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 $E(u_i \mid x_j)=0$?


1 Answer 1


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.)

  • $\begingroup$ I understand all the steps except why $x_j$ should be independent of both $u_i$ and $x_j% $\endgroup$ Nov 12, 2021 at 13:48
  • $\begingroup$ @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 $(y_i-x_i'\beta,\ x_i)=(u_i,x_i).$ $\endgroup$
    – r.e.s.
    Nov 12, 2021 at 18:49

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