}/\sum_{i=1}^{n}\hat{o}(X_{i})(1-Z_{i}),$

with $\hat{o}(X_{i})=\hat{e}(X_{i})/\\{1-\hat{e}(X_{i})\\}$ being the
estimated conditional odds of the treatment. Moreover, we can construct the
doubly robust estimator
$\hat{\tau}_{\textsc{t}}^{\text{dr}}=\hat{\mu}_{\textsc{t}1}-\hat{\mu}_{\textsc{t}0}^{\text{dr}}$
for $\tau_{\textsc{t}}$, where

$\hat{\mu}_{\textsc{t}0}^{\text{dr}}=\hat{\mu}_{\textsc{t}0}^{\text{ht}}-n_{1}^{-1}\sum_{i=1}^{n}\varepsilon_{0}(X_{i})\frac{\hat{e}(X_{i})-Z}{1-\hat{e}(X_{i})}\hat{\mu}_{0}(X_{i}).$

Lu and Ding, (2023) provide more details and also propose a doubly robust
estimator for $\tau_{\textsc{T}}$.

######  18.4 R code

Implement the estimators in Problem 18.3.

######  18.5 Recommended reading

Rosenbaum and Rubin, 1983a  and Imbens, (2003) are two classic papers on
sensitivity analysis which, however, involve more complicated procedures.

###  Chapter 19 Rosenbaum-Style $p$-Values for Matched Observational Studies
with Unmeasured Confounding

Sensitivity Analysis in Matching

Rosenbaum, 1987b  introduced a sensitivity analysis technique for matched
observational studies. Although it works for general settings (Rosenbaum,
2002b, ), the theory is most elegant for one-to-one matching. Different from
Chapters 17 and 18, Rosenbaum-type sensitivity analysis works the best for
matched observational studies for testing the sharp null hypothesis of no
individual treatment effect.

####  95 The model for sensitivity analysis with matched data

Consider exactly matched pairs from an observational study, with $(i,j)$
indexing unit $j$ in pair $i$ $(i=1,\ldots,n;j=1,2)$. Assume iid sampling, and
define the propensity score as

$e_{ij}=\textup{pr}\\{Z_{ij}=1\mid X_{i},Y_{ij}(1),Y_{ij}(0)\\}.$

Let $\mathbb{S}_{i}=\\{Y_{i1}(1),Y_{i1}(0),Y_{i2}(1),Y_{i2}(0)\\}$ denote the
set of all potential outcomes within pair $i$. Conditioning on the event that
$Z_{i1}+Z_{i2}=1$, we have

$\displaystyle\pi_{i1}$ $\displaystyle=$
$\displaystyle\textup{pr}\\{Z_{i1}=1\mid
X_{i},\mathbb{S}_{i},Z_{i1}+Z_{i2}=1\\}$ $\displaystyle=$
$\displaystyle\frac{\textup{pr}\\{Z_{i1}=1,Z_{i2}=0\mid
X_{i},\mathbb{S}_{i}\\}}{\textup{pr}\\{Z_{i1}+Z_{i2}=1\mid
X_{i},\mathbb{S}_{i}\\}}$ $\displaystyle=$
$\displaystyle\frac{\textup{pr}\\{Z_{i1}=1,Z_{i2}=0\mid
X_{i},\mathbb{S}_{i}\\}}{\textup{pr}\\{Z_{i1}=1,Z_{i2}=0\mid
X_{i},\mathbb{S}_{i}\\}+\textup{pr}\\{Z_{i1}=0,Z_{i2}=1\mid
X_{i},\mathbb{S}_{i}\\}}$ $\displaystyle=$
$\displaystyle\frac{e_{i1}(1-e_{i2})}{e_{i1}(1-e_{i2})+(1-e_{i1})e_{i2}}$

Define $o_{ij}=e_{ij}/(1-e_{ij})$ as the odds of the treatment for unit
$(i,j)$, and we have

$\pi_{i1}=\frac{o_{i1}}{o_{i1}+o_{i2}}.$

Under ignorability, $e_{ij}$ is only a function of $X_{i}$, and therefore,
$e_{i1}=e_{i2}$ and $\pi_{i1}=1/2$. Thus the treatment assignment mechanism
conditioning on covariates and potential outcomes is equivalent to that from
an MPE with equal treatment and control probabilities. This is a strategy to
analyze matched observational studies we discussed in Chapter 73.

In general, $e_{ij}$ is also a function of the unobserved potential outcomes,
and it can range from $0$ to $1$. Rosenbaum, 1987b ’s model for sensitivity
analysis imposes bounds on the odds ratio $o_{i1}/o_{i2}$.

######  Assumption 19.1 (Rosenbaum’s sensitivity model)

The odds ratios are bounded by

$o_{i1}/o_{i2}\leq\Gamma,\quad o_{i2}/o_{i1}\leq\Gamma,$

for some pre-specified $\Gamma\geq 1$. Equivalently,

$\frac{1}{1+\Gamma}\leq\pi_{i1}\leq\frac{\Gamma}{1+\Gamma}$

for some pre-specified $\Gamma\geq 1$.

Under Assumption 19.1, we have a biased MPE with unequal and varying treatment
and control probabilities across pairs. When $\Gamma=1$, we have $\pi_{i1}$
and thus a standard MPE. Therefore, $\Gamma>1$ measures the deviation from the
ideal MPE due to the omitted variables in matching.

####  96 Worst-case $p$-values under Rosenbaum’s sensitivity model

Consider testing the sharp null hypothesis

$H_{0\textsc{f}}:Y_{ij}(1)=Y_{ij}(0)\text{ for }i=1,\ldots,n\text{ and }j=1,2$

based on the within-pair differences
$\hat{\tau}_{i}=(2Z_{i1}-1)(Y_{i1}-Y_{i2})\ (i=1,\ldots,n).$ Under
$H_{0\textsc{f}}$, $|\hat{\tau}_{i}|$ is fixed but $S_{i}=I(\hat{\tau}_{i}>0)$
is random if $\hat{\tau}_{i}\neq 0$. Consider the following class of test
statistics:

$T=\sum_{i=1}^{n}S_{i}q_{i},$

where $q_{i}\geq 0$ is a function of
$(|\hat{\tau}_{1}|,\ldots,|\hat{\tau}_{n}|).$ Special cases include the sign
statistic, the pair $t$ statistic (up to some constant shift), and the
Wilcoxon sign rank statistic:

$T=\sum_{i=1}^{n}S_{i},\quad T=\sum_{i=1}^{n}S_{i}|\hat{\tau}_{i}|,\quad
T=\sum_{i=1}^{n}S_{i}R_{i},$

where $(R_{1},\ldots,R_{n})$ are the ranks of
$(|\hat{\tau}_{1}|,\ldots,|\hat{\tau}_{n}|).$

What is the null distribution of the test statistic with general $\Gamma$? It
can be quite complicated because we do not fully specify the exact values of
the $\pi_{i1}$’s. Fortunately, we know that the worse case distribution
correspond to

$S_{i}\stackrel{{\scriptstyle\text{IID}}}{{\sim}}\text{Bernoulli}\left(\frac{\Gamma}{1+\Gamma}\right).$

Here, the FRT with $T$ has the largest $p$-value under the “the worst case”
distribution. The corresponding distribution has mean

$E_{\Gamma}(T)=\frac{\Gamma}{1+\Gamma}\sum_{i=1}^{n}q_{i},$

and variance

$\textup{var}_{\Gamma}(T)=\frac{\Gamma}{(1+\Gamma)^{2}}\sum_{i=1}^{n}q_{i}^{2},$

with a Normal approximation

$\frac{T-\frac{\Gamma}{1+\Gamma}\sum_{i=1}^{n}q_{i}}{\sqrt{\frac{\Gamma}{(1+\Gamma)^{2}}\sum_{i=1}^{n}q_{i}^{2}}}\stackrel{{\scriptstyle\textup{d}}}{{\longrightarrow}}\textsc{N}(0,1).$

In practice, we can report a sequence of $p$-values as a function of $\Gamma$.

####  97 Revisiting the LaLonde data

We conduct Rosenbaum-style sensitivity analysis in the matched LaLonde data.
We consider using the test statistic $T=\sum_{i=1}^{n}S_{i}|\hat{\tau}_{i}|$.
Under the ideal matched pair experiment with $\Gamma=1$, we can simulate the
distribution of $T$ and obtain the $p$-value $0.002$, as shown in the first
subfigure in Figure 21. With a slightly larger $\Gamma=1.1$, the distribution
of $T$ shifts to the right and the $p$-value increases to $0.011$. If we
further increase $\Gamma$ to $1.3$, then the distribution of $T$ shifts
further and the $p$-value exceeds $0.05$. Figure 22 shows the histogram of the
$\hat{\tau}_{i}$’s and the $p$-value as a function of $\Gamma$; $\Gamma=1.233$
measures the maximum confounding that we can still reject the null hypothesis
at level $0.05$.

We can also use the senmw function in the sensitivitymw package to obtain a
sequence of $p$-values against $\Gamma$, as shown in Figure 22.

Figure 21: Distributions of $T=\sum_{i=1}^{n}S_{i}|\hat{\tau}_{i}|$ with
$S_{i}$ iid Bernoulli$(\Gamma/(1+\Gamma))$, based on the LaLonde data. Figure
22: $p$-value as a function of $\Gamma$, based on the LaLonde data.

(a) erpcp data

(b) ead250 data

Figure 23: Two examples

####  98 Homework Problems

######  19.1 Application of Rosenbaum’s approach

Re-analyze Example 10.3 using Rosenbaum’s approach.

######  19.2 Recommended reading

Rosenbaum, (2015) provides a tutorial for his two R packages for sensitivity
analysis with matched observational studies.

