# The Least Squares Assumption 1

Assume the following linear relationship: $Y_i = \beta_0 + \beta_1 X_i + u_i$, where $Y_i$ is the dependent variable, $X_i$ a single independent variable and $u_i$ the error term.

According to Stock & Watson (Introduction to Econometrics; [Chapter 4]), the first least squares assumption is $E[u_i|X_i]=0$. Does this imply (law of iterated expectation): $E[u_i]=0$?

• You appear to answer your question while you ask it! Are there perhaps some undisclosed conditions that cause you to doubt the applicability of your reasoning? – whuber Nov 4 '16 at 17:48
• You said it: by law of iterated expectations, this holds. – Richard Hardy Nov 4 '16 at 19:39

Yes, it does imply that, but it is a stronger assumption: Setting the conditional mean of the error terms to zero is a stronger assumption than just setting their marginal mean to zero. If we have $\mathbb{E}(U | X = x) = 0$ for all allowable $x$ then using the law-of-iterated-expectations we get:
$$\mathbb{E}(U) = \mathbb{E}(\mathbb{E}(U | X)) = \mathbb{E}(0) = 0.$$
\begin{equation} \begin{aligned} \mathbb{Cov}(U,X) &= \mathbb{E}(UX) - \mathbb{E}(U)\mathbb{E}(X) \\[6pt] &= \mathbb{E}(X \cdot\mathbb{E}(U | X)) - \mathbb{E}(U)\mathbb{E}(X) \\[6pt] &= \mathbb{E}(X \cdot 0) - 0 \cdot\mathbb{E}(X) = 0. \end{aligned} \end{equation}
Note that this is a stronger assumption than if we just assume that $\mathbb{E}(U) = 0$. That is, you can have situations where the error terms have a marginal mean of zero, even though they are correlated with the explanatory variable $X$.