Proof. (i) For \(\varepsilon \sim N(0, \sigma ^2)\) and any \(a\), \(E(\varepsilon \mid \varepsilon > -a) = \sigma \phi (a/\sigma )/\Phi (a/\sigma )\). To see this, write \(\varepsilon = \sigma v\) with \(v\) standard normal; then \[ E(v \mid v > -c) = \frac {\int _{-c}^\infty v\phi (v)\,dv}{\Phi (c)} = \frac {\bigl [-\phi (v)\bigr ]_{-c}^\infty }{\Phi (c)} = \frac {\phi (c)}{\Phi (c)} , \] using \(\phi '(v) = -v\phi (v)\) and the symmetry \(\phi (-c) = \phi (c)\), \(P(v > -c) = \Phi (c)\). With \(a = x_i'\beta \), \(E(y_i \mid y_i > 0) = x_i'\beta + E(\varepsilon _i \mid \varepsilon _i > -x_i'\beta )\).

(ii) Since \(y_i = 0\) when it is not positive, \[ E(y_i) = P(y_i > 0)E(y_i \mid y_i > 0) = \Phi (c_i)\bigl [x_i'\beta + \sigma \lambda (c_i)\bigr ] = \Phi (c_i)x_i'\beta + \sigma \phi (c_i) . \] Differentiating with respect to \(x_{ij}\), with \(\partial c_i/\partial x_{ij} = \beta _j/\sigma \), \[ \frac {\partial E(y_i)}{\partial x_{ij}} = \phi (c_i)\frac {\beta _j}{\sigma }x_i'\beta + \Phi (c_i)\beta _j + \sigma \phi '(c_i)\frac {\beta _j}{\sigma } = \Phi (c_i)\beta _j + \beta _j\phi (c_i)(c_i - c_i) = \Phi (c_i)\beta _j , \] since \(\phi '(c_i) = -c_i\phi (c_i)\).

(iii) A positive \(y_i\) contributes its normal density \(\sigma ^{-1}\phi \bigl ((y_i - x_i'\beta )/\sigma \bigr )\); a zero contributes \(P(y_i^* \leq 0) = 1 - \Phi (c_i)\). □