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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.

Trever
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1 Answers1

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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, $\cdots$

Claude Leibovici
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