# A simple proof for expressing $SSR/\sigma^2$ in simple linear regression as the square of a standard normal

I am new to statistics and trying to prove that in simple linear regression, the $$SS_{\text{Regression}}/\sigma^2$$ can be expressed as the square of a standard normal. ( Where $$SS_{\text{Regression}} = \sum_{1}^n(\hat Y_i - \bar Y)^2$$ is the sum of squares due to regression)

I tried searching for Cochran's theorem but the theorem with heavy matrix theory is heavy for me right now. Is there a simpler proof for the case of simple linear regression?

Help is deeply appreciated!

• stats.stackexchange.com/q/362590/119261 Mar 27, 2021 at 6:56
• @StubbornAtom This article seems to show that the $MS_{\text{Residuals}}$ follows a $n-2$ chi square distribution. Whereas actually, my query is for the distribution of $MS_{\text{Regression}}$ Mar 27, 2021 at 16:04

Consider the model $$y_i=\alpha+\beta x_i+\varepsilon_i\quad,\,i=1,2,\ldots,n$$

The least square estimators of $$\alpha,\beta$$ are then given by

$$\hat\alpha=\bar y-\hat\beta \bar x\qquad,\qquad\hat\beta=\frac{s_{xy}}{s_{xx}}$$

where $$s_{xy}=\sum_{i=1}^n (x_i-\bar x)(y_i-\bar y) \quad \text{ and } \quad s_{xx}=\sum_{i=1}^n (x_i-\bar x)^2$$

When the errors $$\varepsilon_i$$ are distributed as independent $$N(0,\sigma^2)$$ variables, we have

$$\frac{\hat\beta \sqrt{s_{xx}}}{\sigma}\sim N\left(\beta \sqrt{s_{xx}},1 \right) \tag{1}$$

Now

$$\hat y_i=\hat\alpha+\hat\beta x_i=\overline y+\hat\beta(x_i-\overline x)$$

Therefore, $$\frac{SSR}{\sigma^2}=\frac1{\sigma^2}\sum_{i=1}^n (\hat y_i-\overline y)^2=\left(\frac{\hat\beta \sqrt{s_{xx}}}{\sigma}\right)^2$$

From $$(1)$$, it follows that under normality of errors, $$SSR/\sigma^2$$ has a noncentral $$\chi^2$$ distribution with $$1$$ degree of freedom and noncentrality parameter $$\beta^2 s_{xx}$$. So it is not true in general that $$SSR/\sigma^2$$ can be expressed as the square of a standard normal variable.