# Why can we assume $\hat{\beta_1}$ is normally distributed?

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 $$x_i$$.

• What is the formula given for beta? What makes you think that it is not linear in y? Apr 4, 2020 at 6:02
• @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. Apr 4, 2020 at 8:28

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

$$b_1 = \sum k_i Y_i$$

where

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

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.

• 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! Apr 4, 2020 at 8:33