In the context of the linear regression model: $$y_i = x_i'\beta + u_i, \quad E(u_i|x_i)=0, \quad i=1,...n.$$ one of the assumptions is strict exogeneity: $$E(u_i|x_1,...,x_n )=0,\quad \forall \, i =1,...,n $$ I would like to understand why this hypothesis not just that? $$E(u_i|x_i )=0,\quad \forall \, i =1,...,n $$ My intuition says that this is possible since the $x's$ are i.i.d. In fact, $(y_i,x_i) \sim f, i.i.d.$
I will try to demonstrate that $$E(u_i| x_1,...x_n) = E(u_i|x_i) $$
$$ E\left[u_{i} \mid x_{1}, \ldots, x_{n}\right]=\int u_{i} f\left(u_{i} \mid x_{1}, \ldots, x_{n}\right) d u_{i}=\int u_{i} \frac{f\left(u_{i}, x_{1}, \ldots, x_{n}\right)}{f\left( x_{1}, \ldots, x_{n}\right)} d u_{i} $$ iid sampling implies: $$ \begin{aligned} \int u_{i} \frac{f\left(u_{i}, x_{1}, \ldots, x_{n}\right)}{f\left( x_{1}, \ldots, x_{n}\right)} d u_{i}=\int u_{i} \frac{f\left(u_{i}, x_{i}\right) f\left(x_{2}, \ldots, x_{i-1},x_{i+1},\ldots, x_{n}\right)}{f\left(x_{i}\right) f\left(x_{2}, \ldots, x_{i-1},x_{i+1},\ldots, x_{n}\right)} d u_{i}=& \\ \int u_{i} f\left(u_{i} \mid x_{i}\right) d u_{i}=\mathrm{E}\left[u_{i} \mid x_{i}\right] \end{aligned} $$
I don't understand the need to condition on all variables $x_1,...,x_n$ on the strict exogeneity assumption! Some help.
