<p><span class="math-container">\begin{align*}
\mathbb{E}\left[ \epsilon_t \vert X \right] &amp;= \mathbb{E}\left[\epsilon_t \vert X_1, \dots, X_t, \dots, X_n \right]\\
&amp;= \mathbb{E} [Y_t - X_t'\beta^0 \, \vert X_1, \dots, X_t, \dots, X_n] \\
&amp;= \mathbb{E} [Y_t - X_t'\beta^0 \, \vert X_t] \quad \text{since $\{Y_t,X_t\}$ is iid.} \\
&amp;=\mathbb{E}\left[ \epsilon_t \vert X_t \right] \\
&amp;= 0 \quad \text{by Assumption A2.}
\end{align*}</span></p>

CC BY-SA 4.0
2019-10-26T14:38:19.860Z
Aren't you appealing to Assumption 4.2 in the second line of this proof?
H Rogers
CC BY-SA 4.0
2019-10-26T19:51:53.693Z
@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
CC BY-SA 4.0
2019-10-26T14:17:21.520Z
2019-10-26T14:17:21.520Z
Walrasian Auctioneer
<p>The proposition is from lecture notes of advanced econometrics of Yongmiao Hong: <a href="https://i.sstatic.net/QinQu.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/QinQu.jpg" alt="notes page"></a></p>

<p>A1: <span class="math-container">$\{Y_t,X_t'\}_{t=1}^n$</span> is an i.i.d. random sample.</p>

<p>A2: <span class="math-container">$E(\varepsilon_t|X_t)=0$</span> almost surely with <span class="math-container">$E(\varepsilon_t^2)=\sigma^2&lt;\infty$</span>.</p>

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

CC BY-SA 4.0
2019-10-26T12:34:40.457Z
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
CC BY-SA 4.0
2019-10-26T15:41:38.557Z
@mark leeds thanks, it's very intuitive.
HXW
CC BY-SA 4.0
2020-01-31T05:48:39.883Z
Your comment is apparently valid. You may answer the question.
Subhash C. Davar
<econometrics><regression><applied-econometrics>
Econometrics: why the i.i.d assumption and weak exogeneity assumption imply strict exogeneity?
HXW
# Econometrics: why the i.i.d assumption and weak exogeneity assumption imply strict exogeneity?

(Asked by: username_0 on 2019-10-26)

The proposition is from lecture notes of advanced econometrics of Yongmiao Hong: [![notes page](https://i.sstatic.net/QinQu.jpg)](https://i.sstatic.net/QinQu.jpg)

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?

**Tags:** econometrics, regression, applied-econometrics
> **Comments:**
> **username_1**: 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.  
> **username_0**: @mark leeds thanks, it's very intuitive.  
> **username_2**: Your comment is apparently valid. You may answer the question.  

---

## 1 answers

### Answer by username_3 on 2019-10-26. Score: 0 | Accepted answer

\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\*}

> **Comments:**
> **username_4**: Aren't you appealing to Assumption 4.2 in the second line of this proof?  
> **username_3**: @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.  

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