In a simple linear regression model, expectation of $u$ is taken to be $0$. Why is that?
Can anyone please verify if the explanation I have given below is valid?
Suppose $y= \beta_0 + \beta_1x +u$ acurately models the relationship between $x$ and $y$. Then, we want to model the relationship between $x$ and $y$ keeping all other factors fixed. To do so, we create a simple linear regression model such that $\hat{y} = \hat{\beta_0} +\hat{\beta_1}x$
Now, $\beta_o \ne \hat{\beta_o}$, as now, our $\hat{\beta_0}$ also accounts for the average of the error term. Are we using the average of the error term to allow for a linear relationship? Because average would be just a number and would allow us to have a linear model? Then, since the average of the error term which doesn't necessarily have to be 0 is being accounted for in our fitted line, we can safely assume that $E[u]=0$
Is the aforementioned reasoning valid? Please let me know if I have made any conceptual error or if I'm missing anything that provides better understanding.
Thank you!
First, you can always define $u\equiv y-\beta_0-\beta_1x$ so if no assumption is made about $u$, $\beta_0$ and $\beta_1$ carry no meaning.
Second, the idea behind the assumptioin $E(u)=0$ is that you're saying on average, the prediction $\beta_0+\beta_1x$ is correct. That is, suppose there are $n$ independent instances of the observations $(x_1,y_1),\ldots,(x_n,y_n)$, then we would like that $\frac{1}{n}\sum_{i=1}^n(\beta_0+\beta_1x_i-y_i)$ converges almost surely to $0$.
Third, once you assume that there are some $\beta_0$ and $\beta_1$ such that $E(y-\beta_0-\beta_1x)=0$, you can move on to the business of the estimation. If you're willing to assume further that $E[x(y-\beta_0-\beta_1x)]=0$ then you can least squares. If not, then something else (IV, GMM) is appropriate.