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In the earlier chapters of my notes, the formula for $\hat{\beta_1}$ in simple linear regression was given as $$\frac{\hat{\sigma}}{\sqrt{\sum_{i=1}^{n}(x_i - \bar{x})^2}}$$.

However, in some later chapters, namely in discussion of the no-intercept model, the formula became $$\frac{\hat{\sigma}}{\sqrt{\sum_{i=1}^{n}x_i^2}}$$ instead.

Is it simply a typo, or is there a difference in formula where the no-intercept model is concerned?

statsguy21
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  • It’s not a typo. –  Mar 04 '19 at 07:07
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    Could you please explain why? Also, how would one attempt to find the denominator in the second expression? The first denominator can be easily found using variance formula, but I can't see an easy way to solve for the second one. – statsguy21 Mar 04 '19 at 07:09
  • I can certainly type something up in the morning . –  Mar 04 '19 at 07:10
  • If you try and derive the least squares estimator, you should notice a difference. In the model with no intercept ($y_i = \beta x_i + \varepsilon_i$), the least squares estimator for $\beta$ is the $b$ that minimises $\sum\limits_{i=1}^{n}(y_i - bx_i)^2$, as a function of $b$ (this is because in this model, $\hat{y}i$ will be $\hat{\beta}x_i$, since we estimate without any intercept). You can try minimising this with respect to $b$ using simple calculus and you should see that a $\sum\limits{i=1}^{n}x_{i}^2$ term appears in your answer. – Minus One-Twelfth Mar 05 '19 at 01:37

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