Solution. (a) The index is \(-0.5\), so \(p = \Lambda (-0.5) = 1/(1 + e^{0.5}) = 0.378\). The marginal effect is \(p(1 - p)\beta = 0.378 \times 0.622 \times 0.5 = 0.118\): another thousand dollars raises the probability by about \(12\) percentage points at this income.

(b) \(e^{0.5} = 1.65\): each thousand dollars multiplies the odds of owning a car by \(1.65\), at every income level, although the change in the probability itself varies.

Solution. The score equation for the intercept, whose regressor is \(1\), is \(\sum (y_i - \hat \Lambda _i) = 0\), so \(\frac 1n\sum \hat \Lambda _i = \bar y\). For the probit the score is \(\sum \frac {(y_i - \Phi _i)\phi _i}{\Phi _i(1 - \Phi _i)}x_i\), a weighted sum with weights \(\phi _i/(\Phi _i(1 - \Phi _i))\) that are not constant, so the average fitted probability is close to \(\bar y\) but not in general equal to it.

Solution. (a) \(c = 1/2\), \(\Phi (0.5) = 0.691\) and \(\phi (0.5) = 0.352\). Then \(P(y > 0) = 0.691\), \(E(y \mid y > 0) = 1 + 2(0.352/0.691) = 2.02\), and \(E(y) = 0.691 \times 1 + 2 \times 0.352 = 1.40\), which also equals \(0.691 \times 2.02\).

(b) By Theorem 17.2.1(ii), \(\Phi (0.5)\beta _j = 0.691 \times 0.3 = 0.207\).

Solution. By Theorem 17.2.1(i), the mean of \(y\) among positive observations is \(x'\beta + \sigma \lambda (x'\beta /\sigma )\). The inverse Mills ratio \(\lambda (c)\) is decreasing in \(c\): units with a low index are observed only if their disturbance is large. So the omitted term \(\sigma \lambda \) is negatively correlated with \(x'\beta \), and for a regressor with a positive coefficient the bias is downward, towards zero. The observed relation is flatter than the latent one.

Solution. The second step regresses \(w\) on \(x\) and \(\lambda (x'\hat \gamma )\). Since \(\lambda \) is a non-linear function, \(\lambda (x'\hat \gamma )\) is not an exact linear combination of the columns of \(x\), so the coefficients are identified. But over the range of \(x'\gamma \) found in most samples, \(\lambda \) is close to linear, so \(\hat \lambda \) is nearly collinear with \(x\), the variance of the estimates is very large, and the results depend heavily on the normality assumption that gives \(\lambda \) its exact shape. A variable that shifts participation but not wages breaks the collinearity.

Solution. The score equation for the intercept is \(\sum (y_i - \hat \mu _i) = 0\), so \(\bar {\hat \mu } = \bar y\). A dummy coefficient of \(0.162\) means that the expected count is \(e^{0.162} = 1.18\) times as large when the dummy is one, \(18\%\) higher, other things equal; in Software 17.5.5 this is the effect of being male on days absent, which the robust standard error shows to be insignificant.