###  Chapter 20 Overlap in Observational Studies: Difficulties and
Opportunities

####  99 Implications of overlap

In Part III of this book, causal inference with observational studies relies
on two critical assumptions: unconfoundedness

$Z\begin{picture}(9.0,8.0)\put(0.0,0.0){\line(1,0){9.0}}
\put(3.0,0.0){\line(0,1){8.0}} \put(6.0,0.0){\line(0,1){8.0}}
\end{picture}\\{Y(1),Y(0)\\}\mid X$

and overlap

$0<e(X)<1.$

D’Amour et al., (2021) pointed out the tension between these two assumptions:
typically, more covariates make the unconfoundedness assumption more plausible
(ignoring M-bias discussed in Chapter 82.1), but more covariates make the
overlap assumption less plausible because the treatment becomes more
predictable.

If some units has $e(X)=0$ or $e(X)=1$, then we have philosophic difficulty of
thinking about the counterfactual potential outcomes (King and Zeng,, 2006).
In particular, if a unit deterministically receives the treatment, then it may
not be meaningful to conceive its potential outcome under control; vice versa.
Even if the true propensity score is not exactly $0$ or $1$, the estimated
propensity score can be very close to $0$ or $1$ in finite sample, which makes
the estimators based on inverse probability weighting numerically unstable.
Many statistical analyses in fact require a strict version of overlap:

######  Assumption 20.1 (strict overlap)

$\eta\leq e(X)\leq 1-\eta$ for some $\eta\in(0,1/2).$

However, D’Amour et al., (2021, Corollary 1) showed that Assumption 20.1 has
very strong implications. For simplicity, I present only one of their results.
Let $X_{k}\ (k=1,\ldots,p)$ be the $k$th component of the covariate
$X=(X_{1},\ldots,X_{p})$, and $e=\textup{pr}(Z=1)$ be the proportion of the
treated units.

###### Theorem 20.1

Assumption 20.1 implies that $\eta\leq e\leq 1-\eta$ and

$\displaystyle p^{-1}\sum_{k=1}^{p}\big{|}E(X_{k}\mid Z=1)-E(X_{k}\mid
Z=0)\Big{|}$ (82) $\displaystyle\leq$ $\displaystyle
p^{-1/2}C^{1/2}\left\\{e\lambda_{1}^{1/2}+(1-e)\lambda_{0}^{1/2}\right\\},$

where

$C=\frac{(e-\eta)(1-e-\eta)}{e^{2}(1-e)^{2}\eta(1-\eta)}$

is a positive constant depending only on $(e,\eta)$, and $\lambda_{1}$ and
$\lambda_{0}$ are the maximum eigenvalues of the covariance matrices
$\textup{cov}(X\mid Z=1)$ and $\textup{cov}(X\mid Z=0)$, respectively.

What is the order of the maximum eigenvalues in Theorem 20.1? D’Amour et al.,
(2021) showed that it is usually smaller than $O(p)$ unless the components of
$X$ are highly correlated. If the components of $X$ are highly correlated,
then some components are redundant after including other components. If the
components of $X$ are not highly correlated, then the right-hand side
converges to zero. So the average difference in means of the covariates is
close to zero, that is, the treatment and control groups are nearly balanced
in means averaged over all dimensions of the covariates. Mathematically, the
left-hand side of (82) converging to zero rules out the possibility that all
dimensions of $X$ have non-vanishing difference in means across treatment and
control groups. It is a strong requirement in observational studies with many
covariates.

#####  99.1 Trimming in the presence of limited overlap

When Assumption 20.1 does not hold, it is common to trim the units based on
the estimated propensity scores (Crump et al.,, 2009; Yang and Ding,, 2018).
Trimming drops units within regions of little overlap, which changes the
population and estimand. The restrictive implications of overlap in Assumption
20.1 suggest that trimming must be employed more often and one may need to
trim a large proportion of units to achieve desirable overlap in high
dimensions.

#####  99.2 Outcome modeling in the presence of limited overlap

The somewhat negative results in D’Amour et al., (2021) also highlight the
limitation of focusing only on the propensity score in the presence of limited
overlap. With high dimensional covariates, outcome modeling becomes more
important. In particular, if the outcome means only depend on a function of
the original covariates in that

$E\\{Y(z)\mid X\\}=f_{z}(r(X)),\quad(z=0,1)$

then it suffices to control for $r(X)$, a lower dimensional summary of the
original covariates. Due to the dimension reduction, the strict overlap
condition on $r(X)$ can be much weaker than the strict overlap condition on
$X$. This is conceptually straightforward, but the corresponding theory and
methods are missing.

####  100 Causal inference with no overlap: regression discontinuity

Starting from the simple case with a univariate $X$. An extreme treatment
assignment mechanism is a deterministic one:

$Z=1(X\geq x_{0}),$

where $x_{0}$ is a predetermined threshold. An interesting consequence of this
assignment is that the unconfoundedness assumption holds automatically:

$Z\begin{picture}(9.0,8.0)\put(0.0,0.0){\line(1,0){9.0}}
\put(3.0,0.0){\line(0,1){8.0}} \put(6.0,0.0){\line(0,1){8.0}}
\end{picture}\\{Y(1),Y(0)\\}\mid X$

because $Z$ is a deterministic function of $X$. However, the overlap
assumption is violated by definition:

$e(X)=\textup{pr}(Z=1\mid X)=1(X\geq x_{0})=\begin{cases}1&\text{ if }X\geq
x_{0},\\\ 0&\text{ if }X<x_{0}.\end{cases}$

So our analytic strategies discussed in Part IV are no longer applicable here.
We must change our perspective.

The discussion here seems contrived, with a deterministic treatment
assignment. Interestingly, it has many applications in practice, and is called
regression discontinuity. Below, I first review some canonical examples and
then give a mathematical formulation of this type of studies.

#####  100.1 Examples and graphical diagnostics

###### Example 20.1

Thistlethwaite and Campbell, (1960) first proposed the idea of regression-
discontinuity analysis. Their motivating example was to study the effect of
students’ winning Certificated of Merit on later career plans, where the
Certificated of Merit was determined by whether the Scholarship Qualifying
Test score was above a certain threshold. Their initial analysis was mainly
graphical. Figure 24 shows one of their graphs.

Figure 24: A graph from Thistlethwaite and Campbell, (1960) with minor
modifications of the unclear text in the original paper

###### Example 20.2

Bor et al., (2014) used regression discontinuity to study the effect of when
to start HIV patients with antiretroviral on their mortality, where the
treatment is determined by whether the patients’ CD4 counts were below 200
cells/$\mu$L.252525CD4 cells are white blood cells that fight infection.

###### Example 20.3

Carpenter and Dobkin, (2009) studied the effect of alcohol consumption on
mortality, which leverages the minimum legal drinking age as a discontinuity
for alcohol consumption. They derived mortality data from the National Center
for Health Statistics, including the decedent’s date of birth and date of
death. They computed age profile of deaths per 100,000 person years with
outcomes measured by the following nine variables:  all all deaths, the sum of
internal and external internal deaths due to internal causes external deaths
due to external causes, the sum of the rest homicide homicides suicide
suicides mva motor vehicle accidents alcohol deaths with a mention of alcohol
drugs deaths with a mention of drug use externalother deaths due to other
external causes

Figure 25 plots the number of deaths per 100,000 person years for nine
measures based on the data used by Angrist and Pischke, (2014). From the jumps
at age 21, it seems obvious that there is an increase of mortality at age 21,
primarily due to motor vehicle accidents. I leave the formal analysis as
Problem 20.3.

Figure 25: Minimum legal drinking age example

#####  100.2 A mathematical formulation of regression discontinuity

The technical term for the variable $X$ that determines the treatment is the
running variable. Intuitively, regression discontinuity can identify a local
average causal effect at the cutoff point $x_{0}$:

$\tau(x_{0})=E\\{Y(1)-Y(0)\mid X=x_{0}\\}.$

In particular, for the potential outcome under treatment, we have

$\displaystyle E\\{Y(1)\mid X=x_{0}\\}$ $\displaystyle=$
$\displaystyle\lim_{\varepsilon\rightarrow 0+}E\\{Y(1)\mid
X=x_{0}+\varepsilon\\}$ (83) $\displaystyle=$
$\displaystyle\lim_{\varepsilon\rightarrow 0+}E\\{Y(1)\mid
Z=1,X=x_{0}+\varepsilon\\}$ (84) $\displaystyle=$
$\displaystyle\lim_{\varepsilon\rightarrow 0+}E(Y\mid
Z=1,X=x_{0}+\varepsilon),$ (85)

where (83) holds if $E\\{Y(1)\mid X=x\\}$ is continuous from the right at
$x_{0}$ and (84) follows by the definition of $Z$. Similarly, for the
potential outcome under control, we have

$E\\{Y(0)\mid X=x_{0}\\}=\lim_{\varepsilon\rightarrow 0+}E(Y\mid
Z=0,X=x_{0}-\varepsilon)$

if $E\\{Y(0)\mid X=x\\}$ is continuous from the left at $x_{0}$. So the local
average causal effect at $x_{0}$ can be identified by the difference of the
two limits. I summarize the key identification result below.

###### Theorem 20.2

Assume that $E\\{Y(1)\mid X=x\\}$ is continuous from the right at $x_{0}$ and
$E\\{Y(0)\mid X=x\\}$ is continuous from the left at $x_{0}$. Then the local
average treatment effect at $X=x_{0}$ is identified by

$\tau(x_{0})=\lim_{\varepsilon\rightarrow 0+}E(Y\mid
Z=1,X=x_{0}+\varepsilon)-\lim_{\varepsilon\rightarrow 0+}E(Y\mid
Z=0,X=x_{0}-\varepsilon).$

Since the right-hand side of the above equation only involves observables, the
parameter $\tau(x_{0})$ is nonparametrically identified. However, the form of
the identification formula is totally different from what we derived before.
In particular, the identification formula involves limits of two conditional
expectation functions.

#####  100.3 Regressions near the boundary

If we are lucky, graphical diagnostic sometimes can clearly show the causal
effect at the cutoff point. However, many outcomes are noisy so graphical
diagnostic is not enough. Figure 26 shows two examples with obvious jumps at
the cutoff point and two examples without obvious jumps, although the
underlying data generating processes all have discontinuities.

