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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 $\Phi(c)$.

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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.

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  • $\begingroup$ I tried to award you the bounty, but it says: "You may award your bounty in 18 hours". Remind me, if I forget it :-) $\endgroup$ – Stat Tistician Jul 16 '13 at 17:48
  • $\begingroup$ Thanks for your award, and I am glad that the post helps. $\endgroup$ – semibruin Jul 16 '13 at 22:08

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