R-squared and F-distribution

Question

I am trying to understand the relations between the coefficient of determination $R^2$ and the $F$-distributions.

Hereafter are the notations I use :

Assume a simple linear regression model $y_i=ax_i+b + \varepsilon_i, 1\leq i\leq n$, where the errors $\varepsilon_i$ follow a normal distribution with mean $0$ and constant variance $\sigma^2$. The regression line calculated by OLS is $y=\hat{a}x+\hat{b}$ and $\hat{y}_i=\hat{a} x_i +\hat{b}$ is the predicted value (at point $x_i$) and $\hat{\varepsilon}_i=y_i-\hat{y}_i$ is the residual.

Let $SSR=\sum(\hat{y}_i - \bar{y})^2$ be the sum of squares explained by the regression, $SSE=\sum(y_i - \hat{y}_i)^2$ the sum of squared errors and $SST=\sum(y_i - \bar{y})^2$ the total sum of squares.

Then $R^2=\frac{SSR}{SST}=1-\frac{SSE}{SST}$.

What I understood from documents about linear regression is that, under the null hypothesis that $a=0$ (the slope of the true regression line is zero), the two statistics $ \frac{SSR}{\sigma^2}$ and $ \frac{SST}{\sigma^2}$ follow (respectively) a chi-squared distribution with $1$ df and a chi-squared distribution with $n-1$ df, and that they are independent.

So, in my mind, the ratio $\frac{\frac{SSR}{\sigma^2} / 1}{\frac{SST}{\sigma^2} / (n-1)}$ should follow $F(1, n-1)$, a Fisher-Snedecor distribution with $df$ $1$ and $n-1$...

Instead of finding this result, what I see written everywhere is that $(n-2) \frac{R^2}{1-R^2}$ follows $F(1, n-2)$...

Why do people use this latter form instead of the former ? Are they equivalent ? Something escapes me…

Answer 1

In a model with a single explanatory variable and an intercept term you have regression degrees-of-freedom $DF_R = 1$ and residual degrees-of-freedom $DF_E = n-2$, so you get:

$$\begin{equation} \begin{aligned} F \equiv \frac{MSR}{MSE} &= \frac{SSR / DF_R}{SSE / DF_E} \\[6pt] &= \frac{DF_E}{DF_R} \cdot \frac{SSR}{SSE} \\[6pt] &= \frac{DF_E}{DF_R} \cdot \frac{SSR}{SST-SSR} \\[6pt] &= (n-2) \cdot \frac{R^2}{1-R^2}. \\[6pt] \end{aligned} \end{equation}$$

In your own calculations, you appear to have used an incorrect value for the residual degrees-of-freedom ($n-1$ instead of $n-2$) so your result is not equivalent to the correct form.