Figure 26: Examples of regression discontinuity. In the first column, the data
generating processes result in visible jumps at the cutoff points; in the
second column, the jumps are not so visible. In the first row, the data
generating processes have constant $\tau(x)$; in the second row, $\tau(x)$
varies with $x$.

Assume that $E(Y\mid Z=1,X=x)=\gamma_{1}+\beta_{1}x$ and $E(Y\mid
Z=0,X=x)=\gamma_{0}+\beta_{0}x$ are linear in $x$. We can run OLS based on the
treated and control data to obtain the fitted lines
$\hat{\gamma}_{1}+\hat{\beta}_{1}x$ and $\hat{\gamma}_{0}+\hat{\beta}_{0}x$,
respectively. We can then estimate the average causal effect at the point
$X=x_{0}$ as

$\hat{\tau}(x_{0})=(\hat{\gamma}_{1}-\hat{\gamma}_{0})+(\hat{\beta}_{1}-\hat{\beta}_{0})x_{0}.$

Numerically, $\hat{\tau}(x_{0})$ is identical to the coefficient of $Z_{i}$ in
the OLS

$\displaystyle Y_{i}\sim\\{1,Z_{i},X_{i}-x_{0},Z_{i}(X_{i}-x_{0})\\},$ (86)

and it is also identical to the coefficient of $Z_{i}$ in the OLS

$\displaystyle Y_{i}\sim\\{1,Z_{i},R_{i},L_{i}\\},$ (87)

where

$R_{i}=\max(X_{i}-x_{0},0),\qquad L_{i}=\min(X_{i}-x_{0},0)$

indicate the right and left parts of $X_{i}-x_{0}$, respectively. I leave the
algebraic details to Problem 20.1.

However, this approach may be sensitive to the violation of the linear model.
Theory suggests running regression using only the local observations near the
cutoff point262626This is called local linear regression in nonparametric
statistics, which belongs to a broader class of local polynomial regression
(Fan and Gijbels,, 1996). . However, the rule for choosing the “local points”
are quite involved. Fortunately, the rdrobust function in the rdrobust package
in R implements various choices of “local points.” Since choosing the “local
points” is the key in regression discontinuity, it seems more sensible to
report estimates and confidence intervals based on various choices of the
“local points.”

#####  100.4 An example

Lee, (2008) gave a famous example of using regression discontinuity to study
the incumbency advantage in the U.S. House. He wrote that “incumbents are, by
definition, those politicians who were successful in the previous election. If
what makes them successful is somewhat persistent over time, they should be
expected to be somewhat more successful when running for re-election.”
Therefore, this is a fundamentally challenging causal inference problem. The
regression discontinuity is a clever study design to study this problem.

The running variable is the lagged vote in the previous election centered at
$0$, and the outcome is the vote in current election, with units being the
congressional districts. The treatment is the binary indicator for being the
current incumbent party in a district, determined by the lagged vote. Figure
27 show the raw data.

Figure 27: Raw data of Lee, (2008)

The rdrobust function gives three sets of the point estimate and confidence
intervals. They all suggest positive incumbency advantage.

⬇

> library(rdrobust)

> library(rddtools)

> data(house)

> RDDest = rdrobust(house$y, house$x)

[1] "Mass points detected in the running variable."

> cbind(RDDest$coef, RDDest$ci)

Coeff CI Lower CI Upper

Conventional 0.06372533 0.04224798 0.08520269

Bias-Corrected 0.05937028 0.03789292 0.08084763

Robust 0.05937028 0.03481238 0.08392818

Figure 28 shows the point estimates and the confidence intervals based on OLS
with different choices of the local points defined by $|X|<x_{0}$. While the
point estimates and the confidence intervals are sensitive to the choice of
$x_{0}$, the qualitative result remains the same as above.

Figure 28: Estimates based on local linear regressions

#####  100.5 Problems of regression discontinuity

What can go wrong with the regression discontinuity analysis? The technical
challenge is to specify the neighborhood near the cutoff point. We have
discussed this issue above.

In addition, Theorem 20.2 holds under a continuity condition. It may be
violated in practice. For instance, if the mortality rate jumps at the age of
21, then the jumps in Figure 25 may not be due to the change of drinking
behavior due to the legal drink age. However, it is hard to check the
violation of the continuity condition empirically.

McCrary, (2008) proposed an indirect test for the validity of the regression
discontinuity. He suggested checking the density of the running variable at
the cutoff point. The discontinuity in the density of the running variable at
the cutoff point may suggest that some units were able to manipulate their
treatment status perfectly.

####  101 Homework Problems

######  20.1 Linear potential outcome models

This problem gives more details for the numerical equivalence in Section
100.3.

Show that $\hat{\tau}(x_{0})$ equals the coefficients of $Z_{i}$ in OLS fits
(86) and (86).

Hint: It is helpful for start with the figures of $Z_{i}(X_{i}-x_{0})$,
$L_{i}$, and $R_{i}$ with $X_{i}$ on the x-axis. The conclusion holds by
reparametrizating the OLS regressions.

######  20.2 Simulation for regression discontinuity

RDD_numerical.R simulates potential outcomes from linear models and generates
Figure 26. Change them to nonlinear models, and compare different point
estimators and confidence intervals, including the biases and variances of the
point estimators, and the coverage properties of confidence intervals.

######  20.3 Re-analysis of the data on the minimum legal drink age

Analyze the data mlda.csv of Carpenter and Dobkin, (2009).

######  20.4 Recommended reading

D’Amour et al., (2021) discussed the implications of overlap with high
dimensional covariates.

Thistlethwaite and Campbell, (1960)’s original paper on regression
discontinuity was re-printed as Thistlewaite and Campbell, (2016) with many
insightful comments. Coincidentally, Thistlethwaite and Campbell, (1960) and
Rubin, (1974) were both published in the Journal of Educational Psychology.

##  Part V Instrumental variables

###  Chapter 21 An Experimental Perspective

The instrumental variable method has been a powerful tool in econometrics. It
identifies causal effects in studies without unconfoundedness between the
treatment and the outcome. It relies on an additional variable, called the
instrumental variable (IV), that satisfies certain conditions. These
conditions may not be easy to digest when you read for the first time. In some
sense, IV is a magic. This chapter presents a not-so-magic perspective based
on the encouragement design. This again echos Dorn, (1953)’s suggestion that
the planner of an observational study should always ask himself the following
question:

> How would the study be conducted if it were possible to do it by controlled
> experimentation?

The experimental analog of the IV method is the encouragement design (Zelen,,
1979; Powers and Swinton,, 1984; Holland,, 1986).

####  102 Encouragement Design and Noncompliance

Consider an experiment with units indexed by $i=1,\ldots,n$. Let $Z_{i}$ be
the treatment assigned, with $1$ for the treatment and $0$ for the control.
Let $D_{i}$ be the treatment received, with $1$ for the treatment and $0$ for
the control. When $Z_{i}\neq D_{i}$ for some unit $i$, the noncompliance
problem arises. Noncompliance is a very common problem especially in
encouragement designs involving human beings as experimental units. In these
cases, the experimenters cannot force the units to take the treatment but
rather only encourage them to do so. Let $Y_{i}$ be the outcome of interest.

Consider complete randomization of $Z$ and ignore covariates $X$ now. We have
the potential values for the treatment received $\\{D_{i}(1),D_{i}(0)\\}$ and
the potential values for the outcome $\\{Y_{i}(1),Y_{i}(0)\\}$, all with
respect to the treatment assignment levels $1$ and $0$. Their observed values
are $D_{i}=Z_{i}D_{i}(1)+(1-Z_{i})D_{i}(0)$ and
$Y_{i}=Z_{i}Y_{i}(1)+(1-Z_{i})Y_{i}(0)$, respectively. For notational
simplicity, we assume
$\\{Z_{i},D_{i}(1),D_{i}(0),Y_{i}(1),Y_{i}(0)\\}_{i=1}^{n}\stackrel{{\scriptstyle\text{IID}}}{{\sim}}\\{Z,D(1),D(0),Y(1),Y(0)\\}$
and sometimes drop the subscript $i$ without causing confusions.

We start with completely randomized experiments.

######  Assumption 21.1 (randomization)

$Z\begin{picture}(9.0,8.0)\put(0.0,0.0){\line(1,0){9.0}}
\put(3.0,0.0){\line(0,1){8.0}} \put(6.0,0.0){\line(0,1){8.0}}
\end{picture}\\{D(1),D(0),Y(1),Y(0)\\}$.

Randomization allows for identification of the average causal effects on $D$
and $Y$:

$\tau_{D}=E\\{D(1)-D(0)\\}=E(D\mid Z=1)-E(D\mid Z=0)$

and

$\tau_{Y}=E\\{Y(1)-Y(0)\\}=E(Y\mid Z=1)-E(Y\mid Z=0).$

We can use simple difference-in-means estimators $\hat{\tau}_{D}$ and
$\hat{\tau}_{Y}$ to estimate $\tau_{D}$ and $\tau_{Y}$, respectively.

Reporting the estimate $\hat{\tau}_{Y}$ with the associated standard error is
called the intention-to-treat (ITT) analysis. It estimates the effect of the
treatment assignment on the outcome, and complete randomization in Assumption
21.1 justifies this analysis. However, it may not answer the scientific
question, that is, the causal effect of the treatment received on the outcome.

