# Derivation of Olsens LS Selectivity Correction

There are many estimation procedures that correct for sample selection. The most famous is Heckman's two-step selectivity correction (in two equations) that assumes bivariate normality of the error terms in both equations.

Olsen (1980) ["A Least Squares Correction For Selectivity Bias", Ecta.] showed that it is sufficient to assume a distributional form for the errors of the selection equation $\nu_i$ and linearity of the conditional expectation of $u_i$ given $\nu_i$, where $u_i$ is the error term of the regression model of interest.

How can the OLS selectivity correction be derived? This should be simple to see, but I would appreciate if someone writes down the derivation step by step.

Olsen starts from the following specification and assumptions:

The regression model of interest is $$y_i=X_i\beta+u_i.$$ $y_i$ is observed if and only if $I_i = 1$ where

$$I_i = 1 \text{ iff } \nu_i<Z_i\gamma,$$ $$I_i = 0 \text{ iff } \nu_i\geq Z_i\gamma.$$

He assumes $X_i$ and $Z_i$ are row vectors which conform to the column vectors of unknown coefficients $\beta$ and $\gamma$, respectively. The $X$'s and $Z$'s are exogenous.

Further, he assumes that the expected value of the error of the intensive regression is zero $$E(u_i)=0,$$ $$E(u_iu_j)=\sigma^2_u \text{ for } i=j,$$ $$E(u_iu_j)=0 \text{ for } i\neq j,$$ $$E(\nu_i)=\mu_\nu,$$ $$E[(\nu_i-\mu_\nu)(\nu_i-\mu_\nu)]=\sigma^2_\nu \text{ for } i=j,$$ $$E[(\nu_i-\mu_\nu)(\nu_i-\mu_\nu)]=0 \text{ for } i\neq j,$$ $$Cov(u_i,\nu_i)=E(u_i\nu_i)-E(u_i)E(\nu_i)=\rho \sigma_\nu \sigma_u \text{ for } i=j,$$ $$Cov(u_i,\nu_i)=0 \text{ for } i\neq j,$$ $$E(u_i|\nu_i)=\rho(\nu_i-\mu_\nu)\sigma_u/\sigma_\nu.$$

Here I got stuck.

Therefore my question 1 is: How can it be shown that $E(u_i|\nu_i)=\rho(\nu_i-\mu_\nu)\sigma_u/\sigma_\nu$ (see also p. 1816 in Olsen, 1980)?

Olsen goes on by assuming the conditional expectation of $u_i$ given $\nu_i$ is linear in $\nu_i$, so we can use the decomposition $$u_i=\rho(\nu_i-\mu_\nu)\sigma_u/\sigma_\nu+\varepsilon_i,$$ where $\varepsilon_i$ and $\nu_i$ are uncorrelated.

This is $$y_i=X_i\beta+\rho(\nu_i-\mu_\nu)\sigma_u/\sigma_\nu+\varepsilon_i.$$

And it follows that $E(y_i|X_i,\nu_i<Z_i\gamma)=X_i\beta+\rho\sigma_uE(\nu_i|\nu_i<Z_i\gamma)/\sigma_\nu-\rho\sigma_u\mu_\nu/\sigma_\nu$.

Now, assuming that the $\nu_i$'s were uniformly distributed over $[0,1]$, he finds $E(y_i|X_i,\nu_i<Z_i\gamma)=X_i\beta+\rho\sigma_u\sqrt{3}(Z_i\gamma-1)$. But how?

Therefore my question 2 is: What are the steps from $E(y_i|X_i,\nu_i<Z_i\gamma)=X_i\beta+\rho\sigma_uE(\nu_i|\nu_i<Z_i\gamma)/\sigma_\nu-\rho\sigma_u\mu_\nu/\sigma_\nu$ to $E(y_i|X_i,\nu_i<Z_i\gamma)=X_i\beta+\rho\sigma_u\sqrt{3}(Z_i\gamma-1)$ (see also p. 1816/1817 Olsen, 1980)?