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:

*

*$(y_i, x_i)$ independent of $(y_j, x_j)$ for $i \neq j$ implies $(\varepsilon_i, x_i)$ is independent of $x_j$ for $j \neq i$;

*$E(\varepsilon_i | x_i,x_j)=E(\varepsilon_i | x_i)$
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
$$E(\varepsilon_i|x_i,x_j)=E(\varepsilon_i|x_i)$$
Quod Erat Demonstrandum!
