Suppose that \(x_{i}\) is endogenous in the regression \(y_{i}=\beta_{1}+\beta_{2} x_{i}+e_{i}\). Suppose that \(z_{i}\) is an instrumental variable that takes two values, one and zero with probabilities \(p\) and \(1-p\), respectively, that is, \(\operatorname{Pr}\left(z_{i}=1\right)=p\) and \(\operatorname{Pr}\left(z_{i}=0\right)=1-p\).

a. Show that \(E\left(z_{i}\right)=p\).

b. Show that \(E\left(y_{i} z_{i}\right)=p E\left(y_{i} \mid z_{i}=1\right)\).

c. Use the law of iterated expectations to show that \(E\left(y_{i}\right)=p E\left(y_{i} \mid z_{i}=1\right)+(1-p) E\left(y_{i} \mid z_{i}=0\right)\).

d. Substitute (a), (b), and (c) results into \(E\left(y_{i} z_{i}\right)-E\left(y_{i}\right) E\left(z_{i}\right)\) to show that \(\operatorname{cov}\left(y_{i}, z_{i}\right)=p(1-p) E\left(y_{i} \mid z_{i}=1\right)-p(1-p) E\left(y_{i} \mid z_{i}=0\right)\).

e. Use the arguments in (a)- (d) to show that \(\operatorname{cov}\left(x_{i}, z_{i}\right)=p(1-p)\left[E\left(x_{i} \mid z_{i}=1\right)-E\left(x_{i} \mid z_{i}=0\right)\right]\).

f. Assuming \(E\left(e_{i}\right)=0\) show \(\left[y_{i}-E\left(y_{i}\right)\right]=\beta_{2}\left[x_{i}-E\left(x_{i}\right)\right]+e_{i}\).

g. Multiply both sides of the expression in (f) by \(z_{i}-E\left(z_{i}\right)\) and take expectations to show \(\operatorname{cov}\left(y_{i}, z_{i}\right)=\beta_{2} \operatorname{cov}\left(x_{i}, z_{i}\right)\) if \(\operatorname{cov}\left(e_{i}, z_{i}\right)=0\).

h. Using (d), (f), and (g) show that \(\beta_{2}=\frac{E\left(y_{i} \mid z_{i}=1\right)-E\left(y_{i} \mid z_{i}=0\right)}{E\left(x_{i} \mid z_{i}=1\right)-E\left(x_{i} \mid z_{i}=0\right)}\)

i. Show that the empirical implementation of (h) leads to \(\hat{\beta}_{W A L D}=\left(\bar{y}_{1}-\bar{y}_{0}\right) /\left(\bar{x}_{1}-\bar{x}_{0}\right)\).