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

Question

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: \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?

Answer 1

\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*}