# OLS with unobservable $\ z_i$

$\ y_i$ = $\ α_1$ + $\ α_2xi_2$ + $\ α_3xi_3$ + $\ α_4z_i$ + $\ α_5 xi_2 z_i$ + $\ ui_4$

with E($\ u_i|xi_2, xi_3, z_i) = 0$

$\ z_i$ is not observable but linearly depending on $\ xi_2 , xi_3$

$\ zi = γ_1 + γ_2xi_2 + γ_3xi_3 + vi$

How can one find the partial effects of $\ xi_2 , xi_3$ on the $\ E(y_i|x_i)$

Can the OLS estimator be of closed form? How can one show this?

Also, how can one show that $\ E(e_i)=0$ ?

What can one conclude about $\ E(εi|xi_2, xi_3, x_2i_2, xi_2xi_3)$?

Is there a way to show that any function for $\ xi_2, xi_3$ is uncorrelated with $\ e_i$ ?

• Added the tag and included my thoughts below – Bonsaibubble Dec 7 '16 at 15:09

$\ yi = β1 + β_2xi_2 + β_3xi_3 + β_4x_2i_2 + β_5xi_2xi_3 + ε_i$ with $\ E(ε_i|xi_2, xi_3) = 0$
• This answer will be complete if you include the correspondence between the betas and the alpha-gammas, as well as showing why mean-independence of $\varepsilon$ from the $x$'s does indeed hold. – Alecos Papadopoulos Dec 3 '16 at 19:31
• You need to assume (which realistic_ that $E(v_i \mid x_1, x_2) = 0$ – Alecos Papadopoulos Dec 3 '16 at 19:44