Consistency of OLS when no intercept

Question

Suppose I have a model $y_i = \beta_0 + \beta_1 x_i + e_i$ but instead I estimate $y_i = \beta_1 x_i + u_i$ using OLS. That is, I ignore the intercept. Working out the algebra, based on this post, we have

$\hat{\beta}_1 = \beta_1 + \beta_0 \frac{\sum_{i=1}^n x_i}{\sum_{i=1}^n x_i^2} + \frac{\sum_{i=1}^n x_i u_i}{\sum_{i=1}^n x_i^2} $

I know how to care care of the third term, but can you help me verify if the following is correct?

I can write

$$ \frac{\sum_{i=1}^n x_i}{\sum_{i=1}^n x_i^2} = \frac{\frac{1}{n}\sum_{i=1}^n x_i}{\frac{1}{n} \sum_{i=1}^n x_i^2} $$

Using the continuous mapping theorem, the numerator converges in probability to $E[x_i]$ and the denominator $E[x_i^2]$.

Suppose I demean the data. Then $E[x_i] = 0$. Does that mean $\hat{\beta}_1$ is consistent even if I ignore the intercept as long as I demean $x_i$ but not $y_i$?

Answer 1

The intercept is represented in regression just as a predictor with the value of 1 for each observation, the maths works exactly as with the other predictor(s). So the proof of consistency is the same, and has the same assumptions, that is, the model is correct. In this case, that is, that the true intercept is zero.

When you demeans $x_i$, you assure that the intercept is truly zero, by construction. So yes, in this case $\hat{\beta}_1$ is consistent, if it is consistent in the model without demeaning and including the intercept.