# Gauss Markov Theorem and zero conditional mean/mean independent assumption

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.

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:

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

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