####  103 Latent Compliance Status and Effects

#####  103.1 Nonparametric identification

Following Imbens and Angrist, (1994) and Angrist et al., (1996), we stratify
the population based on the joint potential values of of
$\\{D_{i}(1),D_{i}(0)\\}$. Because $D$ is binary, we have four possible
combinations:

$\displaystyle U_{i}=\left\\{\begin{array}[]{ll}\textup{a},&\text{if}\hskip
5.69054ptD_{i}(1)=1\hskip 5.69054pt\text{and}\hskip 5.69054ptD_{i}(0)=1;\\\
\textup{c},&\text{if}\hskip 5.69054ptD_{i}(1)=1\hskip
5.69054pt\text{and}\hskip 5.69054ptD_{i}(0)=0;\\\ \textup{d},&\text{if}\hskip
5.69054ptD_{i}(1)=0\hskip 5.69054pt\text{and}\hskip 5.69054ptD_{i}(0)=1;\\\
\textup{n},&\text{if}\hskip 5.69054ptD_{i}(1)=0\hskip
5.69054pt\text{and}\hskip 5.69054ptD_{i}(0)=0,\end{array}\right.$

where “a” is for “always taker,” “c” is for “complier,” “d” is for “defier,”
and “n” is for “never taker.” Because we cannot observe $D_{i}(1)$ and
$D_{i}(0)$ simultaneously, $U_{i}$ is a latent variable for the compliance
behavior of unit $i$.

Based on $U$, we can use the law of total probability to decompose the average
causal effect on $Y$ into four terms:

$\displaystyle\tau_{Y}$ $\displaystyle=$ $\displaystyle E\\{Y(1)-Y(0)\mid
U=\textup{a}\\}\textup{pr}(U=\textup{a})$ (89)
$\displaystyle+E\\{Y(1)-Y(0)\mid U=\textup{c}\\}\textup{pr}(U=\textup{c})$
$\displaystyle+E\\{Y(1)-Y(0)\mid U=\textup{d}\\}\textup{pr}(U=\textup{d})$
$\displaystyle+E\\{Y(1)-Y(0)\mid U=\textup{n}\\}\textup{pr}(U=\textup{n}).$

Therefore, $\tau_{Y}$ is a weighted average of four latent subgroup effects.
We will look into more details of the latent groups below.

Assumption 21.2 below restricts the third term in (89) to be zero.

######  Assumption 21.2 (monotonicity)

$\textup{pr}(U=\textup{d})=0$ or $D_{i}(1)\geq D_{i}(0)$, that is, there are
no defiers.

Assumption 21.2 holds automatically with one-sided noncompliance when the
units assigned to the control arm have no access to the treatment, i.e.,
$D_{i}(0)=0$ for all units. Under randomization, Assumption 21.2 has a
testable implication that

$\textup{pr}(D=1\mid Z=1)\geq\textup{pr}(D=1\mid Z=0).$

But Assumption 21.2 is much stronger than the inequality above. The former
restricts $D_{i}(1)$ and $D_{i}(0)$ at the individual level and the latter
restricts them only on average. Nevertheless, when this testable implication
holds, we cannot use the observed data to refute Assumption 21.2.

Assumption 21.3 below restricts the first and last terms in (89) to be zero
based on the mechanism of the treatment assignment on the outcome through only
the treatment received.

######  Assumption 21.3 (exclusion restriction)

$Y_{i}(1)=Y_{i}(0)$ for always takers with $U_{i}=\textup{a}$ and never takers
with $U_{i}=\textup{n}$.

Assumption 21.3 requires that the treatment assignment affects the outcome
only if it affects the treatment received. In double-blind clinical
trial272727In general, it is better to blind the experiment to avoid various
biases arising from placebo effects, patients’ expectation, etc. In double
blind trials, both doctors and patients do not know the treatment; in single
blind trials, the patients do not know the treatment but the doctors know.
Sometimes, it is impossible to conduct double or even single blind trials.
Those trials are called open trials., it is biologically plausible because the
outcome only depends on the actual treatment received. That is, if the
treatment assignment does not change the treatment received, it does not
change the outcome either. It can be violated if the treatment assignment has
direct effects on the outcome not through the treatment received. For example,
some randomized controlled trials are not double blinded, and the treatment
assignment can have some unknown pathways to the outcome.

Under Assumptions 21.2 and 21.3, the decomposition (89) only has the second
term :

$\displaystyle\tau_{Y}=E\\{Y(1)-Y(0)\mid
U=\textup{c}\\}\textup{pr}(U=\textup{c}).$ (90)

Similarly, we can decompose the average causal effect on $D$ into four terms:

$\displaystyle\tau_{D}$ $\displaystyle=$ $\displaystyle E\\{D(1)-D(0)\mid
U=\textup{a}\\}\textup{pr}(U=\textup{a})$ $\displaystyle+E\\{D(1)-D(0)\mid
U=\textup{c}\\}\textup{pr}(U=\textup{c})$ $\displaystyle+E\\{D(1)-D(0)\mid
U=\textup{d}\\}\textup{pr}(U=\textup{d})$ $\displaystyle+E\\{D(1)-D(0)\mid
U=\textup{n}\\}\textup{pr}(U=\textup{n})$ $\displaystyle=$ $\displaystyle
0\times\textup{pr}(U=\textup{a})+1\times\textup{pr}(U=\textup{c})+(-1)\times\textup{pr}(U=\textup{d})+0\times\textup{pr}(U=\textup{n}),$

which, under Assumption 21.2, reduces to

$\displaystyle\tau_{D}=\textup{pr}(U=\textup{c}).$ (91)

This is an interesting fact that the proportion of the compliers
$\pi_{\textup{c}}$ equals the average causal effect of the treatment assigned
on $D$, an identifiable quantity under complete randomization. Although we do
not know all the compliers based on the observed data, we can identify their
proportion in the whole population based on (91). Combining (90) and (91), we
have the following result.

###### Theorem 21.1

Under Assumptions 21.2–21.3, we have

$E\\{Y(1)-Y(0)\mid U=\textup{c}\\}=\frac{\tau_{Y}}{\tau_{D}}$

if $\tau_{D}\neq 0$.

Following Imbens and Angrist, (1994) and Angrist et al., (1996), we define a
new causal effect below.

######  Definition 21.1 (CACE or LATE)

Define

$\tau_{\textup{c}}\equiv E\\{Y(1)-Y(0)\mid U=\textup{c}\\}$

as the “complier average causal effect (CACE)” or the “local average treatment
effect (LATE)”. It has alternative forms:

$\displaystyle\tau_{\textup{c}}$ $\displaystyle=$ $\displaystyle
E\\{Y(1)-Y(0)\mid D(1)=1,D(0)=0\\}$ $\displaystyle=$ $\displaystyle
E\\{Y(1)-Y(0)\mid D(1)>D(0)\\}.$

Based on Definition 21.1, we can rewrite Theorem 21.1 as

$\tau_{\textup{c}}=\frac{\tau_{Y}}{\tau_{D}},$

that is, the CACE or LATE equals the ratio of the average causal effects on
$Y$ over that on $D$. Under Assumption 21.1, we further identify the CACE
below.

###### Corollary 21.1

Under Assumptions 21.1–21.3, we have

$\tau_{\textup{c}}=\frac{E(Y\mid Z=1)-E(Y\mid Z=0)}{E(D\mid Z=1)-E(D\mid
Z=0)}.$

Therefore, under randomization, monotonicity, and exclusion restriction, we
can nonparametrically identify the CACE as the ratio of the difference in
means of the outcome over the difference in means of the treatment received.

#####  103.2 Estimation

Based on Corollary 21.1, we can estimate $\tau_{\textup{c}}$ by a simple ratio

$\hat{\tau}_{\textup{c}}=\frac{\hat{\tau}_{Y}}{\hat{\tau}_{D}},$

which is called the Wald estimator (Wald,, 1940) or the IV estimator. In the
above discussion, $Z$ acts as the IV for $D$.

We can obtain the variance estimator based on the following heuristics (see
Example A1.3):

$\hat{\tau}_{\textup{c}}-\tau_{\textup{c}}=(\hat{\tau}_{Y}-\tau_{\textup{c}}\hat{\tau}_{D})/\hat{\tau}_{D}\approx(\hat{\tau}_{Y}-\tau_{\textup{c}}\hat{\tau}_{D})/\tau_{D}=\hat{\tau}_{A}/\tau_{D},$

where $\hat{\tau}_{A}$ is the difference-in-means of the adjusted outcome
$A_{i}=Y_{i}-\tau_{\textup{c}}D_{i}$. So the asymptotic variance of
$\hat{\tau}_{\textup{c}}$ is close to the variance of $\hat{\tau}_{A}$ divided
by $\tau_{D}^{2}$. The variance estimation proceeds in the following steps:

  1. 1.

obtain the adjusted outcomes $\hat{A}_{i}=Y_{i}-\hat{\tau}_{\textup{c}}D_{i}\
(i=1,\ldots,n)$;

  2. 2.

obtain the Neyman-type variance estimate based on the adjusted outcomes:

$\hat{V}_{\hat{A}}=\frac{\hat{S}_{\hat{A}}^{2}(1)}{n_{1}}+\frac{\hat{S}_{\hat{A}}^{2}(0)}{n_{0}},$

where $\hat{S}_{\hat{A}}^{2}(1)$ and $\hat{S}_{\hat{A}}^{2}(0)$ are the sample
variances of the $\hat{A}_{i}$’s under treatment and control, respectively;

  3. 3.

obtain the final variance estimator $\hat{V}_{\hat{A}}/\hat{\tau}_{D}^{2}.$

Under the null hypothesis that $\tau_{\textup{c}}=0$, we can simply
approximate the variance by $\hat{V}_{Y}/\hat{\tau}_{D}^{2}$, where
$\hat{V}_{Y}$ is the Neyman-type variance estimate for the difference in means
of $Y$. This variance estimator is inconsistent if the true
$\tau_{\textup{c}}$ is not zero. Therefore, it works for testing but not for
estimation. Nevertheless, it gives interesting insights for the ITT estimator
and the Wald estimator. The ITT estimator $\hat{\tau}_{Y}$ has estimated
standard error $\sqrt{\hat{V}_{Y}}$. The Wald estimator
$\hat{\tau}_{Y}/\hat{\tau}_{D}$ essentially equals the ITT estimator
multiplied by $1/\hat{\tau}_{D}>1$, which is larger in magnitude but at the
same time its estimated standard error increases by the same factor. The
confidence intervals for $\tau_{Y}$ and $\tau_{\textup{c}}$ are

$\hat{\tau}_{Y}\pm z_{1-\alpha/2}\sqrt{\hat{V}_{Y}}$

and

$\hat{\tau}_{Y}/\hat{\tau}_{D}\pm
z_{1-\alpha/2}\sqrt{\hat{V}_{Y}}/\hat{\tau}_{D}=\left(\hat{\tau}_{Y}\pm
z_{1-\alpha/2}\sqrt{\hat{V}_{Y}}\right)/\hat{\tau}_{D}.$

These confidence intervals give the same qualitative conclusions since they
will both cover zero or not. In some sense, the IV analysis provides the same
qualitative information as the ITT analysis of $Y$ although it involves more
complicated procedures.

