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

Question

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?

Answer 1

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!