The correct condition for OLS estimates to be unbiased?

Question

For the ordinary least square (OLS) estimates of regression ($\vec{y} =\mathbf{X} \cdot \vec{\beta} + \vec{\epsilon}$) to be unbiased (without considering the efficiency), which one of the three conditions is required (sufficient):

$$\mathbb{E}(\vec{\epsilon}) = \vec{0} \tag{Assumption 1}$$

$$\mathbb{E}(\vec{\epsilon} | \mathbf{X}) = \vec{0} \tag{Assumption 2}$$

$$\mathbb{E}(\mathbf{X}^T \cdot \vec{\epsilon}) = \vec{0} \tag{Assumption 3}$$

I came to the first one when deriving by myself:

The expectation of $\hat{\vec{\beta}}$ is:

$$ \begin{align*} \mathbb{E}(\hat{\vec{\beta}}) & = \mathbb{E}[\vec{\beta} + (\mathbf{X}^T \cdot \mathbf{X})^{-1} \cdot \mathbf{X}^T \cdot\vec{\epsilon}] \\ & = \mathbb{E}(\vec{\beta}) + \mathbb{E}[(\mathbf{X}^T \cdot \mathbf{X})^{-1} \cdot \mathbf{X}^T \cdot\vec{\epsilon}] \\ & as\ \vec{\beta} \ is \ constant\ and\ \mathbf{X}\ is\ also\ constant\\ & = \vec{\beta} + (\mathbf{X}^T \cdot \mathbf{X})^{-1} \cdot \mathbf{X}^T \cdot \mathbb{E}(\vec{\epsilon}) \\ \end{align*} $$

But some materials in econometrics stated the Assumption 2 or Assumption 3. I am confused as $\mathbf{X}$ is considered observed or fixed in OLS, conditioned expectation in Assumption 2 doesn't seem to make sense. And Assumption 3 should easily be simplified to Assumption 1 given $\mathbf{X}$ fixed.

Answer 1

If you regard $X$ as fixed then assumptions 1 and 2 are the same. Since $X$ often isn't fixed, just conditioned on, it's helpful to have the assumption in a form that doesn't rely on $X$ being fixed. That's the point of assumption 2. Both assumptions 1 and 2 are saying that $E[Y|X=x]$ is truly is $x\beta$.

Assumption 3 is importantly weaker. Assumptions 1 and 2 require each residual individually to have zero mean; assumption 3 only gives conditions on sums of residuals.

For example, suppose $E[Y|X=x]$ is actually a curve, so that $E[Y|X=x]\neq x\beta$. There will still be a best-fitting true line, a value of $\beta$ that makes $E[X^Te]=0$, but we will not have $E[e_i|X_i=x]=0$ for any individual point. In this case it is still true that $\hat\beta$ is unbiased for the slope of the true best-fitting line, but that line is not $E[Y|X=x]$.