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$\ y_i$ = $\ 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 $ ?

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  • $\begingroup$ Added the tag and included my thoughts below $\endgroup$ – Bonsaibubble Dec 7 '16 at 15:09
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I so far got the following model - a rewritten form of the above:

$\ 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 $

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  • $\begingroup$ 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. $\endgroup$ – Alecos Papadopoulos Dec 3 '16 at 19:31
  • $\begingroup$ How is that done in a accurate way. Struggling to mention all necessary assumptions... $\endgroup$ – Bonsaibubble Dec 3 '16 at 19:43
  • $\begingroup$ You need to assume (which realistic_ that $E(v_i \mid x_1, x_2) = 0$ $\endgroup$ – Alecos Papadopoulos Dec 3 '16 at 19:44
  • $\begingroup$ I am struggling to see why minor issues such a closed form are important. Or do you mean my solution was wrong to begin with? Sorry for asking again. Many thanks! $\endgroup$ – Bonsaibubble Dec 3 '16 at 19:50
  • $\begingroup$ What do you mean by closed form? How can it not be? $\endgroup$ – Alecos Papadopoulos Dec 3 '16 at 19:52

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