meaning of ratio of density function to distribution function of the standard normal? Consider the model:
$ y_i = \beta' X_i + u_i $ if RHS $> 0.$
$ y_i = 0$     otherwise.
So the expected value of $y$ under condition of $y_i > 0$ is:
$$ E(y_i| y_i >0) = \beta' X_i + E(u_i|u_i > -\beta'X_i)$$
$  E(y_i| y_i >0)=  \beta' X_i + \sigma \dfrac{\phi_i}{\Phi_i}$,
where $ \phi_i$ and  $\Phi_i$ are the density function and the distribution of the standard normal evaluated at $\beta'X_i/\sigma$. $\sigma$ is variance of $u_i$.
My question is:
why $E(u_i|u_i > -\beta'X_i)= \sigma \dfrac{\phi_i}{\Phi_i}$  and what is the meaning of ratio $\dfrac{\phi_i}{\Phi_i}$? 
 A: Assume $u_i\in N(0,\sigma^2)$, $\sigma$ the standard deviation. 
I think that one should use large deviations to analyse this. Note that. So assume $a$ positive and large, then
$$
E(u_i|u_i > a) = a + \int_a^\infty P(u_i>x|u_i>a) \,dx = a + \int_a^\infty P(u_i>x)/P(u_i>a) \,dx=a + \frac{\int_a^\infty P(u_i>x)\,dx}{P(u_i>a)}
$$
Now large deviations means that
$$
P(u_i>x) \approx \frac{\sigma}{x\sqrt{2\pi}}\exp(-x^2/(2\sigma^2))
$$
Now, for the bulk of this integral of this is $\frac{1}{x}\approx 1/a$ hence another application of the approximation lead to,
$$
E(u_i|u_i > a) \approx a + \frac{\frac{\sigma^3}{a^2}\phi(a)}{P(u_i>a)}
$$ 
Now use the following identity
$$
0 = E(u) = E(u|u>-a)P(u>-a) + E(u|u<-a)P(u<-a)
$$ And hence conclude that (with a play of the symmetry and another application of the large deviation approximation),
$$
E(u|u>-a) = a\frac{P(u<-a)}{P(u>-a)} + \frac{\frac{\sigma^3}{a^2}\phi(a/\sigma)}{\Phi(a/\sigma)} \approx \left ( \sigma + \frac{\sigma^3}{a^2} \right )\frac{\phi(a/\sigma)}{\Phi(a/\sigma)}
$$ Now for $a^2>>\sigma^3$ the second term in the paranthesis can be removed and you get the expression, assuming the offset large enugh so that the approximations are ok, but with $\sigma$ standard deviation.
