E(ui-E(ui I X1i, X2i) I X1i,X2i) = E(ui I X1i, X2i)- E(ui I X1i, X2i) = 0 i cant
ID: 3204732 • Letter: E
Question
E(ui-E(ui I X1i, X2i) I X1i,X2i) = E(ui I X1i, X2i)- E(ui I X1i, X2i) = 0
i cant really do the conditional-on line thats suppose to be here properly, i hope that makes sense, the question is regarding conditional mean independence. As a second question i need to learn the rules that are in play regarding conditional expectations, i cant seem to find them collected anywhere. How to manuever them and what you allowed to do. do you have any recommendations of book or link where i could find this?
E(ui-E(ui)%; Xe.)| -W . E (ui IXú-E (ui)%ikir 0 U,-E/uilXcXZi UilXii2:Explanation / Answer
Given X1i,X2i, E(ui I X1i, X2i) is a constant.
Let it be denoted by a.
a = E(ui I X1i, X2i) , given X1i, X2i
So,
E(ui-E(ui I X1i, X2i) I X1i,X2i) = E(ui-a I X1i,X2i)
You can simply take 'a' outside of the expectation as it is a constant..
= E(ui I X1i,X2i) - a
= E(ui I X1i, X2i)- E(ui I X1i, X2i) = 0
You might consider reading about conditional probability from the book of W. Hines (Wiley publications)
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