Properties of bivariate standard normal and implied conditional probability in the Roy model

Question

Sorry for the long title, but my problem is quite specific and hard to explain in one title.

I am currently learning about the Roy Model (treatment effect analysis).

There is one derivation step at my slides, which I do not understand.

We calculate the expected outcome with treatment in the tretment group (dummy D is treatment or not treatment). This is written as

\begin{align} E[Y_1|D=1] \end{align}

since $Y_1=\mu_1 + U_1$ this can be rewritten as \begin{align} E[Y_1|D=1] &= E[\mu_1+U_1|D=1]\\ &=\mu_1+ E[U_1|D=1] \end{align} before we also said, that $D=1$ if $Y_1>Y_0$ so it follows:

$Y_1-Y_0>0$

$\mu_1+U_1-(\mu_0-U_0)>0$

$(\mu_1+U_1)/\sigma-(\mu_0-U_0)/ \sigma >0$

$Z-\epsilon>0$

so $D=1$ if $\epsilon<Z$

Therefore it holds, that \begin{align} E[Y_1|D=1] &=\mu_1 + E[U_1|\epsilon<Z] \end{align}

It is further known, that \begin{align} \begin{bmatrix} U_1 \\ U_0 \\ \epsilon \end{bmatrix}=N\left( \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} \sigma_1^2 & \sigma_{10} & \sigma_{1\epsilon} \\ \sigma_{10} & \sigma_{0}^2 & \sigma_{0\epsilon} \\ \sigma_{1\epsilon} & \sigma_{0\epsilon} & \sigma_{\epsilon}^2 \end{bmatrix}\right) \end{align}

therefore it follows: $P(D=1)=P(\epsilon<Z)=\Phi(Z)$

So now comes my question, the slides say, that \begin{align} \mu_1 - E[U_1|\epsilon<Z] =\mu_1 - \sigma_{1\epsilon} \frac{\phi(Z)}{\Phi(Z)} \end{align} And I do not understand why?

I know, that if two random variables follow a standard bivariate normal distribution: $E[u_1|u_2)=\rho u_2$

so $E[u_1|u_2>c)=E[\rho u_2|u_2>c]=\rho E[u_2|u_2>c)=\rho\frac{\phi(c)}{1-\Phi(c)}$

Therefore I would have expected a "plus" and not a minus sign? Also why do we use the covariance $\sigma_{1\epsilon}$ and not the correlation $\rho$? So I would have expected something like

\begin{align} \mu_1 - E[U_1|\epsilon<Z] =\mu_1 + \rho \frac{\phi(Z)}{\Phi(Z)} \end{align}

I am aware of the fact, that if I do the truncation from above the $1-\Phi(c)$ becomes a $\Phi(c)$.

Answer 1

First, in the Roy model, $\sigma_{\varepsilon}^{2}$ is normalized to be $1$ for identification reason (c.f. Cameron and Trivedi: Microeconometrics: methods and applications). I will maintain this normalization hereafter. To answer your question, let's show $$ \mathrm{{E}}\left(U_{1}\mid\varepsilon<Z\right)=-\sigma_{1\varepsilon}\frac{\phi\left(Z\right)}{\Phi\left(Z\right)} $$ first. Here $\phi$ and $\Phi$ are the pdf and cdf of a standard normal distribution, respectively. Note that $$ \mathrm{E}\left(U_{1}\mid\varepsilon<Z\right)=\mathrm{E}\left(\mathrm{E}\left(U_{1}\mid\varepsilon\right)\mid\varepsilon<Z\right) $$ by the law of iterated expectation. The vector $\left(U_{1},\varepsilon\right)$ is a bivariate normal with mean $\left(0,0\right)'$ and covariance matrix $$ \left[\begin{array}{cc} \sigma_{1}^{2} & \sigma_{1\epsilon}\\ & 1 \end{array}\right]. $$ The conditional mean $\mathrm{{E}}\left(U_{1}\mid\varepsilon\right)=\sigma_{1\varepsilon}\varepsilon$ (note that covariance not correlation arises here because $\sigma_{\varepsilon}^{2}=1$). Thus, $$ \mathrm{E}\left(U_{1}\mid\varepsilon<Z\right)=\sigma_{1\varepsilon}\mathrm{E}\left(\varepsilon\mid\varepsilon<Z\right). $$ The density function of $\varepsilon\mid\varepsilon<Z$ is $$ f\left(\varepsilon\mid\varepsilon<Z\right)=\begin{cases} \frac{\phi\left(\varepsilon\right)}{\Phi\left(Z\right)}, & -\infty<\varepsilon<Z;\\ 0, & \varepsilon\geq Z. \end{cases} $$ The conditional mean $\mathrm{E}\left(\varepsilon\mid\varepsilon<Z\right)$ is \begin{eqnarray*} \mathrm{E}\left(\varepsilon\mid\varepsilon<Z\right) & = & \int_{-\infty}^{Z}t\frac{\phi\left(t\right)}{\Phi\left(Z\right)}\,\mathrm{{d}}t\\ & = & \frac{1}{\Phi\left(Z\right)}\int_{-\infty}^{Z}t\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{1}{2}t^{2}\right)\,\mathrm{{d}}t\\ & = & -\frac{1}{\Phi\left(Z\right)}\int_{-\infty}^{Z}\frac{\partial}{\partial t}\left\{ \frac{1}{\sqrt{2\pi}}\exp\left(-\frac{1}{2}t^{2}\right)\right\} \,\mathrm{{d}}t\\ & = & -\frac{1}{\Phi\left(Z\right)}\left(\phi\left(Z\right)-\phi\left(-\infty\right)\right). \end{eqnarray*} Note how the negative sign comes out. Thus, $\mathrm{E}\left(\varepsilon\mid\varepsilon<Z\right)=-\phi\left(Z\right)/\Phi\left(Z\right)$, and the conclusion follows.