Gauss Markov Theorem and zero conditional mean/mean independent assumption

Question

So I read online that one of the assumptions of Gauss Markov Theorem is: $$E[\epsilon_i]=0$$However, we also know that one of the assumptions for linear regression is the zero conditional mean: $$E[\epsilon|X]=0$$ So I'm wondering, why the difference here? Do we have to further assume mean independence for the first case, i.e. $$E[\epsilon|X]=E[\epsilon ]$$ to arrive at the strict exogeneity criteria? Also, whats the benefit of making these 2 assumptions, as opposed to just assuming $E[\epsilon|X]=0$?

A related discussion is here: What's the difference between "mean independent" and independent? where carlos says "mean independence is not an assumption for linear regression" which further confuses me.

Answer 1

Comment:

The concept of mean independence is often used in statistics with two implications:

• strong assumption of independent random variables $\left(X_{1} \perp X_{2}\right)$
• weak assumption of uncorrelated random variables $\left(\operatorname{Cov}\left(X_{1}, X_{2}\right)=0\right)$.

There's nothing outside of these two implications.

So what is?

$\mathbb E[X_1 X_2]=\mathbb E[X_1] \mathbb E[X_2]$

This is uncorrelated random variables.

How we know this?

$\operatorname{Cov}(X_1 X_2)=\mathbb E[X_1 X_2]-\mathbb E[X_1] \mathbb E[X_2]=0$

So know we know two RVs are not correlated what is independent?

Let's say first: RVs that are independent are by definition uncorrelated. So being independent assumes uncorrelated. But what is independent?

Two RVs $X_1$ and $X_2$ are independent if their joint distribution is equal to the product of their marginal distributions:

$\mathbb P(x_1,x_2) = \mathbb P(x_1)\mathbb P(x_2)$ , where $x_1 \sim X_1$ and $x_2 \sim X_2$

In other words, having information on $X_1$ provides no additional information on $X_2$ and vice versa:

$\mathbb P(X_1 \mid X_2) = \mathbb P(X_1)$

$\mathbb P(X_2 \mid X_1) = \mathbb P(X_2)$