A question about the strict exogeneity assumption on the linear regression model

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.

• You are quite right that the two conditions amount to the same thing under iid sampling. The thing is that there are cases where iid sampling cannot work, see e.g. stats.stackexchange.com/questions/323695/… Commented Jan 31, 2022 at 13:27