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

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    $\begingroup$ Independence survives functions. I.e., if $X$ and $Y$ are independent, so are $f(X)$ and $g(Y)$. And $u_i=y_i-x_i'\beta$ is a function of $(y_i, x_i)$. $\endgroup$
    – Christoph Hanck
    Commented Nov 11, 2021 at 15:48
  • $\begingroup$ @Christoph Hanck I see no reason not simply to post that as an answer. $\endgroup$
    – Jesper for President
    Commented Nov 11, 2021 at 16:47
  • $\begingroup$ OK :-). I added a reference, too. $\endgroup$
    – Christoph Hanck
    Commented Nov 12, 2021 at 9:08

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Independence survives functions. I.e., if random variables $X$ and $Y$ are independent, so are $f(X)$ and $g(Y)$. See this thread for a precise discussion.

And $u_i=y_i−x_i'\beta$ is a function of $(y_i,x_i)$.

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  • $\begingroup$ So I understand that since $u_i$ is a function of yi, then $(u_i,x_i)$ must also be independent over $i=1,...,n$, but it's not obvious to me why $(u_i,x_i)$ is then independent from $x_j$ $\endgroup$
    – Giacomo Oliva
    Commented Nov 12, 2021 at 13:52

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