# Econometrics: why the i.i.d assumption and weak exogeneity assumption imply strict exogeneity?

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

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: $$$$E(\varepsilon_t|X)=E(\varepsilon_t|X_1,...,X_t,...,X_n) =E(\varepsilon_t|X_t)=0$$$$ My question is about the second equality: why we have $$E(\varepsilon_t|X_1,...,X_t,...,X_n) =E(\varepsilon_t|X_t)$$ under assumptions given above?

• 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. – mark leeds Oct 26 '19 at 12:34
• @mark leeds thanks, it's very intuitive. – Huaixin Oct 26 '19 at 15:41
• Your comment is apparently valid. You may answer the question. – Subhash C. Davar Jan 31 at 5:48

\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*}
• @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. – Walrasian Auctioneer Oct 26 '19 at 19:51