####  104 Covariates

#####  104.1 Covariate adjustment in complete randomization

We now consider completely randomized experiments with covariates, and assume
$Z\begin{picture}(9.0,8.0)\put(0.0,0.0){\line(1,0){9.0}}
\put(3.0,0.0){\line(0,1){8.0}} \put(6.0,0.0){\line(0,1){8.0}}
\end{picture}\\{D(1),D(0),Y(1),Y(0),X\\}$. With covariates $X$, we can obtain
Lin, (2013)’s estimators $\hat{\tau}_{D,\textsc{L}}$ and
$\hat{\tau}_{Y,\textsc{L}}$ for both $D$ and $Y$, resulting in
$\hat{\tau}_{\textup{c},\textsc{L}}=\hat{\tau}_{Y,\textsc{L}}/\hat{\tau}_{D,\textsc{L}}$.
Recall that

$\displaystyle\hat{\tau}_{D,\textsc{L}}$ $\displaystyle=$
$\displaystyle\left\\{\hat{\bar{D}}(1)-\hat{\beta}_{D1}^{\mkern-1.5mu\mathsf{T}}\hat{\bar{X}}(1)\right\\}-\left\\{\hat{\bar{D}}(0)-\hat{\beta}_{D0}^{\mkern-1.5mu\mathsf{T}}\hat{\bar{X}}(0)\right\\},$
$\displaystyle\hat{\tau}_{Y,\textsc{L}}$ $\displaystyle=$
$\displaystyle\left\\{\hat{\bar{Y}}(1)-\hat{\beta}_{Y1}^{\mkern-1.5mu\mathsf{T}}\hat{\bar{X}}(1)\right\\}-\left\\{\hat{\bar{Y}}(0)-\hat{\beta}_{Y0}^{\mkern-1.5mu\mathsf{T}}\hat{\bar{X}}(0)\right\\},$

where $\hat{\beta}_{D1}$ and $\hat{\beta}_{Y1}$ are the coefficients of $X$ in
the OLS fits of $D$ and $Y$ in the treated group, and $\hat{\beta}_{D0}$ and
$\hat{\beta}_{Y0}$ are the coefficients of $X$ in the OLS fits of $D$ and $Y$
in the control group. We can approximate the standard error of
$\hat{\tau}_{\textup{c},\textsc{L}}$ based on the following heuristics (again
see Example A1.3):

$\hat{\tau}_{\textup{c},\textsc{L}}-\tau_{\textup{c}}=(\hat{\tau}_{Y,\textsc{L}}-\tau_{\textup{c}}\hat{\tau}_{D,\textsc{L}})/\hat{\tau}_{D,\textsc{L}}\approx(\hat{\tau}_{Y,\textsc{L}}-\tau_{\textup{c}}\hat{\tau}_{D,\textsc{L}})/\tau_{D}=\hat{\tau}_{A}/\tau_{D},$

where $\hat{\tau}_{A}$ is the difference-in-means of $A$, defined as

$A_{i}=\left\\{\begin{array}[]{ll}(Y_{i}-\hat{\beta}_{Y1}^{\mkern-1.5mu\mathsf{T}}X_{i})-\tau_{\textup{c}}(D_{i}-\hat{\beta}_{D1}^{\mkern-1.5mu\mathsf{T}}X_{i}),&\text{if}\hskip
5.69054ptZ_{i}=1,\\\
(Y_{i}-\hat{\beta}_{Y0}^{\mkern-1.5mu\mathsf{T}}X_{i})-\tau_{\textup{c}}(D_{i}-\hat{\beta}_{D0}^{\mkern-1.5mu\mathsf{T}}X_{i}),&\text{if}\hskip
5.69054ptZ_{i}=0.\end{array}\right.$

The variance estimation proceeds in the following steps:

  1. 1.

obtain the adjusted outcomes $\hat{A}_{i}\ (i=1,\ldots,n)$ with

$\hat{A}_{i}=\left\\{\begin{array}[]{ll}(Y_{i}-\hat{\beta}_{Y1}^{\mkern-1.5mu\mathsf{T}}X_{i})-\hat{\tau}_{\textup{c},\textsc{L}}(D_{i}-\hat{\beta}_{D1}^{\mkern-1.5mu\mathsf{T}}X_{i}),&\text{if}\hskip
5.69054ptZ_{i}=1,\\\
(Y_{i}-\hat{\beta}_{Y0}^{\mkern-1.5mu\mathsf{T}}X_{i})-\hat{\tau}_{\textup{c},\textsc{L}}(D_{i}-\hat{\beta}_{D0}^{\mkern-1.5mu\mathsf{T}}X_{i}),&\text{if}\hskip
5.69054ptZ_{i}=0;\end{array}\right.$

  2. 2.

obtain the Neyman-type variance estimate based on the adjusted outcomes:

$\hat{V}_{\hat{A}}=\frac{\hat{S}_{\hat{A}}^{2}(1)}{n_{1}}+\frac{\hat{S}_{\hat{A}}^{2}(0)}{n_{0}},$

where $\hat{S}_{\hat{A}}^{2}(1)$ and $\hat{S}_{\hat{A}}^{2}(0)$ are the sample
variances of the $\hat{A}_{i}$’s under the treatment and control,
respectively;

  3. 3.

obtain the final variance estimator
$\hat{V}_{\hat{A}}/\hat{\tau}_{D,\textsc{L}}^{2}.$

Again under the null with $\tau_{\textup{c}}=0$, we can approximate the
estimated standard error for $\hat{\tau}_{\textup{c},\textsc{L}}$ by the
estimated standard error of $\hat{\tau}_{Y,\textsc{L}}$ (e.g., the EHW
standard error in the fully interacted linear model) divided by
$\hat{\tau}_{D,\textsc{L}}$.

#####  104.2 Covariates in conditional randomization or unconfounded
observational studies

If randomization holds conditionally, i.e.,

$Z\begin{picture}(9.0,8.0)\put(0.0,0.0){\line(1,0){9.0}}
\put(3.0,0.0){\line(0,1){8.0}} \put(6.0,0.0){\line(0,1){8.0}}
\end{picture}\\{D(1),D(0),Y(1),Y(0)\\}\mid X,$

then we must adjust for covariates to avoid bias. The analysis is also
straightforward since we already have discussed many estimators in Part III
for estimating the effects of $Z$ on $D$ and $Y$, respectively. We can just
use them in the ratio formula
$\hat{\tau}_{\textup{c}}=\hat{\tau}_{Y}/\hat{\tau}_{D}$ and use the bootstrap
to approximate the asymptotic variance.

####  105 Weak IV

Even $\tau_{D}>0$, there is a positive probability that $\hat{\tau}_{D}$ is
zero, so the variance of $\hat{\tau}_{\textup{c}}$ is infinity. The variance
from the Normal approximation discussed before is not the variance of
$\hat{\tau}_{\textup{c}}$ but rather the variance of its asymptotic
distribution. This is a subtle technical point. When $\tau_{D}$ is close to
$0$, which is referred to as the weak IV case, the ratio estimator
$\hat{\tau}_{\textup{c}}=\hat{\tau}_{Y}/\hat{\tau}_{D}$ has poor finite-sample
properties. Under this scenario, $\hat{\tau}_{\textup{c}}$ has finite sample
bias and non-Normal asymptotic distribution, and the corresponding Wald-type
confidence intervals have poor coverage properties282828The theory often
assumes that $\tau_{D}$ has the order $n^{-1/2}$. Under this regime, the
proportion of compliers goes to 0 as $n$ goes to infinity. The IV method can
only identify a subgroup average causal effect with the proportion shrinking
to 0. This is a contrived regime for theoretical analysis. It is hard to
justify this assumption in practice. The follow discussion does not assume
it.. In the simple case with a binary outcome $Y$, we know that $\tau_{Y}$
must be bounded between $-1$ and $1$, but there is no guarantee that
$\hat{\tau}_{\textup{c}}$ is bounded between $-1$ and $1$. How do we deal with
a weak IV?

From a testing perspective, there is an easy solution. Because
$\tau_{\textup{c}}=\tau_{Y}/\tau_{D}$, so the following two null hypotheses
are equivalent:

$H_{0}:\tau_{\textup{c}}=0\Longleftrightarrow H_{0}^{\prime}:\tau_{Y}=0.$

Therefore, we simply test $H_{0}^{\prime}$, i.e., the average causal effect of
$Z$ on $Y$ is zero. This echos our discussion in Section 103.2.

