# Least Squares Criterion Problem

Problem: Show that the least squares criterion applied to the "intercept-only" model, i.e. $$y_i = \beta_0 + \epsilon_i$$, $$i = 1, 2, ..., n$$ results in the least squares estimator of $$\beta_0: \hat \beta = \bar y$$

I'm not sure where to begin with this problem, but doesn't it have something to do with minimizing. Also, I believe this formula matters in this problem: $$S =$$ $$\sum_{i=0}^n \epsilon_i^2$$ $$=$$ $$\sum_{i=0}^n$$ $$(y_i - \beta_0)^2$$

Help will be appreciated.

You are correct when thinking that you need to consider $$S=\sum_{i=1}^n \epsilon_i^2=\sum_{i=1}^n (y_i-\beta_0)^2$$ Since you want $$S$$ to be minimum, its derivative with respect to $$\beta_0$$ must be zero. $$\frac{dS}{d \beta_0}=-2\sum_{i=1}^n (y_i-\beta_0)=0\implies \sum_{i=1}^n y_i=\beta_0\sum_{i=1}^n 1$$ Then, $$\cdots$$