I'm reading trough my textbook (“A Modern Approach to Regression” Sheather. Page 35) and it says that we can assume $\hat{\beta_1}|X$ is normally distributed because $Y_i|X$ is normally distributed and $\hat{\beta_1}|X$ is a linear combination of $y_i$'s.

Clearly, $y_i$ is a linear combination of $\beta_0$, $\beta_1x_i$ and $e_i$ but I don't think that implies the parameters are therefore linear combinations of $y_i$. The parameters are dependent on both $y_i$ and $x_i$.

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    $\begingroup$ What is the formula given for beta? What makes you think that it is not linear in y? $\endgroup$ Apr 4, 2020 at 6:02
  • $\begingroup$ @conjectures it’s a simple bivariate regression. We’re assuming that y conditioned on x is normally distributed. I’m not questioning that, I’m wondering why that fact alone implies Beta must be normally distributed. $\endgroup$ Apr 4, 2020 at 8:28

1 Answer 1


First, it is important that $Y_i |X$ is normally distributed. Also, a linear combination of independent normal random variables is normal. With that knowledge, all that is left is to show that $\hat{\beta_1}$ is a linear combination of $Y_i | X$.

From the textbook Applied Linear Regression Models (4th ed) by Kutner, Nachtsheim, and Neter it states

$$ b_1 = \frac{\sum (X_i - \bar{X})(Y_i - \bar{Y})}{\sum(X_i - \bar{X})^2} $$

where and $b_1$ is a sampling distribution. Written as a linear combination of $Y_i$ we have

$$ b_1 = \sum k_i Y_i $$


$$ k_i = \frac{X_i - \bar{X}}{\sum(X_i - \bar{X})^2}. $$

The text gives a lengthy proof for how to go from the first equation to the second equation which I won't include here (pg. 42 for reference) but the conclusion is that $\hat{\beta_1}$ can indeed be written as a linear combination of the $Y_i$ independent normally distributed random variables. This implies $\hat{\beta_1}$ is normal.

  • $\begingroup$ Ohhhh makes sense. Since we’re conditioning on x, x doesn’t influence the distribution of Beta1. I don’t know why it didn’t click before, thank you! $\endgroup$ Apr 4, 2020 at 8:33

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