From an estimation perspective, we can focus on the confidence interval
although the point estimator has poor finite-sample properties. Because
$\tau_{\textup{c}}=\tau_{Y}/\tau_{D}$, this is similar to the classical
Fieller–Creasy problem in statistics. Below we discuss a strategy for
constructing confidence interval for $\tau_{\textup{c}}$ motivated by Fieller,
(1954); see Section 155.2. Given the true value $\tau_{\textup{c}}$, we have

$\tau_{Y}-\tau_{\textup{c}}\tau_{D}=0.$

So we can construct a confidence set for $\tau_{\textup{c}}$ by inverting a
sequence of null hypotheses

$H_{0}(b):\tau_{\textup{c}}=b$

This null hypothesis is equivalent to the null hypothesis of zero average
causal effect on the outcome $A_{i}(b)=Y_{i}-bD_{i}$:

$H_{0}(b):\tau_{A(b)}=0.$

Let $\hat{\tau}_{A}(b)$ be a generic estimator for $\tau_{A(b)}$ with the
associated variance estimator $\hat{V}_{A}(b)$. In the CRE without covariates,
$\hat{\tau}_{A}(b)$ is the difference in means of the outcome $A_{i}(b)$ and
$\hat{V}_{A}(b)$ is the Neyman-type variance estimator. In the CRE with
covariates, $\hat{\tau}_{A}(b)$ is Lin, (2013)’s estimator for the outcome
$A_{i}(b)$ and $\hat{V}_{A}(b)$ is the EHW variance estimator in the
associated OLS fit of $Y_{i}-bD_{i}$ on $(Z_{i},X_{i},Z_{i}X_{i})$. In
unconfounded observational studies, we can obtain the estimator for the
average causal effect on $A_{i}(b)$ and the associated variance estimator
based on many existing strategies in Part III.

Based on $\hat{\tau}_{A}(b)$ and $\tau_{A(b)}$, we can construct a Wald-type
test for $H_{0}(b)$. Inverting tests, we can construct the following
confidence set for $\tau_{\textup{c}}$:

$\left\\{b:\frac{\hat{\tau}_{A}^{2}(b)}{\hat{V}_{A}(b)}\leq
z_{\alpha}^{2}\right\\}.$

This is close to the Anderson–Rubin-type confidence interval in econometrics
(Anderson and Rubin,, 1950). Due to its connection to Fieller, (1954), I will
call it the Fieller–Anderson–Rubin confidence interval. These weak-IV
confidence intervals reduce to the asymptotic confidence intervals when the IV
is strong. But they have additional guarantees when the IV is weak. I
recommend using them in practice.

###### Example 21.1

To gain intuition about the Fieller–Anderson–Rubin confidence interval, we
look into the simple case of the CRE without covariates. The quadratic
inequality in the confidence interval reduces to

$\displaystyle(\hat{\tau}_{Y}-b\hat{\tau}_{D})^{2}$ $\displaystyle\leq$
$\displaystyle
z_{\alpha}^{2}\left[n_{1}^{-1}\\{\hat{S}_{Y}^{2}(1)+b^{2}\hat{S}_{D}^{2}(1)-2b\hat{S}_{YD}(1)\\}\right.$
$\displaystyle\left.+n_{0}^{-1}\\{\hat{S}_{Y}^{2}(0)+b^{2}\hat{S}_{D}^{2}(0)-2b\hat{S}_{YD}(0)\\}\right],$

where $\\{\hat{S}_{Y}^{2}(1),\hat{S}_{D}^{2}(1),\hat{S}_{YD}(1)\\}$ and
$\\{\hat{S}_{Y}^{2}(0),\hat{S}_{D}^{2}(0),\hat{S}_{YD}(0)\\}$ are the sample
variances and covariances of $Y$ and $D$ under treatment and control,
respectively. The confidence set can be a close interval, two disconnected
intervals, an empty set, or the whole real line. I relegate the detailed
discussion to Problem 21.3.

####  106 Application

The mediation package contains a dataset jobs from Job Search Intervention
Study (JOBS II), which was a randomized field experiment that investigates the
efficacy of a job training intervention on unemployed workers. The variable
treat is the indicator for whether a participant was randomly selected for the
JOBS II training program, and the variable comply is the indicator for whether
a participant actually participated in the JOBS II program. An outcome of
interest is job_seek for measuring the level of job-search self-efficacy with
values from 1 to 5. A few standard covariates are sex, age, marital, nonwhite,
educ, and income.

Without using covariates, the confidence intervals based on the delta method
and the bootstrap are

    > est
    [1] 0.1087904
    > c(est - 1.96*dse, est + 1.96*dse)
    [1] -0.05002163  0.26760235
    > c(est - 1.96*bse, est + 1.96*bse)
    [1] -0.04657384  0.26415455

Adjusting for covariates, the confidence intervals based on the delta method
and the bootstrap are

    > est
    [1] 0.1176332
    > c(est - 1.96*dse, est + 1.96*dse)
    [1] -0.03638421  0.27165070
    > c(est - 1.96*bse, est + 1.96*bse)
    [1] -0.03926737  0.27453386

We can also construct confidence interval by inverting tests. Without using
covariates, it is

    > ARCI
    [1] -0.050  0.267

adjusting for covariates, it is

    > ARCI
    [1] -0.046  0.281

Figure 29 plots the $p$-values for a sequence of tests.

Figure 29: Confidence interval of $\tau_{\textup{c}}$ by inverting tests

####  107 Interpreting the Complier Average Causal Effect

The notation for potential outcomes $\\{D(1),D(0),Y(1),Y(0)\\}$ is with
respect to the hypothetical intervention of the treatment assigned $Z$. So
$\tau_{\textup{c}}$ is the average causal effect of the treatment assigned on
the outcome for compliers. Fortunately, $D=Z$ for compliers, so we can also
interpret $\tau_{\textup{c}}$ as the average causal effect of the treatment
received on the outcome for compliers. This partially answers the scientific
question.

Some papers (e.g., Angrist et al.,, 1996) use different notation. They use
$Y_{i}(z,d)$ for the potential outcome of unit $i$ under a $2\times 2$
factorial experiment with the treatment assigned $z$ and treatment received
$d$. The exclusion restriction assumption has the following form.

######  Assumption 21.4 (exclusion restriction)

$Y_{i}(z,d)=Y_{i}(d)$ for all $i$, that is, the potential outcome is only a
function of $d$.

Based on the causal graph below, Assumption 21.4 rules out the direct arrow
from $Z$ to $Y$. In such case, $Z$ is an IV for $D$.

$\textstyle{U\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{Z\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{D\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{Y}$

Under Assumption 21.4, the augmented notation $Y_{i}(z,d)$ reduces to
$Y_{i}(d)$, which justifies the name of “exclusion restriction.” Therefore,
$Y_{i}(1,d)=Y_{i}(0,d)$ for $d=0,1$, which, coupled with Assumption 21.2,
implies that

$\displaystyle Y_{i}(z=1)-Y_{i}(z=0)$ $\displaystyle=$ $\displaystyle
Y_{i}(1,D_{i}(1))-Y_{i}(0,D_{i}(0))$ $\displaystyle=$
$\displaystyle\left\\{\begin{array}[]{ll}0,&\text{if}\hskip
5.69054ptU_{i}=\textup{a},\\\ 0,&\text{if}\hskip 5.69054ptU_{i}=\textup{n},\\\
Y_{i}(d=1)-Y_{i}(d=0),&\text{if}\hskip
5.69054ptU_{i}=\textup{c}.\end{array}\right.$

In the above, we emphasize the potential outcomes are with respect to $z$, $d$
or both, to avoid confusions. The previous decomposition of $\tau_{Y}$ holds
and we have the following result from Imbens and Angrist, (1994) and Angrist
et al., (1996).

Recall the average causal effect on $D$, $\tau_{D}=E\\{D(1)-D(0)\\},$ define
the average causal effect on $Y$ as $\tau_{Y}=E\\{Y(D(1))-Y(D(0))\\}$, and
define the complier average causal effect as

$\tau_{\textup{c}}=E\\{Y(d=1)-Y(d=0)\mid U=\textup{c}\\}.$

###### Theorem 21.2

Under Assumptions 21.2–21.4, we have

$Y(D(1))-Y(D(0))=\\{D(1)-D(0)\\}\times\\{Y(d=1)-Y(d=0)\\}$

and $\tau_{\textup{c}}=\tau_{Y}/\tau_{D}.$

The proof is almost identical to the proof of Theorem 21.1 with modifications
of the notation. I leave it as Problem 21.2. From the notation $Y_{i}(d)$, it
is more convenient to interpret $\tau_{\textup{c}}$ as as the average causal
effect of the treatment received on the outcome for compliers.

####  108 Homework problems

######  21.1 Variance of the Wald estimator

Show that $\textup{var}(\hat{\tau}_{\textup{c}})=\infty$.

######  21.2 Proof of the main theorem of Imbens and Angrist, (1994) and
Angrist et al., (1996)

Prove Theorem 21.2.

######  21.3 More on the Fieller–Anderson–Rubin confidence set

The confidence set in Example 21.1 can be a close interval, two disconnected
intervals, an empty set, or the whole real line. Find the precise conditions
for each case.

######  21.4 Binary IV and ordinal treatment received

Angrist and Imbens, (1995) discussed a more general setting with a binary IV
$Z$, an ordinal treatment received $D\in\\{0,1,\ldots,J\\}$, and an outcome
$Y$. The ordinal treatment received has potential outcomes $D(1)$ and $D(0)$
with respect to the binary IV, and the outcome has potential outcomes $Y(z,d)$
with respect to both the binary IV and the ordinal treatment received. Extend
the discussion in Section 107 and the corresponding IV assumptions as below.

###### Assumption 21.5

We have (1) randomization that
$Z\begin{picture}(9.0,8.0)\put(0.0,0.0){\line(1,0){9.0}}
\put(3.0,0.0){\line(0,1){8.0}} \put(6.0,0.0){\line(0,1){8.0}}
\end{picture}\\{D(z),Y(z,d):z=0,1;d=0,1,\ldots,J\\}$; (2) monotonicity that
$D(1)\geq D(0)$; and (3) exclusion restriction that $Y(z,d)=Y(d)$ for all
$z=0,1$ and $d=0,1,\ldots,J$.

They proved Theorem 21.3 below.

###### Theorem 21.3

Under Assumption 21.5, we have

$\frac{E(Y\mid Z=1)-E(Y\mid Z=0)}{E(D\mid Z=1)-E(D\mid
Z=0)}=\sum_{j=1}^{J}w_{j}E\\{Y(j)-Y(j-1)\mid D(1)\geq j>D(0)\\}$

where

$w_{j}=\frac{\textup{pr}\\{D(1)\geq
j>D(0)\\}}{\sum_{j^{\prime}=1}^{J}\textup{pr}\\{D(1)\geq j^{\prime}>D(0)\\}}.$

Prove Theorem 21.3.

