# Consequences of Strict Exogeneity in a i.i.d. random sample

In the context of Strict Exogeneity on the classical regression model $$y_i = \beta x_i+ \varepsilon_j$$ we have: $$E(\varepsilon_i | x_i , x_j )=0,\quad \forall i,j$$ Under the linearity assumption, we have that $$\varepsilon_i$$ is a function of $$(y_i, x_i)$$

But, if $$(y_i,x_i)_{i=1}^n$$ is an i.i.d. I want to show that:

1. $$(y_i, x_i)$$ independent of $$(y_j, x_j)$$ for $$i \neq j$$ implies $$(\varepsilon_i, x_i)$$ is independent of $$x_j$$ for $$j \neq i$$;
2. $$E(\varepsilon_i | x_i,x_j)=E(\varepsilon_i | x_i)$$

I don't know how to prove the first item. As for the second one, I'm trying to use the following property, $$E(\varepsilon_i | x_i) = E( E( \varepsilon_i |x_i,x_j) |x_i)$$ but I don't know how to complete it.

Any help for my two items above?

First if $$X$$ and $$Y$$ are independent random vectors, then $$f(X)$$, any function of $$X$$, is independent of $$g(Y)$$, any other function of $$Y$$. So if $$(y_i,x_i)$$ is independent of $$(y_j,x_j)$$, then taking $$f(x,y)= (y- \beta x,x)$$ and $$g(x,y)=x$$ we have that $$(\varepsilon_i, x_i) \hbox{ is independent of } x_j.$$ This shows the first item.
For the second item, if $$(x,y)$$ is independent of $$z$$, then $$E(x|y,z)=E(x|y).$$ Thus, using the first item, we conclude that $$E(\varepsilon_i|x_i,x_j)=E(\varepsilon_i|x_i)$$