I have been reading Heckman (1979) and have tried to prove some result used (the paper points to a book which does not show the work either). I alter the notation a bit for clarity. Assume we have: $$\left(\left.\begin{matrix}u_{1}\\ u_{2} \end{matrix}\right|x\right)\sim\mathcal{N}\left(\mathbf{0},\left[\begin{matrix}\sigma_{1}^{2} & \rho\cdot\sigma_{1}\\ \rho\cdot\sigma_{1} & 1 \end{matrix}\right]\right)$$ with $x\in \mathbb{R}^{k}$ a constant column. I want to compute: $$\mathbb{E}\left[{u_{1}|x,u_{2}\ge-c\cdot x}\right]$$ with $c\in\mathbb{R}^{k}$ a constant row. It turns out that it equals $\rho\cdot\sigma_{1}\cdot\lambda\left(c\cdot x\right)$ with $\lambda\left(\cdot \right)=\dfrac{\phi\left({\cdot}\right)}{\Phi\left({\cdot}\right)}$ the ratio of the standard normal PDF and CDF (the so-called "inverse Mills ratio").

I haven't been able to do much work. Help would be appreciated.


1 Answer 1


In the bivariate normal case (and given the zero-mean assumption and the unit variance of $u_2$ here) we have

$$E(u_1 \mid u_2) = \rho\sigma_1u_2$$

Using the law of iterated expectations we can write

$$E(u_1 \mid u_2 > -cx) = E\big[E(u_1 \mid u_2 ) \mid u_2> -cx\big]$$

and inserting the first relation we have

$$E(u_1 \mid u_2 > -cx) = E\big[\rho\sigma_1u_2 \mid u_2> -cx\big]$$

$$=\rho\sigma_1E\big[u_2 \mid u_2> -cx\big]$$

We now have an expected value from a truncated normal, which is (always using the specific assumptions of the question)

$$E\big[u_2 \mid u_2> -cx\big] = \frac {\phi(-cx)}{1-\Phi(-cx)} = \frac {\phi(cx)}{\Phi(cx)}$$

the last equality by the properties/symmetries of $\phi$ and $\Phi$.

So we arrive at

$$E(u_1 \mid u_2 > -cx) = \rho\sigma_1\frac {\phi(cx)}{\Phi(cx)}$$

Note: the existence of the conditioning $x$ does not play a part in the above derivation other than fixing the truncation boundary.


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