Remark: When $J=1$, Theorem 21.3 reduces to Theorem 21.2. It states that the
standard IV formula identifies a weighted average of some latent subgroup
effects. The weights are proportional to the probability of the latent groups
defined by $D(1)\geq j>D(0)$, and the latent subgroup effects
$E\\{Y(j)-Y(j-1)\mid D(1)\geq j>D(0)\\}$ compare the adjacent levels of the
treatment received. However, this weighted average may not be easy to
interpret because the latent groups overlap.

The proof can be tedious. A trick is to write the treatment received and
outcome under treatment assignment $z$ as

$D(z)=\sum_{j=1}^{J}j1\\{D(z)=j\\},\quad
Y(D(z))=\sum_{j=1}^{J}Y(j)1\\{D(z)=j\\}$

to obtain

$D(1)-D(0)=\sum_{j=0}^{J}j[1\\{D(1)=j\\}-1\\{D(0)=j\\}]$

and

$Y(D(1))-Y(D(0))=\sum_{j=0}^{J}Y(j)[1\\{D(1)=j\\}-1\\{D(0)=j\\}].$

Then use the following Abel’s lemma, also called summation by parts:

$\sum_{j=0}^{J}f_{j}(g_{j+1}-g_{j})=f_{J}g_{J+1}-f_{0}g_{0}-\sum_{j=1}^{J}g_{j}(f_{j}-f_{j-1})$

for appropriately specified sequences $(f_{j})$ and $(g_{j})$.

######  21.5 Data analysis: a flu shot encouragement design (McDonald et al.,,
1992)

The dataset in fludata.txt is from a randomized encouragement design of
McDonald et al., (1992), which was also re-analyzed by Hirano et al., (2000).
It contains the following variables:

assign | binary encouragement to receive the flu shot
---|---
receive | binary indicator for receiving the flu shot
outcome | binary outcome for flu related hospitalization
age | age of the patient
sex | sex of the patient
race | race of the patient
copd, dm, heartd, renal, liverd | various disease background covariates

Analyze the data with and without adjusting for the covariates.

######  21.6 Data analysis: the Karolinska data

Rubin, (2008) used the Karolinska data as an example for the IV method. In
karolinska.txt, whether a patient was diagnosed at large volume hospital can
be viewed as an IV for whether a patient was treated at a large volume
hospital. This is plausible at least conditioning on other observed
covariates. See Rubin, (2008)’s analysis for more details.

Reanalyze the data assuming that the IV is randomly assigned conditional on
observed covariates.

######  21.7 Data analysis: a job training program (Schochet et al.,, 2008)

jobtraining.rtf contains the description of the data files X.csv and Y.csv.

X.csv contains the pretreatment covariates; you can view the sampling weight
variable wgt as a covariate too. It is generally difficult to deal with
sampling weights. Many previous analyses made this simplification. Conduct
analyses with and without covariates.

Y.csv contains the sampling weight, treatment assigned, treatment received,
and many post-treatment variables. Therefore, this data contains many outcomes
depending on your questions of interest. The data also have many
complications. First, some outcomes are missing. Second, unemployed
individuals do not have wages or incomes. Third, the outcomes are repeatedly
observed over time. When you do the data analysis, please give details about
your choice of the questions of interest and estimators.

######  21.8 Recommended reading

Angrist et al., (1996) bridged the econometric IV perspective and statistical
causal inference based on potential outcomes and demonstrated its usefulness
with an application.

Some other early references on IV are Permutt and Hebel, (1989), Sommer and
Zeger, (1991), Baker and Lindeman, (1994), and Cuzick et al., (1997).

###  Chapter 22 Disentangle Mixture Distributions and Instrumental Variable
Inequalities

The IV model in Chapter 21 imposes Assumptions 21.1–21.3:

  1. 1.

$Z\begin{picture}(9.0,8.0)\put(0.0,0.0){\line(1,0){9.0}}
\put(3.0,0.0){\line(0,1){8.0}} \put(6.0,0.0){\line(0,1){8.0}}
\end{picture}\\{D(1),D(0),Y(1),Y(0)\\}$;

  2. 2.

$\textup{pr}(U=\textup{d})=0$;

  3. 3.

$Y(1)=Y(0)$ for $U=\textup{a}$ or n.

Table 7 summarizes the observed groups and the corresponding latent groups.

Table 7: Observed groups and latent groups under Assumption 21.2 $Z=1$ | $D=1$ | $D(1)=1$ |  $U=\textup{c}$ or a
---|---|---|---
$Z=1$ | $D=0$ | $D(1)=0$ | $U=\textup{n}$
$Z=0$ | $D=1$ | $D(0)=1$ | $U=\textup{a}$
$Z=0$ | $D=0$ | $D(0)=0$ |  $U=\textup{c}$ or n

Interestingly, Assumptions 21.1–21.3 together have some testable implications.
Balke and Pearl, (1997) called them the instrumental variable inequalities.
This chapter will give an intuitive derivation of a special case of these
inequalities. The proof is a direct consequence of identifying the means of
the potential outcomes for all latent groups defined by $U$.

####  109 Disentangle Mixture Distributions and Instrumental Variable
Inequalities

We summarize the main results in Theorem 22.1 below. Recall $\pi_{u}$ as the
proportion of type $U=u$, and define

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

###### Theorem 22.1

Under Assumptions 21.1–21.3, we can identify the proportions of the latent
types by

$\displaystyle\pi_{\textup{n}}$ $\displaystyle=$
$\displaystyle\textup{pr}(D=0\mid Z=1),$ $\displaystyle\pi_{\textup{a}}$
$\displaystyle=$ $\displaystyle\textup{pr}(D=1\mid Z=0),$
$\displaystyle\pi_{\textup{c}}$ $\displaystyle=$ $\displaystyle E(D\mid
Z=1)-E(D\mid Z=0),$

and the type-specific means of the potential outcomes by

$\displaystyle\mu_{1\textup{n}}=\mu_{0\textup{n}}\equiv\mu_{\textup{n}}$
$\displaystyle=$ $\displaystyle E(Y\mid Z=1,D=0),$
$\displaystyle\mu_{1\textup{a}}=\mu_{0\textup{a}}\equiv\mu_{\textup{a}}$
$\displaystyle=$ $\displaystyle E(Y\mid Z=0,D=1),$
$\displaystyle\mu_{1\textup{c}}$ $\displaystyle=$
$\displaystyle\pi_{\textup{c}}^{-1}\left\\{E(DY\mid Z=1)-E(DY\mid
Z=0)\right\\},$ $\displaystyle\mu_{0\textup{c}}$ $\displaystyle=$
$\displaystyle\pi_{\textup{c}}^{-1}\big{[}E\\{(1-D)Y\mid Z=0\\}-E\\{(1-D)Y\mid
Z=1\\}\big{]}.$

Proof of Theorem 17.1: Part I: We first identify the proportions of the latent
compliance types. We can identify the proportion of the never takers by

$\displaystyle\textup{pr}(D=0\mid Z=1)$ $\displaystyle=$
$\displaystyle\textup{pr}(U=\textup{n}\mid Z=1)$ $\displaystyle=$
$\displaystyle\textup{pr}(U=\textup{n})=\pi_{\textup{n}},$

and the proportion of the always takes by

$\displaystyle\textup{pr}(D=1\mid Z=0)$ $\displaystyle=$
$\displaystyle\textup{pr}(U=\textup{a}\mid Z=0)$ $\displaystyle=$
$\displaystyle\textup{pr}(U=\textup{a})=\pi_{\textup{a}}.$

Therefore, the proportion of compliers is

$\displaystyle\pi_{\textup{c}}$ $\displaystyle=$
$\displaystyle\textup{pr}(U=\textup{c})=1-\pi_{\textup{n}}-\pi_{\textup{a}}$
$\displaystyle=$ $\displaystyle 1-\textup{pr}(D=0\mid Z=1)-\textup{pr}(D=1\mid
Z=0)$ $\displaystyle=$ $\displaystyle E(D\mid Z=1)-E(D\mid Z=0)=\tau_{D},$

which is coherent with our discussion before. Although we do not know
individual latent compliance types for all units, we can identify the
proportions of never takers, always takers, and compliers.

Part II: We then identify the means of the potential outcomes within latent
compliance types. Under Assumption 21.3,

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

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

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

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

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

The observed group $(Z=1,D=1)$ has both compliers and always takers, so

$\displaystyle E(Y\mid Z=1,D=1)$ $\displaystyle=$ $\displaystyle E\\{Y(1)\mid
Z=1,D(1)=1\\}$ $\displaystyle=$ $\displaystyle E\\{Y(1)\mid D(1)=1\\}$
$\displaystyle=$ $\displaystyle\textup{pr}\\{D(0)=1\mid D(1)=1\\}E\\{Y(1)\mid
D(1)=1,D(0)=1\\}$ $\displaystyle+\textup{pr}\\{D(0)=0\mid
D(1)=1\\}E\\{Y(1)\mid D(1)=1,D(0)=0\\}$ $\displaystyle=$
$\displaystyle\frac{\pi_{\textup{c}}}{\pi_{\textup{c}}+\pi_{\textup{a}}}\mu_{1\textup{c}}+\frac{\pi_{\textup{a}}}{\pi_{\textup{c}}+\pi_{\textup{a}}}\mu_{\textup{a}}.$

Solve the linear equation above to obtain

$\displaystyle\mu_{1\textup{c}}$ $\displaystyle=$
$\displaystyle\pi_{\textup{c}}^{-1}\left\\{(\pi_{\textup{c}}+\pi_{\textup{a}})E(Y\mid
Z=1,D=1)-\pi_{\textup{a}}E(Y\mid Z=0,D=1)\right\\}$ $\displaystyle=$
$\displaystyle\pi_{\textup{c}}^{-1}\left\\{\textup{pr}(D=1\mid Z=1)E(Y\mid
Z=1,D=1)\right.$ $\displaystyle\left.\qquad-\textup{pr}(D=1\mid Z=0)E(Y\mid
Z=0,D=1)\right\\}$ $\displaystyle=$
$\displaystyle\pi_{\textup{c}}^{-1}\left\\{E(DY\mid Z=1)-E(DY\mid
Z=0)\right\\}.$

