# Random sampling in linear model

Consider the linear model

$$y_i = x_i' \beta+u_i$$ for $$i=1,\ldots,n$$

with $$E(y_i \mid x_i)=x_i' \beta \iff E(u_i \mid x_i)=0$$. Assume that the observations on $$(y_i, x_i')$$ are independent over $$i=1,...,n$$

The textbook claims that $$E(u_i \mid x_i,\ldots,x_n)=E(u_i \mid x_i)$$. Why is this? How does knowing that $$(y_i, x_i')$$ is independent from $$(y_j, x_j')$$ tell us that $$E(u_i \mid x_j)=0$$?

We're given that $$(y_1,x_1'),\ldots,(y_n,x_n')$$ are independent, and $$u_i=y_i-x_i'\beta.$$ Now let $$X_{-i}$$ be any collection of the $$x_j$$ that does not include $$x_i,$$ so $$X_{-i}$$ is independent of both $$u_i$$ and $$x_i.$$ Then we have the following probability density functions (assuming they all exist): \begin{align} f(u_i\mid x_i, X_{-i}) &={f(u_i,x_i,X_{-i})\over f(x_i, X_{-i})}\quad\text{by definition of the conditional p.d.f.}\\[2ex] &={f(u_i,x_i)f(X_{-i})\over f(x_i)f(X_{-i})}\quad\text{because X_{-i} is independent of both u_i and x_i}\\[2ex] &={f(u_i,x_i)\over f(x_i)}\\[2ex] &=f(u_i\mid x_i) \end{align} Therefore, $$\mathsf E(u_i\mid x_i,X_{-i})=\mathsf E(u_i\mid x_i).$$
(The notation is sloppy, writing the same letter $$f$$ for all the different p.d.f.s, which are to be distinguished by the symbols used in their arguments; also, the same letters denote both random variables and their values.)
• I understand all the steps except why $x_j$ should be independent of both $u_i$ and $x_j% Nov 12, 2021 at 13:48 • @JacopoOlivieri Your comment has "$x_j$" twice -- it should say "$x_j$is independent of both$u_i$and$x_i$" (assuming$j\ne i$). We're given that$(y_j,x_j)$is independent of$(y_i,x_i)$when$j\ne i$, in which case$x_j$is independent of$(y_i,x_i),$hence$x_j$is independent of$(y_i-x_i'\beta,\ x_i)=(u_i,x_i).\$ Nov 12, 2021 at 18:49