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The proposition is from lecture notes of advanced econometrics of Yongmiao Hong: notes page

A1: $\{Y_t,X_t'\}_{t=1}^n$ is an i.i.d. random sample.

A2: $E(\varepsilon_t|X_t)=0$ almost surely with $E(\varepsilon_t^2)=\sigma^2<\infty$.

Then A1 and A2 imply the strict exogeneity holds: \begin{equation} E(\varepsilon_t|X)=E(\varepsilon_t|X_1,...,X_t,...,X_n) =E(\varepsilon_t|X_t)=0 \end{equation} My question is about the second equality: why we have $E(\varepsilon_t|X_1,...,X_t,...,X_n) =E(\varepsilon_t|X_t)$ under assumptions given above?

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    $\begingroup$ Hi: $\epsilon_t$ is only in the same equation as $X_t$. Therefore, strict exogeneity of $\epsilon_t$ only depends on $X_t$ and not the $X$'s at other times. $\endgroup$
    – mark leeds
    Oct 26, 2019 at 12:34
  • $\begingroup$ @mark leeds thanks, it's very intuitive. $\endgroup$
    – HXW
    Oct 26, 2019 at 15:41
  • $\begingroup$ Your comment is apparently valid. You may answer the question. $\endgroup$ Jan 31, 2020 at 5:48

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\begin{align*} \mathbb{E}\left[ \epsilon_t \vert X \right] &= \mathbb{E}\left[\epsilon_t \vert X_1, \dots, X_t, \dots, X_n \right]\\ &= \mathbb{E} [Y_t - X_t'\beta^0 \, \vert X_1, \dots, X_t, \dots, X_n] \\ &= \mathbb{E} [Y_t - X_t'\beta^0 \, \vert X_t] \quad \text{since $\{Y_t,X_t\}$ is iid.} \\ &=\mathbb{E}\left[ \epsilon_t \vert X_t \right] \\ &= 0 \quad \text{by Assumption A2.} \end{align*}

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  • $\begingroup$ Aren't you appealing to Assumption 4.2 in the second line of this proof? $\endgroup$
    – H Rogers
    Oct 26, 2019 at 14:38
  • $\begingroup$ Got it, thank you so much! $\endgroup$
    – HXW
    Oct 26, 2019 at 15:39
  • $\begingroup$ @HRogers I am indeed, but even if the relationship is non-linear (i.e. $Y_t = g(X_t) + \epsilon_t$), the result would be equivalent. $\endgroup$ Oct 26, 2019 at 19:51

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