The observed group $(Z=0,D=0)$ has both compliers and never takers, so we have

$\displaystyle E(Y\mid Z=0,D=0)$ $\displaystyle=$ $\displaystyle E\\{Y(0)\mid
Z=0,D(0)=0\\}$ $\displaystyle=$ $\displaystyle E\\{Y(0)\mid D(0)=0\\}$
$\displaystyle=$ $\displaystyle\textup{pr}\\{D(1)=1\mid D(0)=0\\}E\\{Y(0)\mid
D(1)=1,D(0)=0\\}$ $\displaystyle+\textup{pr}\\{D(1)=0\mid
D(0)=0\\}E\\{Y(0)\mid D(1)=0,D(0)=0\\}$ $\displaystyle=$
$\displaystyle\frac{\pi_{\textup{c}}}{\pi_{\textup{c}}+\pi_{\textup{n}}}\mu_{0\textup{c}}+\frac{\pi_{\textup{n}}}{\pi_{\textup{c}}+\pi_{\textup{n}}}\mu_{\textup{n}}.$

Solve the linear equation above to obtain

$\displaystyle\mu_{0\textup{c}}$ $\displaystyle=$
$\displaystyle\pi_{\textup{c}}^{-1}\left\\{(\pi_{\textup{c}}+\pi_{\textup{n}})E(Y\mid
Z=0,D=0)-\pi_{\textup{n}}E(Y\mid Z=1,D=0)\right\\}$ $\displaystyle=$
$\displaystyle\pi_{\textup{c}}^{-1}\left\\{\textup{pr}(D=0\mid Z=0)E(Y\mid
Z=0,D=0)\right.$ $\displaystyle\left.\qquad-\textup{pr}(D=0\mid Z=1)E(Y\mid
Z=1,D=0)\right\\}$ $\displaystyle=$
$\displaystyle\pi_{\textup{c}}^{-1}\big{[}E\\{(1-D)Y\mid Z=0\\}-E\\{(1-D)Y\mid
Z=1\\}\big{]}.$

$\square$

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

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

which is the same as the formula in Theorem 21.1 before.

Theorem 22.1 focuses on identifying the means of the potential outcomes,
$\mu_{zu}$. Imbens and Rubin, (1997) derived more general identification
formulas for the distribution of the potential outcomes; I leave the details
to Problem 22.2.

####  110 Testable implications

Is there any additional value of the this detour for deriving the formula of
$\tau_{\textup{c}}$? The answer is yes. For binary outcome, the following
inequalities must be true:

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

which implies four inequalities

$\displaystyle E(DY\mid Z=1)-E(DY\mid Z=0)$ $\displaystyle\geq$ $\displaystyle
0,$ $\displaystyle E(DY\mid Z=1)-E(DY\mid Z=0)$ $\displaystyle\leq$
$\displaystyle E(D\mid Z=1)-E(D\mid Z=0),$ $\displaystyle E\\{(1-D)Y\mid
Z=0\\}-E\\{(1-D)Y\mid Z=1\\}$ $\displaystyle\geq$ $\displaystyle 0,$
$\displaystyle E\\{(1-D)Y\mid Z=0\\}-E\\{(1-D)Y\mid Z=1\\}$
$\displaystyle\leq$ $\displaystyle E(D\mid Z=1)-E(D\mid Z=0).$

Rearranging terms, we obtain the following unified inequalities.

######  Theorem 22.2 (Instrumental Variable Inequalities)

With a binary outcome $Y$, Assumptions 21.1–21.3 imply

$\displaystyle E(Q\mid Z=1)-E(Q\mid Z=0)\geq 0,$ (93)

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

Under the IV assumptions 21.1–21.3, the difference in means for
$Q=DY,D(1-Y),(D-1)Y$ and $D+Y-DY$ 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. The above proving strategy is due to Jiang and Ding, (2020) for
a slightly more complex setting. Theorem 22.2 states the testable implications
only for a binary outcome. Problem 22.3 gives an equivalent form, and Problem
22.4 gives the result for a general outcome.

####  111 Examples

For a binary outcome, we can estimate all the parameters by the method of
moment as below.

⬇

## function for binary data (Z, D, Y)

## n_{zdy}’s are the counts from 2X2X2 table

IVbinary = function(n111, n110, n101, n100,

n011, n010, n001, n000){

n_tr = n111 + n110 + n101 + n100

n_co = n011 + n010 + n001 + n000

n = n_tr + n_co

## proportions of the latent strata

pi_n = (n101 + n100)/n_tr

pi_a = (n011 + n010)/n_co

pi_c = 1 - pi_n - pi_a

## four observed means of the outcomes (Z=z,D=d)

mean_y_11 = n111/(n111 + n110)

mean_y_10 = n101/(n101 + n100)

mean_y_01 = n011/(n011 + n010)

mean_y_00 = n001/(n001 + n000)

## means of the outcomes of two strata

mu_n1 = mean_y_10

mu_a0 = mean_y_01

## ER implies the following two means

mu_n0 = mu_n1

mu_a1 = mu_a0

## stratum (Z=1,D=1) is a mixture of c and a

mu_c1 = ((pi_c + pi_a)*mean_y_11 - pi_a*mu_a1)/pi_c

## stratum (Z=0,D=0) is a mixture of c and n

mu_c0 = ((pi_c + pi_n)*mean_y_00 - pi_n*mu_n0)/pi_c

## identifiable quantities from the observed data

list(pi_c = pi_c,

pi_n = pi_n,

pi_a = pi_a,

mu_c1= mu_c1,

mu_c0= mu_c0,

mu_n1= mu_n1,

mu_n0= mu_n0,

mu_a1= mu_a1,

mu_a0= mu_a0)

}

We then re-visit two canonical examples.

###### Example 22.1

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 $\tau_{\textup{c}}$
is $0.131$ with 95% confidence interval $(-0.036,0.298)$ including $0$. Using
the function above, we can obtain

⬇

$mu_c1

[1] 0.7086064

$mu_c0

[1] 0.6292042

There is no evidence of violating the IV assumptions.

###### Example 22.2

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 $\tau_{\textup{c}}$ is $0.116$ with 95% confidence interval
$(-0.061,0.293)$ including $0$. Using the function above, we can obtain

⬇

$mu_c1

[1] -0.004548064

$mu_c0

[1] 0.1200094

Since $\hat{\mu}_{1\textup{c}}<0$, there is evidence of violating the IV
assumptions.

Table 8: Binary data and IV inequalities

(a) Investigators et al., (2014)’s study | $Z=1$ |  | $Z=0$
---|---|---|---
| $D=1$ | $D=0$ |  | $D=1$ | $D=0$
$Y=1$ | 107 | 68 |  | 24 | 131
$Y=0$ | 42 | 42 |  | 8 | 79

(b) Hirano et al., (2000)’s study | $Z=1$ |  | $Z=0$
---|---|---|---
| $D=1$ | $D=0$ |  | $D=1$ | $D=0$
$Y=1$ | 31 | 85 |  | 30 | 99
$Y=0$ | 424 | 944 |  | 237 | 1041

####  112 Homework problems

######  22.1 Risk ratio for compliers

With binary outcome, we can define the risk ratio for compliers as

$\textsc{rr}_{\textup{c}}=\frac{\textup{pr}\\{Y(1)=1\mid
U=\textup{c}\\}}{\textup{pr}\\{Y(0)=1\mid U=\textup{c}\\}}.$

Show that under Assumptions 21.1–21.3, we can identify it by

$\textsc{rr}_{\textup{c}}=\frac{E(DY\mid Z=1)-E(DY\mid Z=0)}{E\\{(D-1)Y\mid
Z=1\\}-E\\{(D-1)Y\mid Z=0\\}}.$

Remark: Using Theorem 22.1, we can identify any comparisons between
$E\\{Y(1)\mid U=\textup{c}\\}$ and $E\\{Y(0)\mid U=\textup{c}\\}$.

######  22.2 Disentangle the mixtures: distributional results

This problem extends Theorem 22.1. Define

$f_{zu}(y)=\textup{pr}\\{Y(z)=y\mid
U=u\\},\quad(d=0,1;u=\textup{a},\textup{n},\textup{c})$

as the density of $Y(z)$ for latent stratum $U=u$, and define

$g_{zd}(y)=\textup{pr}(Y=y\mid Z=z,D=d)$

as the density of the outcome within the observed group $(Z=z,D=d)$. Show
Theorem 22.3 below.

###### Theorem 22.3

Under Assumptions 21.1–21.3, we can identify the type-specific densities of
the potential outcomes by

$\displaystyle f_{1\textup{n}}(y)=f_{0\textup{n}}(y)\equiv f_{\textup{n}}(y)$
$\displaystyle=$ $\displaystyle g_{10}(y),$ $\displaystyle
f_{1\textup{a}}(y)=f_{0\textup{a}}(y)\equiv f_{\textup{a}}(y)$
$\displaystyle=$ $\displaystyle g_{01}(y),$ $\displaystyle f_{1\textup{c}}(y)$
$\displaystyle=$ $\displaystyle\pi_{\textup{c}}^{-1}\\{\textup{pr}(D=1\mid
Z=1)g_{11}(y)-\textup{pr}(D=1\mid Z=0)g_{01}(y)\\},$ $\displaystyle
f_{0\textup{c}}(y)$ $\displaystyle=$
$\displaystyle\pi_{\textup{c}}^{-1}\\{\textup{pr}(D=0\mid
Z=0)g_{00}(y)-\textup{pr}(D=0\mid Z=1)g_{10}(y)\\}.$

######  22.3 Alternative form of Theorem 22.2

The inequalities in (93) can be re-written as