Create an example like Example 16.2 in which $\alpha > 0$ and $\theta < 0$.
Solution:
Consider the data:
| $X$ | $Y$ | $C_0$ | $C_1$ |
|---|---|---|---|
| 0 | 0 | 0 | 0* |
| 0 | 1 | 1 | 0* |
| 1 | 1 | 1* | 1 |
| 1 | 1 | 1* | 1 |
Then, $\alpha = \mathbb{E}(Y \mid X = 1) - \mathbb{E}(Y \mid X = 0) = 1 - \frac{1}{2} = \frac{1}{2} > 0$. Meanwhile, $\theta = \mathbb{E}(C_1) - \mathbb{E}(C_0) = \frac{1}{2} - \frac{3}{4} = -\frac{1}{4} < 0$.
Prove Theorem 16.4:
In general, $\theta(x) \ne r(x)$. However, when $X$ is randomly assigned, $\theta(x) = r(x)$.
Solution:
Use example 16.5 to show equality does not hold in general.
Suppose $X$ is randomly assigned. Then $X \amalg C(X)$, and we have
\begin{align*} \theta(x) &= \mathbb{E}[C(x)] \\ &= \mathbb{E}[C(x) \mid X = x] \tag{independence}\\ &= \mathbb{E}[Y \mid X = x] = r(x) \end{align*}Suppose you are given data $(X_1, Y_1), \dots, (X_n, Y_n)$ from an observational study, where $X_i \in \{0, 1\}$ and $Y_i \in \{0, 1\}$. Although it is not possible to estimate the causal effect $\theta$, it is possible to put bounds on $\theta$. Find upper and lower bounds on $\theta$ that can be consistently estimated from the data. Show that the bounds have width 1.
Hint: Note that $\mathbb{E}(C_1) = \mathbb{E}(C_1 \mid X = 1)\mathbb{P}(X = 1) + \mathbb{E}(C_1 \mid X = 0) \mathbb{P}(X = 0)$.
Solution:
We have:
\begin{align*} \theta &= \mathbb{E}(C_1) - \mathbb{E}(C_0) \\ &= \mathbb{E}(C_1 \mid X = 1)\mathbb{P}(X = 1) + \mathbb{E}(C_1 \mid X = 0) \mathbb{P}(X = 0) - \mathbb{E}(C_0 \mid X = 1)\mathbb{P}(X = 1) - \mathbb{E}(C_0 \mid X = 0) \mathbb{P}(X = 0) \\ &= \mathbb{E}(Y \mid X = 1)\mathbb{P}(X = 1) - \mathbb{E}(Y \mid X = 0) \mathbb{P}(X = 0) + \mathbb{E}(C_1 \mid X = 0) \mathbb{P}(X = 0) - \mathbb{E}(C_0 \mid X = 1)\mathbb{P}(X = 1) \\ &= \delta + \mathbb{E}(C_1 \mid X = 0) \mathbb{P}(X = 0) - \mathbb{E}(C_0 \mid X = 1)\mathbb{P}(X = 1) \end{align*}where $\delta = \mathbb{E}(Y \mid X = 1)\mathbb{P}(X = 1) - \mathbb{E}(Y \mid X = 0) \mathbb{P}(X = 0)$. Then,
\begin{align*} \theta &\le \delta + \mathbb{E}(C_1 \mid X = 0) \mathbb{P}(X = 0) \le \delta + \mathbb{P}(X=0), \end{align*}and
\begin{align*} \theta &\ge \delta - \mathbb{E}(C_) \mid X = 1) \mathbb{P}(X = 1) \le \delta - \mathbb{P}(X=1). \end{align*}Define the upper and lower bounds $U = \delta + \mathbb{P}(X=0)$ and $L = \delta - \mathbb{P}(X=1)$, respectively. Observe the bounds have with 1.
A consistent estimator of $U$ is given by:
$$\bar{Y}_1 \bar{X} + \bar{Y}_0 (1 - \bar{X}) + (1 - \bar{X})$$where
$$\bar{Y}_1 = \frac{1}{n_1} \sum_{i=1}^n Y_i X_i, \qquad \bar{Y}_0 = \frac{1}{n_0} \sum_{i=1}^n Y_i(1 - X_i),$$$$n_1 = \sum_{i=1}^n X_i \text{ and } n_0 \sum_{i=1}^n (1 - X_i).$$Consistency follows from the law of large numbers and Theorems 5.5a and 5.5d. Similarly, a consistent estimator of $L$ is given by:
$$\bar{Y}_1 \bar{X} + \bar{Y}_0 (1 - \bar{X}) - \bar{X}$$Suppose that $X \in \mathbb{R}$ and that, for each subject $i$, $C_i(x) = \beta_{1i}x$. Each subject has their own slope $\beta_{1i}$. Construct a joint distribution on $(\beta_1, X)$ such that $\mathbb{P}(\beta_1 > 0) = 1$ but $\mathbb{E}(Y | X = x)$ is a decreasing function of $x$, where $Y = C(X)$. Interpret.
Solution:
Let $X$ be any distribution with support on $[0,\infty)$ (e.g., Poisson), and let $\beta_1 = \frac{1}{X^2}$. Then, $\mathbb{P}(\beta_1 > 0) = 1$, and $\mathbb{E}(Y \mid X =x)) = \mathbb{E}(1 / X \mid X = x) = 1 / x$, which is decreasing on $[0, \infty)$.
Interpretation: The experiment is such that subjects with lower observed values of $x$ necessarily have a stronger response to treatment (i.e., slope) than subjects with larger observed values of $x$.
Let $X \in \{0, 1\}$ be a binary treatment variable and let $(C_0, C_1)$ denote the corresponding potential outcomes. Let $Y = C_X$ denote the observed response. Let $F_0$ and $F_1$ be the cumulative distribution functions for $C_0$ and $C_1$. Assume that $F_0$ and $F_1$ are both continuous and strictly increasing. Let $\theta = m_1 - m_0$ where $m_0 = F_0^{-1}(1/2)$ is the median of $C_0$ and $m_1 = F_1^{-1}(1/2)$ is the median of $C_1$. Suppose that the treatment $X$ is assigned randomly. Find an expression for $\theta$ involving only the joint distribution of $X$ and $Y$.
Solution:
Since $X$ is assigned randomly, $X \amalg (C_0, C_1)$. Therefore, $F_0(y) = P(C_0 \le y) = P(C_0 \le y \mid X=0) = P(Y \le y \mid X=0) = F_{Y\mid X=0}(y)$. Similarly, $F_1(y) = F_{Y \mid X=1}(y)$. Since $F_0$ and $F_1$ are continuous and strictly increasing, their inverses exist, and thus do those of $F_{Y\mid X=0}$ and $F_{Y\mid X=1}$.
Therefore, we have
$\theta = m_1 - m_0 = F^{-1}_1(1/2) - F^{-1}_0(1/2) = F^{-1}_{Y \mid X=1}(1/2) - F^{-1}_{Y \mid X=0}(1/2)$.