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

For this problem, suppose that unconfoundedness holds.

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

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.

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

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.

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.