You do not need to turn these in. These are just some additional practice questions for the midterm. You shouldn’t treat this as a practice midterm though as these questions are only for the new material since the previous question.

1. For this problem, suppose that unconfoundedness holds.

1. Define $$CATE(X) = \mathbb{E}[Y(1) - Y(0) | X]$$. Is $$CATE(X)$$ identified in this setting? If yes, provide an expression for it.

2. Define $$CATT(X) = \mathbb{E}[Y(1) - Y(0) | X, D=1]$$. Is $$CATT(X)$$ identified in this setting? If yes, provide an expression for it.

3. Provide an explanation/comparison for the results you get in parts (a) and (b).

2. In class, we mentioned that, for some function $$g(X)$$, it held that

\begin{align*} \mathbb{E}[g(X)|D=1] = \mathbb{E}\left[ \frac{p(X)(1-p)}{(1-p(X))p} g(X) \big| D=0 \right] \end{align*}

Provide a mathematical explanation for why this holds. For simplicity, you can suppose that $$X$$ is continuously distributed in your explanation.

3. Consider the regression $$Y_i = \alpha D_i + X_i'\beta_0 + e_i$$. Suppose that unconfoundedness and overlap both hold. Show that \begin{align*} \alpha = \mathbb{E}[w(D,X) CATE(X)] \end{align*} where $$CATE(X)$$ is defined above and $$w(D,X) = \displaystyle \frac{D(1-\textrm{L}(D|X))}{\mathbb{E}\big[(D-\textrm{L}(D|X))^2\big]}$$ if either (i) $$p(X) = \textrm{L}(D|X)$$ or (ii) $$\mathbb{E}[Y|X,D=0] = \textrm{L}_0(Y|X)$$. To show this result, you can use as a starting part either of the two decompositions for $$\alpha$$ that we discussed in class.