Solution. (a) By Theorem 13.1.1, \(\lambda = 36/(36 + 12) = 0.75\), so \(\plim b = 0.75 \times 0.15 = 0.1125\).

(b) Use the second measurement \(Z\) as an instrument for the first. It is correlated with true income, so it is relevant, and its error is independent of the first measurement’s error and of the disturbance, so it is uncorrelated with \(\varepsilon - \beta m\). Then \(b_{\text {IV}} = \sum zy/\sum zx \xrightarrow {p} 0.15\).

Solution. (a) By Theorem 13.1.2, \(\plim b = 0.8 + 0.2 \times 4/40 = 0.82\).

(b) \(I_t\) is uncorrelated with \(u_t\) by assumption and correlated with \(Y_t\) through \(Y_t = (\alpha + I_t + u_t)/(1 - \beta )\). So \(b_{\text {IV}} = \sum i_tc_t/\sum i_ty_t\), in deviations, and \(\plim b_{\text {IV}} = \Cov (I, C)/\Cov (I, Y)\). Since \(C = Y - I\), \(\Cov (I, C) = \Cov (I, Y) - \sigma _I^2\), with \(\Cov (I, Y) = \sigma _I^2/(1 - \beta )\), so \(\plim b_{\text {IV}} = 1 - (1 - \beta ) = \beta \).

Solution. The means are \(\bar z = 3\), \(\bar x = 4\) and \(\bar y = 7.4\). In deviations, \(\sum zx = 9\), \(\sum zy = 17\), \(\sum x^2 = 10\) and \(\sum xy = 18\). So \(b = 18/10 = 1.8\) with intercept \(7.4 - 1.8(4) = 0.2\), and \(b_{\text {IV}} = 17/9 = 1.889\) with intercept \(7.4 - 1.889(4) = -0.156\). Both lines pass through the means; they differ in the slope, which IV computes from the covariation with \(z\) instead of with \(x\) itself.

Solution. (a) There is one endogenous regressor and three excluded instruments, so the order condition holds with \(\ell - k = 2\) overidentifying restrictions. (Experience and its square are treated as exogenous and are their own instruments.) The rank condition requires the instruments to be jointly significant in the first stage.

(b) \(H = (0.101 - 0.074)^2/(0.013^2 - 0.004^2) = 0.000729/0.000153 = 4.76\), which exceeds the \(\chi ^2_1\) \(5\%\) point \(3.84\): exogeneity of schooling is rejected at \(5\%\) (but not at \(1\%\), \(6.63\)).

(c) Against \(\chi ^2_2\), with \(5\%\) point \(5.99\) and \(1\%\) point \(9.21\), the overidentifying restrictions are rejected at \(5\%\) but not at \(1\%\): there is some evidence that at least one instrument is invalid. Parents’ education may affect earnings directly, through networks or inherited ability, and so is suspect. But the test cannot say which instrument is at fault, and it would not detect a problem common to all three.

Solution. With \(\bar z = n_1/n\), \begin {align*} \sum (z_i - \bar z)y_i &= \sum _{z_i = 1}y_i - \bar z\sum y_i = n_1\bar y_1 - \frac {n_1}{n}(n_1\bar y_1 + n_0\bar y_0)\\ &= \frac {n_1n_0}{n}(\bar y_1 - \bar y_0) . \end {align*}

The same calculation for \(x\) gives \(\frac {n_1n_0}{n}(\bar x_1 - \bar x_0)\), and the ratio is the Wald estimator. Here \(b_{\text {IV}} = (12.4 - 11.2)/(14.1 - 12.6) = 1.2/1.5 = 0.8\).

Solution. (a) \(y - \hat xb = (y - xb) + (x - \hat x)b = \hat u + \hat vb\).

(b) As \(n \to \infty \), \(b \to \beta \), \(\hat \pi \to \pi \), \(\hat u_i\) behaves like \(u_i\) and \(\hat v_i\) like \(v_i\), so \[ \frac 1n\sum (y_i - \hat x_ib)^2 \xrightarrow {p} E(u + \beta v)^2 = \sigma _u^2 + 2\beta \sigma _{uv} + \beta ^2\sigma _v^2 . \] This is the variance of \(u + \beta v = y - \beta \pi z\), the disturbance of the reduced form, not \(\sigma _u^2\). The hand-run second stage therefore uses the wrong residual variance; the right one uses \(\hat u = y - xb\), because the structural disturbance is defined with the actual regressor \(x\). In Software 13.5.1 the error made the standard error a quarter too small, but it can go either way, depending on the sign of \(\sigma _{uv}\).